Rd Sharma 2020 2021 Solutions for Class 9 Maths Chapter 6 Factorization Of Polynomials are provided here with simple step-by-step explanations. These solutions for Factorization Of Polynomials are extremely popular among Class 9 students for Maths Factorization Of Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 2021 Book of Class 9 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 6.14:

#### Question 1:

In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the result by actual division: (1−8)

1. *f*(*x*) = *x*^{3} + 4*x*^{2} − 3*x* + 10, *g*(*x*) = *x* + 4

#### Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will show by actual division

So the remainder by actual division is 22

#### Page No 6.14:

#### Question 2:

*f*(*x*) = 4*x*^{4} − 3*x*^{3} − 2*x*^{2} + *x* − 7, *g*(*x*) = *x* − 1

#### Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will show remainder by actual division

So the remainder by actual division is −7

#### Page No 6.14:

#### Question 3:

*f*(*x*) = 2*x*^{4} − 6*x*^{3} + 2*x*^{2} − *x* + 2, *g*(*x*) = *x* + 2

#### Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate the remainder by actual division

So the remainder by actual division is 92

#### Page No 6.14:

#### Question 4:

*f*(*x*) = 4*x*^{3} − 12*x*^{2} + 14*x* − 3, *g*(*x*) 2*x* − 1

#### Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate the remainder by actual division

So the remainder by actual division is

#### Page No 6.14:

#### Question 5:

*f*(*x*) = *x*^{3} − 6*x*^{2} + 2*x* − 4, *g*(*x*) = 1 − 2*x*

#### Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate remainder by actual division

So the remainder is

#### Page No 6.14:

#### Question 6:

*f*(*x*) = *x*^{4} − 3*x*^{2} + 4, *g*(*x*) = *x* − 2

#### Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

We will calculate remainder by actual division

So the remainder is 8

#### Page No 6.14:

#### Question 7:

*f*(*x*) = 9*x*^{3} − 3*x*^{2} + *x* − 5, *g*(*x*) = $x-\frac{2}{3}$

#### Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Remainder by actual division

Remainder is −3

#### Page No 6.14:

#### Question 8:

$f\left(x\right)=3{x}^{4}+2{x}^{3}-\frac{{x}^{2}}{3}-\frac{x}{9}+\frac{2}{27},g\left(x\right)=x+\frac{2}{3}$

#### Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Remainder by actual division

Remainder is 0

#### Page No 6.14:

#### Question 9:

If the polynomials 2*x*^{3} + *ax*^{2} + 3*x* − 5 and *x*^{3} + *x*^{2} − 4*x* +*a* leave the same remainder when divided by *x* −2, find the value of *a.*

#### Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

By the remainder theorem, whenis divided bythe remainder is

By the given condition, the two remainders are same. Then we have,

#### Page No 6.14:

#### Question 10:

If the polynomials *ax*^{3} + 3*x*^{2} − 13 and 2*x*^{3} − 5*x* + a, when divided by (*x* − 2) leave the same remainder, find the value of *a.*

#### Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

By the remainder theorem, whenis divided bythe remainder is

By the given condition, the two remainders are same. Then we have, *R*_{1} = *R*_{2}

#### Page No 6.14:

#### Question 11:

Find the remainder when *x*^{3} + 3*x*^{2} + 3*x* + 1 is divided by

(i) *x *+ 1

(ii) $x-\frac{1}{2}$

(iii) *x*

(iv) $x+\mathrm{\pi}$

(v) 5 + 2*x*

#### Answer:

Let us denote the given polynomials as

(i) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(ii) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(iii) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(iv) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(v) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

#### Page No 6.15:

#### Question 12:

The polynomials *ax*^{3} + 3*x*^{2} − 3 and 2*x*^{3} − 5*x* + *a* when divided by (*x* − 4) leave the remainders *R*_{1} and *R*_{2} respectively. Find the values of a in each of the following cases, if

(a) *R*_{1} = *R*_{2}

(b) *R*_{1} + *R*_{2} = 0

(c) 2*R*_{1} − *R*_{2} = 0

#### Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

By the remainder theorem, whenis divided bythe remainder is

(i) By the given condition,

*R*_{1} = *R*_{2}

(ii) By the given condition,

*R*_{1} + *R*_{2} = 0

(iii) By the given condition,

2*R*_{1} − *R*_{2} = 0

#### Page No 6.2:

#### Question 1:

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:

(i) 3*x*^{2} − 4*x* + 15

(ii) *y*^{2} + 2$\sqrt{3}$

(iii) $3\sqrt{x}+\sqrt{2}x$

(iv) $x-\frac{4}{x}$

(v) *x*^{12} + *y*^{3} + *t*^{50}

#### Answer:

(i) is a polynomial of degree 2.i.e Quadratic polynomial.

(ii) is a polynomial of degree 2 in *y* variable. i.e. Quadratic polynomial.

(iii)

It is not a polynomial because exponent of *x* is 1/2 which is not a positive integer.

(iv)

It is not a polynomial because is fractional part.

(v)

It is a polynomial in three variables *x*, *y* and *t*.

#### Page No 6.2:

#### Question 2:

Write the coefficient of *x*^{2} in each of the following:

(i) 17 − 2*x* + 7*x*^{2}

(ii) 9 − 12*x* + *x*^{3}

(iii) $\frac{\mathrm{\pi}}{6}{x}^{2}-3x+4$

(iv) $\sqrt{3}x-7$

#### Answer:

(i)

Coefficient of

(ii)

Coefficient of

(iii)

Coefficient of

(iv)

Coefficient of

#### Page No 6.24:

#### Question 1:

In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1−7)

*f*(*x*) = *x*^{3} − 6*x*^{2} + 11*x* − 6; *g*(*x*) = *x* − 3

#### Answer:

Given that:

By the factor theorem,

If *g*(*x*) is a factor of *f*(*x*)

i.e.

Then

As is zero therefore *g*(*x*), is the factor of polynomial *f*(*x*).

#### Page No 6.24:

#### Question 2:

*f*(*x*) = 3*x*^{4} + 17*x*^{3} + 9*x*^{2} − 7*x* − 10; g(x) = x + 5

#### Answer:

It is given that and

By the factor theorem, *g*(*x*) is a factor of polynomial *f*(*x*)

i.e.

Therefore,

$f(-5)=3{\left(-5\right)}^{4}+17{\left(-5\right)}^{3}+9{\left(-5\right)}^{2}-7\left(-5\right)-10\phantom{\rule{0ex}{0ex}}=3\times 625+17\times \left(-125\right)+225+35-10\phantom{\rule{0ex}{0ex}}=1875-2125+250\phantom{\rule{0ex}{0ex}}=0$

Hence, *g*(*x*) is the factor of polynomial *f*(*x*).

#### Page No 6.24:

#### Question 3:

*f*(*x*) = *x*^{5} + 3*x*^{4} − *x*^{3} − 3*x*^{2} + 5*x* + 15,* g*(*x*) =* x* + 3

#### Answer:

It is given that and

By the factor theorem, *g*(*x*) is the factor of polynomial *f*(*x*).

i.e.

*
f *(−3) = 0

Hence, *g*(*x*) is the factor of polynomial *f *(*x*).

#### Page No 6.24:

#### Question 4:

*f*(*x*) = *x*^{3} −6*x*^{2} − 19*x* + 84, *g*(*x*) = *x* − 7

#### Answer:

It is given that and

By the factor theorem, *g*(*x*) is the factor of polynomial *f*(*x*), if *f* (7) = 0.

Therefore, in order to prove that (*x* − 7) is a factor of *f*(*x*).

It is sufficient to show that *f*(7) = 0

Now,

Hence, (*x* − 7) is a factor of polynomial *f*(*x*).

#### Page No 6.24:

#### Question 5:

*f*(*x*) = 3*x*^{3} + *x*^{2} − 20*x* +12, *g*(*x*) = 3*x* − 2

#### Answer:

It is given that and

By the factor theorem,

(3*x* − 2) is the factor of *f(x)*, if

Therefore,

In order to prove that (3*x* − 2) is a factor of *f*(*x*).

It is sufficient to show that

Now,

Hence, (3*x* − 2) is the factor of polynomial *f*(*x*).

#### Page No 6.24:

#### Question 6:

*f*(*x*) = 2*x*^{3} − 9*x*^{2} + *x* + 12, *g*(*x*) = 3 − 2*x*

#### Answer:

It is given that and

By factor theorem, (3 − 2*x*) is the factor of *f*(*x*), if = 0

Therefore,

In order to prove that (3 − 2*x*) is a factor of *f*(*x*). It is sufficient to show that

Now,

Hence, (3 − 2*x*), is the factor of polynomial *f*(*x*).

#### Page No 6.24:

#### Question 7:

*f*(*x*) = *x*^{3} − 6*x*^{2} + 11*x* − 6, *g*(*x*) =* **x*^{2} − 3*x* + 2

#### Answer:

It is given that and

We have

$\Rightarrow \left(x-2\right)$ and (*x* − 1) are factor of *g*(*x*) by the factor theorem.

To prove that (*x* − 2) and (*x* − 1) are the factor of *f*(*x*).

It is sufficient to show that *f*(2) and *f*(1) both are equal to zero.

And

Hence, *g*(*x*) is the factor of the polynomial *f*(*x*).

#### Page No 6.24:

#### Question 8:

Show that (*x* − 2), (*x* + 3) and (*x* − 4) are factors of *x*^{3} − 3*x*^{2} − 10*x* + 24.

#### Answer:

Let be the given polynomial.

By factor theorem,

and are the factor of *f*(*x*).

If and *f*(4) are all equal to zero.

Now,

also

And

Hence, and are the factor of polynomial *f*(*x*).

#### Page No 6.24:

#### Question 9:

Show that (*x* + 4) , (*x* − 3) and (*x* − 7) are factors of *x*^{3} − 6*x*^{2} − 19*x* + 84

#### Answer:

Let be the given polynomial.

By the factor theorem,

and are the factor of *f*(*x*).

If and *f*(7) are all equal to zero.

Therefore,

Also

And

Hence, and are the factor of the polynomial *f*(*x*).

#### Page No 6.24:

#### Question 10:

For what value of a is (*x* − 5) *a* factor of *x*^{3}^{ }− 3*x*^{2} + *ax *− 10?

#### Answer:

Let be the given polynomial.

By factor theorem, is the factor of *f*(*x*), if *f* (5) = 0

Therefore,

Hence, *a* = − 8.

#### Page No 6.24:

#### Question 11:

Find the value of *a* such that (*x* − 4) is a factors of 5*x*^{3} − 7*x*^{2} − *ax* − 28.

#### Answer:

Let be the given polynomial.

By the factor theorem,

(*x* − 4) is a factor of *f*(*x*).

Therefore *f*(4) = 0

Hence ,

$\Rightarrow 320-112-4a-28=0\phantom{\rule{0ex}{0ex}}\Rightarrow 180-4a=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{180}{4}=45\phantom{\rule{0ex}{0ex}}$

Hence,

#### Page No 6.24:

#### Question 12:

Find the value of *a,* if *x* + 2 is a factor of 4*x*^{4} + 2*x*^{3} − 3*x*^{2} + 8*x* + 5*a*.

#### Answer:

Let 4*x*^{4} + 2*x*^{3} − 3*x*^{2} + 8*x* + 5*a *be the polynomial.

By the factor theorem,

is a factor of *f*(*x*) if *f*(−2) = 0.

Therefore,

Hence,

#### Page No 6.24:

#### Question 13:

Find the value k if *x* − 3 is a factor of *k*^{2}*x*^{3} −* **kx*^{2} + 3*kx* − *k*.

#### Answer:

Let be the given polynomial.

By the factor theorem,

(*x* − 3) is a factor of *f*(*x*) if f (3) = 0

Therefore,

$\Rightarrow 27{k}^{2}-9k+9k-k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 27{k}^{2}-k=0\phantom{\rule{0ex}{0ex}}\Rightarrow k\left(27k-1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=0ork=\frac{1}{27}$

Hence, the value of *k* is 0 or .

#### Page No 6.24:

#### Question 14:

Find the values of *a* and *b*, if *x*^{2} − 4 is a factor of *ax*^{4} + 2*x*^{3} − 3*x*^{2} + *bx* − 4

#### Answer:

Let and be the given polynomial.

We have,

are the factors of *g*(*x*).

By factor theorem, if and both are the factor of *f*(*x*)

Then *f*(2) and *f*(−2) are equal to zero.

Therefore,

and

Adding these two equations, we get

Putting the value of a in equation (i), we get

Hence, the value of *a* and *b* are 1, − 8 respectively.

#### Page No 6.24:

#### Question 15:

Find α and β, if *x* + 1 and *x* + 2 are factors of *x*^{3} + 3*x*^{2}^{ }− 2α*x** *+ β.

#### Answer:

Let be the given polynomial.

By the factor theorem, and are the factor of the polynomial *f*(*x*) if and both are equal to zero.

Therefore,

$f(-1)=(-1{)}^{3}+3(-1{)}^{2}-2\alpha \left(-1\right)+\beta =0\phantom{\rule{0ex}{0ex}}\Rightarrow f(-1)=-1+3+2\alpha +\beta =0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\alpha +\beta =-2...\left(\mathrm{i}\right)$

and

Subtracting (i) from (ii)

We get,

$\alpha =-1$

Putting the value of $\alpha $ in equation (i), we get

Hence, the value of α and β are −1, 0 respectively.

#### Page No 6.24:

#### Question 16:

If *x* − 2 is a factor of each of the following two polynomials, find the values of *a* in each case

(i) *x*^{3} − 2*ax*^{2} + *ax* − 1

(ii) *x*^{5} − 3*x*^{4} − *ax*^{3} + 3*ax*^{2} + 2*ax* + 4

#### Answer:

(i) Let be the given polynomial.

By factor theorem, if (*x* − 2) is a factor of * f*(

*x*), then

*f*(2) = 0

Therefore,

Thus the value of *a* is 7/6.

(ii) Let* f*(*x*) = *x*^{5} − 3*x*^{4} − *ax*^{3} + 3*ax*^{2} + 2*ax* + 4 be the given polynomial.

By the factor theorem, (*x* − 2) is a factor of *f*(*x*), if *f *(2) = 0

Therefore,

Thus, the value of *a* is .

#### Page No 6.25:

#### Question 17:

In each of the following two polynomials, find the value of *a*, if *x* −* a* is factor:

(i) *x*^{6} − *ax*^{5} + *x*^{4} − *ax*^{3} + 3*x* − *a* + 2

(ii *x*^{5} − *a*^{2}*x*^{3} + 2*x* + *a* + 1)

#### Answer:

(i) Let be the given polynomial.

By factor theorem, (*x* − *a*) is a factor of the polynomial if *f*(*a*) = 0

Therefore,

Thus, the value of *a* is − 1.

(ii) Let be the given polynomial.

By factor theorem, (*x* − *a*) is a factor of *f*(*x*), if *f*(*a*) = 0.

Therefore,

Thus, the value of *a* is − 1/3.

#### Page No 6.25:

#### Question 18:

In each of the following two polynomials, find the value of a, if x + a is a factor.

(i) *x*^{3} + *ax*^{2} − 2*x* +*a* + 4

(ii)* **x*^{4} − *a*^{2}*x*^{2} + 3*x* −*a*

#### Answer:

(i) Let be the given polynomial.

By the factor theorem, (+ *a*) is the factor of *f*(*x*), if *f*(− *a*) = 0, i.e.,

Thus, the value of *a* is − 4/3.

(ii) Let be the polynomial. By factor theorem, (*x* + *a*) is a factor of the *f*(*x*), if *f*(− *a*) = 0, i.e.,

Thus, the value of *a* is 0.

#### Page No 6.25:

#### Question 19:

Find the values of *p* and *q* so that *x*^{4} + *px*^{3} + 2*x*^{3} − 3*x* + *q* is divisible by (*x*^{2} − 1).

#### Answer:

Let and be the given polynomials.

We have,

Here, are the factor of *g*(*x*).

If *f*(*x*) is divisible by and , then and are factor of *f*(*x*).

Therefore, *f*(1) and *f*(−1) both must be equal to zero.

Therefore,

and

Adding both the equations, we get,

Putting this value in (i)

Hence, the value of *p* and *q* are 3, −3 respectively.

#### Page No 6.25:

#### Question 20:

Find the values of a and b so that (*x* + 1) and (*x* − 1) are factors of *x*^{4} + *ax*^{3} − 3*x*^{2} + 2*x* + *b*.

#### Answer:

Let be the given polynomial.

By factor theorem, and are the factors of *f*(*x*) if *f*(−1) and *f*(1) both are equal to zero.

Therefore,

and

Adding equation (i) and (ii), we get

Putting this value in equation (i), we get,

Hence, the value of *a* and *b* are – 2 and 2 respectively.

#### Page No 6.25:

#### Question 21:

If *x*^{3} + *ax*^{2} − *bx*+ 10 is divisible by *x*^{2} − 3*x** *+ 2, find the values of *a* and *b*.

#### Answer:

Let and be the given polynomials.

We have,

Here, and are the factors of *g*(*x*),

Now,

By factor theorem,

and

Subtracting (ii) by (i), we get,

$\left(2a-b\right)-\left(a-b\right)=-9-\left(-11\right)\phantom{\rule{0ex}{0ex}}a=2$

Putting this value in equation (ii), we get,

Hence, the value of *a* and *b* are 2 and 13 respectively.

#### Page No 6.25:

#### Question 22:

If both *x* + 1 and *x* − 1 are factors of *ax*^{3} + *x*^{2} − 2*x* + *b*, find the values of *a* and *b*.

#### Answer:

Let be the given polynomial.

By factor theorem, if and both are factors of the polynomial f (x). if *f*(−1) and *f*(1) both are equal to zero.

Therefore,

And

Adding (i) and (ii), we get

And putting this value in equation (ii), we get,

*a* = 2

Hence, the value of *a* and *b* are 2 and −1 respectively.

#### Page No 6.25:

#### Question 23:

What must be added to *x*^{3} − 3*x*^{2} − 12*x* + 19 so that the result is exactly divisibly by *x*^{2} + *x* - 6 ?

#### Answer:

Let and be the given polynomial.

When *p*(*x*) is divided by *q*(*x*), the reminder is a linear polynomial in *x*.

So, let *r*(*x*) = *ax* + *b* is added to *p*(*x*), so that *p*(*x*) + *r*(*x*) is divisible by *q*(*x*).

Let

Then,

We have,

Clearly, *q*(*x*) is divisible by and i.e., and are the factors of *q*(*x*).

Therefore, *f *(*x*) is divisible by *q*(*x*), if and are factors of *f*(*x*), i.e.,

and *f*(2) = 0

Now, *f*(-3) = 0

$\Rightarrow $ *f*(-3) = (-3)^{3} -3(-3)^{2} + (*a*-12)(-3)+19+b = 0

$\Rightarrow $ -27 - 27 - 3*a* + 36 + 19 + *b *= 0

$\Rightarrow $ -27 - 27 - 3*a* + 36 + 19 + *b *= 0

$\Rightarrow $ -54 - 3*a** *+ *b *+ 55 = 0

$\Rightarrow $ -3*a** *+ *b* + 1 = 0 ---- (i)

And

Subtracting (i) from (ii), we get,

Putting this value in equation (ii), we get,

Hence, *p*(*x*) is divisible by *q*(*x*) if added to it.

#### Page No 6.25:

#### Question 24:

What must be subtracted from *x*^{3} − 6*x*^{2} − 15*x* + 80 so that the result is exactly divisible by x^{2} + x − 12?

#### Answer:

By divisible algorithm, when is divided by the reminder is a linear polynomial

Let be subtracted from *p*(*x*) so that the result is divisible by *q*(*x*).

Let

We have,

Clearly, and are factors of *q*(*x*), therefore, *f*(*x*) will be divisible by *q*(*x*) if and are factors of *f*(*x*), i.e. *f *(−4) and *f *(3) are equal to zero.

Therefore,

and

Adding (i) and (ii), we get,

Putting this value in equation (i), we get,

Hence, will be divisible by if 4* x* − 4 is subtracted from it

#### Page No 6.25:

#### Question 25:

What must be added to 3*x*^{3} + *x*^{2} − 22x + 9 so that the result is exactly divisible by 3*x*^{2} + 7*x* − 6?

#### Answer:

By division algorithm, when is divided by the reminder is a linear polynomial. So, let *r*(*x*) = *ax* + *b* be added to *p*(*x*) so that the result is divisible by *q*(*x*)

Let

We have,

$q\left(x\right)=3{x}^{2}+7x-6\phantom{\rule{0ex}{0ex}}=3{x}^{2}+9x-2x-6\phantom{\rule{0ex}{0ex}}=3x\left(x+3\right)-2\left(x+3\right)\phantom{\rule{0ex}{0ex}}=\left(3x-2\right)\left(x+3\right)$

Clearly, $\left(3x-2\right)$ and $\left(x+3\right)$ are factors of *q*(*x*).

Therefore, *f*(*x*) will be divisible by *q*(*x*) if and are factors of *f*(*x*), i.e.,

and *f*(−3) are equal to zero.

Now,

$f\left(\frac{2}{3}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{\left(\frac{2}{3}\right)}^{3}+{\left(\frac{2}{3}\right)}^{2}+\left(a-22\right)\left(\frac{2}{3}\right)+9+b=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3\times \frac{8}{27}+\frac{4}{9}+\frac{2a}{3}-\frac{44}{3}+9+b=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{8}{9}+\frac{4}{9}-\frac{44}{3}+9+\frac{2a}{3}+b=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{8+4-132+81}{9}+\frac{2a}{3}+b=0\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{39}{9}+\frac{2a}{3}+b=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2a}{3}+b=\frac{13}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+3b=13........\left(\mathrm{i}\right)$

And

$f\left(-3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{\left(-3\right)}^{3}+{\left(-3\right)}^{2}+\left(a-22\right)\left(-3\right)+9+b=0\phantom{\rule{0ex}{0ex}}\Rightarrow -81+9-3a+66+9+b=0\phantom{\rule{0ex}{0ex}}\Rightarrow -3a+b=-3\phantom{\rule{0ex}{0ex}}\Rightarrow b=-3+3a.........\left(\mathrm{ii}\right)$

Substituting the value of*b*from (ii) in (i), we get,

$2a+3\left(3a-3\right)=13\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+9a-9=13\phantom{\rule{0ex}{0ex}}\Rightarrow 11a=13+9\phantom{\rule{0ex}{0ex}}\Rightarrow 11a=22\phantom{\rule{0ex}{0ex}}\Rightarrow a=2$

Now, from (ii), we get

$b=-3+3\left(2\right)=-3+6=3$

So, we have

*a*= 2 and

*b*= 3

Hence, *p*(*x*) is divisible by *q*(*x*), if is added to it.

#### Page No 6.3:

#### Question 3:

Write the degrees of each of the following polynomials:

(i) 7*x*^{3} + 4*x*^{2} − 3*x* + 12

(ii) 12 − *x* + 2*x*^{3}

(iii) $5y-\sqrt{2}$

(iv) 7

(v) 0

#### Answer:

(i) 7*x*^{3} + 4*x*^{2} − 3*x* + 12

Degree of the polynomial = 3

Because the highest power of *x* is 3.

(ii)

Degree of the polynomial = 3. Because the highest power of *x* is 3.

(iii)

Degree of the polynomial = 1. Because the highest power of *y* is 1.

(iv) 7

Degree of the polynomial = 0. Because there is no variable term in the expression

(v) 0

Degree of the polynomial is not defined. As there is no variable or constant term

#### Page No 6.3:

#### Question 4:

Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:

(i) *x* + *x*^{2} + 4

(ii) 3*x* − 2

(iii) 2*x* + *x*^{2}

(iv) 3*y*

(v) t^{2} + 1

(vi) 7*t*^{4} + 4*t*^{3} + 3*t* − 2

#### Answer:

(i)

The degree of the polynomial is 2. It is quadratic in *x*.

So, it is quadratic polynomial.

(ii)

The degree of the polynomial is 1. It is a linear polynomial in *x*.

(iii)

The degree of the polynomial is 1.

It is linear a polynomial in *x*.

(iv) 3*y*

The degree of the polynomial is 1. It is linear in *y*.

(v)

The degree of the polynomial is 2. It is quadratic polynomial in *t*.

(vi)

The degree of the polynomial is 4. Therefore, it is bi-quadratic polynomial in *t*.

#### Page No 6.3:

#### Question 5:

Classify the following polynomials as polynomials in one-variable, two variables etc.:

(i) *x*^{2} − *xy* + 7*y*^{2}

(ii) *x*^{2} − 2*tx* + 7*t*^{2} − *x* + t

(iii) *t*^{3} − 3*t*^{2} + 4*t* − 5

(iv) *xy* + *yz* + *zx*

#### Answer:

(i)

Here, *x* and *y* are two variables.

So, it is polynomial in two variables.

(ii)

Here, *x* and *t* are two variables.

So, it is polynomial in two variables.

(iii)

Here, only *t* is variable.

So, it is polynomial in one variable.

(iv)

Here, *x*, *y* and *z* are three variables

So, it is polynomial in three variables.

#### Page No 6.3:

#### Question 6:

Identify polynomials in the following:

(i) *f*(*x*) = 4*x*^{3} − *x*^{2} − 3*x* + 7

(ii) *g*(*x*) = 2*x*^{3} − 3*x*^{2}^{ }+ $\sqrt{x}$ − 1

(iii) *p*(*x*) = $\frac{2}{3}{x}^{2}-\frac{7}{4}x+9$

(iv) *q*(*x*) = 2*x*^{2} − 3*x* + $\frac{4}{x}$+ 2

(v) *h*(*x*) = ${x}^{4}-{x}^{\frac{3}{2}}+x-1$

(vi) *f*(*x*) = $2+\frac{3}{x}+4x$

#### Answer:

(i)

It is cubic in *x*, so, it is cubic polynomial in *x* variable.

(ii)

Here, exponent of *x* in is not a positive integer, so, it is not a polynomial.

(iii)

It is a quadratic polynomial.

(iv) *q*(*x*) = 2*x*^{2} − 3*x* + $\frac{4}{x}$+ 2

Here, exponent of *x* in is not a positive integer. So it is not a polynomial.

(v)

Here, exponent of *x* in x^{3}^{/2} is not a positive integer. So, it is not a polynomial.

(vi)

Here, exponent of *x* in is not a positive integer, so, it not a polynomial.

#### Page No 6.3:

#### Question 7:

Identify constant, linear, quadratic and cubic polynomials from the following polynomials:

(i) *f*(*x*) = 0

(ii) *g*(*x*) = 2*x*^{3} − 7*x* + 4

(iii) *h*(*x*) = -$3x+\frac{1}{2}$

(iv) *p*(*x*) = 2*x*^{2} − *x* + 4

(v) *q*(*x*) = 4*x* + 3

(vi) *r*(*x*) = 3*x*^{3} + 4*x*^{2} + 5*x* − 7

#### Answer:

(i)

The given expression is a Constant polynomial as there is no variable term in it.

(ii)

The given expression is Cubic polynomial as the highest exponent of *x* is 3.

(iii)

The given expression is linear polynomial as the highest exponent of *x* is 1.

(iv)

The given expression is Quadratic polynomial as the highest exponent of *x* is 2.

(v)

The given polynomial is an linear polynomial as the highest exponent of *x* is 1.

(vi)

The given polynomial is Cubic polynomial as the highest exponent of *x* is 3.

#### Page No 6.3:

#### Question 8:

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

#### Answer:

An example of binomial of degree 35 is $f\left(t\right)=4{t}^{35}-\frac{1}{2}$. It is a binomial as it has two terms and degree is 35 because highest exponent of *t* is 35.

An example of monomial of degree 100 is . It is a monomial as it has only one term and degree is 100 because highest exponent of *x *is 100

#### Page No 6.32:

#### Question 1:

*Using factor theorem, factorize each of the following polynomials:*

*x*^{3} + 6*x*^{2} + 11*x* + 6

#### Answer:

Let *f*(*x*) = *x*^{3} + 6*x*^{2} + 11*x* + 6 be the given polynomial.

Now, put the we get

Therefore, is a factor of *f*(*x*).

Now,

$f\left(x\right)={x}^{3}+5{x}^{2}+{x}^{2}+5x+6x+6$

Hence, are the factors of *f*(*x*).

#### Page No 6.32:

#### Question 2:

*x*^{3} + 2*x*^{2} − *x* − 2

#### Answer:

Let be the given polynomial.

Now, put the *x* = -1, we get

$f(-1)={\left(-1\right)}^{3}+2{\left(-1\right)}^{2}-\left(-1\right)-2\phantom{\rule{0ex}{0ex}}=0$

Therefore, is a factor of polynomial *f*(*x*).

Now, *x*^{3} + 2*x*^{2} − *x* − 2 can be written as,

$f\left(x\right)={x}^{3}+3{x}^{2}-{x}^{2}-3x+2x-2$

Hence, are the factors of the polynomial *f(x).*

#### Page No 6.32:

#### Question 3:

*x*^{3} − 6*x*^{2} + 3*x* + 10

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

$f\left(x\right)={x}^{3}-7{x}^{2}+{x}^{2}+10x-7x+10$

Hence, are the factors of the polynomial *f*(*x*).

#### Page No 6.32:

#### Question 4:

*x*^{4} − 7*x*^{3}_{ }+ 9*x*^{2} + 7*x** *− 10

#### Answer:

Let be the given polynomial.

Now, putting we get

$f\left(1\right)={\left(1\right)}^{4}-7{\left(1\right)}^{3}+9{\left(1\right)}^{2}+7\left(1\right)-10\phantom{\rule{0ex}{0ex}}=1-7+9+7-10=0$

Therefore, is a factor of polynomial *f*(*x*).

Now,

$f\left(x\right)={x}^{4}-{x}^{3}-6{x}^{3}+6{x}^{2}+3{x}^{2}-3x+10x-10$

Where

Putting we get

Therefore, is a factor of g(*x*).

Now,

$g\left(x\right)={x}^{3}-7{x}^{2}+{x}^{2}-7x+10x+10$

From equation (i) and (ii), we get

Hence are the factors of polynomial *f*(*x*).

#### Page No 6.32:

#### Question 5:

3*x*^{3} − *x*^{2} − 3*x* + 1

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

Hence are the factors of polynomial *f*(*x*).

#### Page No 6.32:

#### Question 6:

*x*^{3} − 23*x*^{2} + 142*x* − 120

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

Hence are the factors of polynomial *f*(*x*).

#### Page No 6.33:

#### Question 7:

*y*^{3} − 7*y* + 6

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*y*).

Now,

Hence are the factors of polynomial f* *(*y*).

#### Page No 6.33:

#### Question 8:

*x*^{3} − 10*x*^{2} − 53*x* − 42

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

Hence are the factors of polynomial f(*x*).

#### Page No 6.33:

#### Question 9:

*y*^{3} − 2*y*^{2} − 29*y* − 42

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*y*).

Now,

Hence are the factors of polynomial f(*y*).

#### Page No 6.33:

#### Question 10:

2*y*^{3} − 5*y*^{2} − 19*y* + 42

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*y*).

Now,

Hence are the factors of polynomial f(*y*).

#### Page No 6.33:

#### Question 11:

*x*^{3} + 13*x*^{2} + 32*x* + 20

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

Hence are the factors of polynomial f(*x*).

#### Page No 6.33:

#### Question 12:

*x*^{3} − 3*x*^{2} − 9*x* − 5

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

Hence are the factors of polynomial f(*x*).

#### Page No 6.33:

#### Question 13:

2*y*^{3} + *y*^{2} − 2*y* − 1

#### Answer:

Let be the given polynomial.

Now, putting we get

Now,

Hence are the factors of polynomial f(*y*).

#### Page No 6.33:

#### Question 14:

*x*^{3} − 2*x*^{2} − *x* + 2

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

Hence are the factors of polynomial f(*x*).

#### Page No 6.33:

#### Question 15:

Factorize each of the following polynomials:

(i) *x*^{3} + 13*x*^{2} + 31*x* − 45 given that x + 9 is a factor

(ii) 4*x*^{3} + 20*x*^{2} + 33*x* + 18 given that 2*x* + 3 is a factor.

#### Answer:

(i) Let be the given polynomial.

is a factor of the polynomial *f*(*x*).

Now,

Hence are the factors of polynomial f(*x*).

(ii) Let be the given polynomial.

Therefore is a factor of the polynomial *f*(*x*).

Now,

Hence are the factors of polynomial f(*x*).

#### Page No 6.33:

#### Question 16:

*x*^{4} − 2*x*^{3} − 7*x*^{2} + 8*x* + 12

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

$f\left(x\right)={x}^{4}-3{x}^{3}+{x}^{3}-3{x}^{2}-4{x}^{2}+12x-4x+12$

Where

Putting we get

Therefore, is the factor of *g*(*x*).

Now,

$g\left(x\right)={x}^{3}-2{x}^{2}-{x}^{2}-6x+2x+12$

From equation (i) and (ii), we get

Hence are the factors of polynomial *f*(*x*).

#### Page No 6.33:

#### Question 17:

*x*^{4} + 10*x*^{3} + 35*x*^{2} + 50*x* + 24

#### Answer:

Let $f\left(x\right)={x}^{4}+10{x}^{3}+35{x}^{2}+50x+24$ be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

Where

Putting we get

Therefore, is the factor of *g*(*x*).

Now,

From equation (i) and (ii), we get

Hence are the factors of polynomial *f*(*x*).

#### Page No 6.33:

#### Question 18:

2*x*^{4} − 7*x*^{3} − 13*x*^{2} + 63*x* − 45

#### Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial *f*(*x*).

Now,

Where

Putting we get

Therefore, is a factor of *g*(*x*).

Now,

From equation (i) and (ii), we get

Hence are the factors of polynomial *f*(*x*).

#### Page No 6.33:

#### Question 1:

Which one of the following is a polynomial?

(a) $\frac{{x}^{2}}{2}-\frac{2}{{x}^{2}}$

(b) $\sqrt{2x}-1$

(c) ${x}^{2}+\frac{3{x}^{3/2}}{\sqrt{x}}$

(4) $\frac{x-1}{x+1}$

#### Answer:

(a) $\frac{{x}^{2}}{2}-\frac{2}{{x}^{2}}=\frac{{x}^{2}}{2}-2{x}^{-2}$

Exponent of *x* cannot be negative.

Therefore, it is not a polynomial.

(b) $\sqrt{2x}-1$

Exponent of *x* must be a whole number.

Therefore, it is not a polynomial.

(c) ${x}^{2}+\frac{3{x}^{3/2}}{\sqrt{x}}={x}^{2}+3{x}^{\frac{3}{2}-\frac{1}{2}}={x}^{2}+3x$

It is a polynomial.

(d) $\frac{x-1}{x+1}$

It is not a polynomial.

Hence, the correct option is (c).

#### Page No 6.33:

#### Question 2:

Degree of the polynomial *f*(*x*) = 4*x*^{4} + 0*x*^{3 }+ 0*x*^{5} + 5*x* + 7 is

(a) 4

(b) 5

(c) 3

(d) 7

#### Answer:

Given: *f*(*x*) = 4*x*^{4} + 0*x*^{3 }+ 0*x*^{5} + 5*x* + 7 = 4*x*^{4} + 5*x* + 7

Degree is the highest power of *x* in the polynomial.

Here, the highest power of *x* is 4.

Thus, degree of the polynomial is 4.

Hence, the correct option is (a).

#### Page No 6.34:

#### Question 3:

Degree of the zero polynomial is

(a) 0

(b) 1

(c) any natural number

(d) not defined

#### Answer:

The zero polynomial is a polynomial in which all the coefficients are zero.

We can't say about powers of *x*.

Thus, degree of the zero polynomial is not defined.

Hence, the correct option is (d).

#### Page No 6.34:

#### Question 4:

$\sqrt{2}$ is a polynomial of degree

(a) 2

(b) 0

(c) 1

(d) $\frac{1}{2}$

#### Answer:

Given: *f*(*x*) = $\sqrt{2}$

It can also be written as $f\left(x\right)=\sqrt{2}{x}^{0}$.

Degree is the highest power of *x* in the polynomial.

Here, the highest power of *x* is 0.

Thus, degree of the polynomial is 0.

Hence, the correct option is (b).

#### Page No 6.34:

#### Question 5:

Zero of the zero polynomial is

(a) 0

(b) 1

(c) any real number

(d) not defined

#### Answer:

Given: zero polynomial i.e., *f*(*x*) = 0

Zero of a polynomial is the value of *x* at which the polynomial becomes zero.

Here, any value of *x* can be a zero of the zero polynomial.

Hence, the correct option is (c).

#### Page No 6.34:

#### Question 6:

If *f*(*x*) = *x *+ 3, then *f*(*x*) + *f*(–*x*) is equal to

(a) 3

(b) 2*x*

(c) 0

(d) 6

#### Answer:

Given: *f*(*x*) = *x *+ 3

*f*(–*x*) = –*x *+ 3

Thus,

*f*(*x*) + *f*(–*x*) = *x *+ 3 + (–*x *+ 3)

= *x *+ 3 – *x *+ 3

= 6

Hence, the correct option is (d).

#### Page No 6.34:

#### Question 7:

Zero of the polynomial* f*(*x*) = 3*x* + 7 is

(a) $\frac{7}{3}$

(b) $\frac{-3}{7}$

(c) $-\frac{7}{3}$

(d) –7

#### Answer:

Given: *f*(*x*) = 3*x* + 7

Zero of a polynomial is the value of *x* at which the polynomial becomes zero.

$f\left(x\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x+7=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=-7\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{7}{3}$

Therefore, zero of the polynomial* f*(*x*) = 3*x* + 7 is $-\frac{7}{3}$.

Hence, the correct option is (c).

#### Page No 6.34:

#### Question 8:

On of the zeros of the polynomial *f*(*x*) = 2*x*^{2} + 7*x* – 4 is

(a) 2

(b) $\frac{1}{2}$

(c) $-\frac{1}{2}$

(d) –2

#### Answer:

Given: *f*(*x*) = 2*x*^{2} + 7*x* – 4

Zero of a polynomial is the value of *x* at which the polynomial becomes zero.

$f\left(x\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+7x-4=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+8x-x-4=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x\left(x+4\right)-1\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(2x-1\right)\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(2x-1\right)=0\mathrm{or}\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=1\mathrm{or}x=-4\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{1}{2}\mathrm{or}x=-4$

Therefore, zeroes of the polynomial* f*(*x*) = 2*x*^{2} + 7*x* – 4 are $\frac{1}{2}\mathrm{and}-4.$

â€‹

Hence, the correct option is (b).

#### Page No 6.34:

#### Question 9:

If $f\left(x\right)={x}^{2}-2\sqrt{2}x+1,\mathrm{then}f\left(2\sqrt{2}\right)$ is equal to

(a) 0

(b) 1

(c) $4\sqrt{2}$

(d) $8\sqrt{2}+1$

#### Answer:

Given: *f*(*x*) = ${x}^{2}-2\sqrt{2}x+1$

$f\left(2\sqrt{2}\right)={\left(2\sqrt{2}\right)}^{2}-2\sqrt{2}\left(2\sqrt{2}\right)+1\phantom{\rule{0ex}{0ex}}={\left(2\sqrt{2}\right)}^{2}-{\left(2\sqrt{2}\right)}^{2}+1\phantom{\rule{0ex}{0ex}}=1$

â€‹

Hence, the correct option is (b).

#### Page No 6.34:

#### Question 10:

*x *+ 1 is a factor of the polynomial

(a) *x*^{3} + *x*^{2 }–* x* + 1

(b) *x*^{3} + *x*^{2} + *x* + 1

(c) *x*^{4} + *x*^{3} + *x*^{2} + 1

(d) *x*^{4} + 3*x*^{3} + 3*x*^{2} + *x* + 1

#### Answer:

The polynomial *f*(*x*) which becomes zero when *x* = −1 is the required polynomial whose factor is *x* + 1.

(a) Let *f*(*x*) = *x*^{3} + *x*^{2 }–* x* + 1

*f*(−1) = (−1)^{3} + (−1)^{2} − (−1) + 1

= −1 + 1 + 1 + 1 = 2 ≠ 0

Therefore, *x *+ 1 is not a factor of the polynomial,

(b) Let *f*(*x*) = *x*^{3} + *x*^{2} + *x* + 1

*f*(−1) = (−1)^{3} + (−1)^{2} + (−1) + 1

= −1 + 1 − 1 + 1 = 0

Therefore, *x *+ 1 is a factor of the polynomial,

(c) Let *f*(*x*) = *x*^{4} + *x*^{3} + *x*^{2} + 1

*f*(−1) = (−1)^{4} + (−1)^{3} + (−1)^{2} + 1

= 1 − 1 + 1 + 1 = 2 ≠ 0

Therefore, *x *+ 1 is not a factor of the polynomial,

(d) Let *f*(*x*) = *x*^{4} + 3*x*^{3} + 3*x*^{2} + *x* + 1

*f*(−1) = (−1)^{4} + 3(−1)^{3} + 3(−1)^{2} + (−1) + 1

= 1 − 3 + 3 − 1 + 1 = 1 ≠ 0

Therefore, *x *+ 1 is not a factor of the polynomial,

Hence, the correct option is (b).

#### Page No 6.34:

#### Question 11:

If* x*^{2} + *kx *+ 6 = (*x* + 2) (*x* + 3) for all *x*, then the value of *k* is

(a) 1

(b) –1

(c) 5

(d) 3

#### Answer:

Given: *x*^{2} + *kx *+ 6 = (*x* + 2) (*x* + 3)

${x}^{2}+kx+6=\left(x+2\right)\left(x+3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+kx+6={x}^{2}+2x+3x+6\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+kx+6={x}^{2}+5x+6\phantom{\rule{0ex}{0ex}}\Rightarrow kx=5x\phantom{\rule{0ex}{0ex}}\Rightarrow k=5$

Hence, the correct option is (c).

#### Page No 6.34:

#### Question 12:

If *x* − 2 is a factor of *x*^{2} + 3*ax* − 2*a*, then *a* =

(a) 2

(b) − 2

(c) 1

(d) −1

#### Answer:

As is a factor of

i.e.

Hence, correct option is (d).

#### Page No 6.34:

#### Question 13:

If *x*^{3} + 6*x*^{2} + 4*x* + k is exactly divisible by *x* + 2, then k =

(a) −6

(b) −7

(c) −8

(d) −10

#### Answer:

As is exactly divisible by

i.e., is a factor of *f*(*x*).

Therefore,

Hence, the correct option is (c).

#### Page No 6.34:

#### Question 14:

If *x* − *a* is a factor of *x*^{3} −3*x*^{2}*a* + 2*a*^{2}*x* + *b*, then the value of* b* is

(a) 0

(b) 2

(c) 1

(d) 3

#### Answer:

As is a factor of

i.e.

Thus,* b *= 0

Thus, the correct option is (a).

#### Page No 6.34:

#### Question 15:

If *x*^{140} + 2*x*^{151} + k is divisible by *x* + 1, then the value of *k* is

(a) 1

(b) −3

(c) 2

(d) −2

#### Answer:

As is divisible by

i.e., is a factor of *f*(*x*).

Therefore,

Hence, the correct option is (a).

#### Page No 6.34:

#### Question 16:

If *x* + 2 is a factor of *x*^{2} + *mx* + 14, then *m* =

(a) 7

(b) 2

(c) 9

(d) 14

#### Answer:

As is a factor of

Therefore,

Hence, the correct option is (c).

#### Page No 6.34:

#### Question 17:

If *x* − 3 is a factor of *x*^{2} − *ax* − 15, then a =

(a) −2

(b) 5

(c) −5

(d) 3

#### Answer:

As is a factor of polynomial

i.e.

Therefore,

Hence, the correct option is (a)

#### Page No 6.35:

#### Question 23:

Let *f*(*x*) be a polynomial such that $f\left(-\frac{1}{2}\right)$ = 0, then a factor of* f*(*x*) is

(a) 2*x* − 1

(b) 2*x* + 1

(c)* x* − 1

(d) *x* + 1

#### Answer:

Let *f*(*x*) be a polynomial such that

i.e., is a factor.

On rearranging can be written as

Thus, is a factor of *f*(*x*).

Hence, the correct option is (b).

#### Page No 6.35:

#### Question 24:

When x^{3} − 2x^{2} + ax − b is divided by x^{2} − 2x − 3, the remainder is x − 6. The values of *a* and *b *are respectively

(a) −2, −6

(b) 2 and −6

(c) −2 and 6

(d) 2 and 6

#### Answer:

If the reminder (*x* −6) is subtracted from the given polynomial then rest of part of this polynomial is exactly divisible by x^{2} − 2x − 3.

Therefore,

Now,

Therefore, are factors of polynomial *p*(*x*).

Now,

And

and

Solving (i) and (ii) we get

Hence, the correct option is (c).

#### Page No 6.35:

#### Question 25:

One factor of *x*^{4} + *x*^{2} − 20 is *x*^{2} + 5. The other factor is

(a) *x*^{2} − 4

(b) *x* − 4

(c) *x*^{2} − 5

(d) *x* + 2

#### Answer:

It is given that is a factor of the polynomial

Here, reminder is zero. Therefore, is a factor of polynomial.

Thus, the correct option is (a).

#### Page No 6.35:

#### Question 26:

If (x − 1) is a factor of polynomial f(x) but not of g(x) , then it must be a factor of

(a) f(x) g(x)

(b) −f(x) + g(x)

(c) f(x) − g(x)

(d) $\left\{f\left(x\right)+g\left(x\right)\right\}g\left(x\right)$

#### Answer:

Asis a factor of polynomial *f*(*x*) but not of *g*(*x*)

Therefore

Now,

Let

Now

Therefore (*x* − 1) is also a factor of *f*(*x*).*g*(*x*).

Hence, the correct option is (a).

#### Page No 6.35:

#### Question 27:

(*x*+1) is a factor of *x ^{n}* + 1 only if

(i)

*n*is an odd integer

(ii)

*n*is an even integer

(iii)

*n*is a negative integer

(iv)

*n*is a positive integer

#### Answer:

The linear polynomial is a factor of only if

If *n* is odd integer, then

Hence, the correct option is (a).

#### Page No 6.35:

#### Question 28:

If *x*^{2} + *x* + 1 is a factor of the polynomial 3*x*^{3} + 8*x*^{2}^{ }+ 8*x* + 3 + 5*k*, then the value of *k* is

(a) 0

(b) 2/5

(c) 5/2

(d) −1

#### Answer:

Let be the given polynomial,

Since is the factor of *f*(*x*). Therefore, re`maider will be zero.

Now,

Now,

Hence, the correct option is (b).

#### Page No 6.35:

#### Question 29:

If (3*x* − 1)^{7} = *a*_{7}*x*^{7} + *a*_{6}*x*^{6} +* **a*_{5}*x*^{5} +...+ *a*_{1}*x* +* **a*_{0}, then *a*_{7} + *a*_{5}_{ }+ ...+*a*_{1} +* **a*_{0 }*=*

(a) 0

(b) 1

(c) 128

(d) 64

#### Answer:

Given that,

Putting

We get

Hence, the correct option is (c).

#### Page No 6.35:

#### Question 30:

If both *x* − 2 and $x-\frac{1}{2}$ are factors of *px*^{2} + 5*x* + *r,* then

(a) p = *r*

(b)* p* + *r* = 0

(c) 2*p* + *r* = 0

(d) *p* + 2*r* = 0

#### Answer:

As and are the factors of the polynomial

i.e., and

Now,

And

From equation (i) and (ii), we get

Hence, the correct option is (a).

#### Page No 6.35:

#### Question 18:

If *x*^{51} + 51 is divided by *x* + 1, the remainder is

(a) 0

(b) 1

(c) 49

(d) 50

#### Answer:

As is divided by

The remainder will be

Hence, the correct option is (d).

#### Page No 6.35:

#### Question 19:

If *x* + 1 is a factor of the polynomial 2*x*^{2} + *kx*, then *k* =

(a) −2

(b) −3

(c) 4

(d) 2

#### Answer:

As is a factor of polynomial Therefore,

So,

Hence, the correct option is (d).

#### Page No 6.35:

#### Question 20:

If *x* + *a* is a factor of *x*^{4} − *a*^{2}*x*^{2} + 3*x* − 6*a*, then *a* =

(a) 0

(b) −1

(c) 1

(d) 2

#### Answer:

Asis a factor of polynomial

Therefore,

Hence, the correct option is (a).

#### Page No 6.35:

#### Question 21:

The value of *k* for which *x* − 1 is a factor of 4*x*^{3} + 3*x*^{2} − 4*x* + *k*, is

(a) 3

(b) 1

(c) −2

(d) −3

#### Answer:

As is a factor of polynomial *f*(*x*) = 4*x*^{3} + 3*x*^{2} − 4*x* + *k*

Therefore,

Hence, the correct option is (d).

#### Page No 6.35:

#### Question 22:

If *x* + 2 and *x *− 1 are the factors of *x*^{3} + 10*x*^{2} + *mx* + *n*, then the values of *m* and *n* are respectively

(a) 5 and −3

(b) 17 and −8

(c) 7 and −18

(d) 23 and −19

#### Answer:

It is given and are the factors of the polynomial

i.e., and

Now

$-8+40-2m+n=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2m+n=-32\phantom{\rule{0ex}{0ex}}\Rightarrow 2m-n=32...\left(i\right)$

And

Solving equation (i) and (ii) we get

*m* = 7 and *n* = − 18

Hence, the correct option is (c)

#### Page No 6.36:

#### Question 31:

If *x*^{2} − 1 is a factor of *ax*^{4} + *bx*^{3} + *cx*^{2} + *dx* + *e*, then

(a) *a* + *c* + *e* = *b* + *d*

(b) *a* + *b* +*e* = *c* + *d*

(c) *a* + *b* + *c* = *d* + *e*

(d) *b* + *c* + *d* = *a* + *e*

#### Answer:

Asis a factor of polynomial

Therefore,

And

*f*(1) = 0

$a{\left(1\right)}^{4}+b{\left(1\right)}^{3}+c{\left(1\right)}^{2}+d\left(1\right)+e=0\phantom{\rule{0ex}{0ex}}\Rightarrow a+b+c+d+e=0$

And

Hence,

The correct option is (a).

#### Page No 6.36:

#### Question 1:

7 – 9*x *+ 2*x*^{2} is called a _________ polynomial.

#### Answer:

Polynomial with degree 2 is known as a quadratic polynomial.

Hence, 7 – 9*x *+ 2*x*^{2} is called a __quadratic__ polynomial.

#### Page No 6.36:

#### Question 2:

12*x* – 7*x*^{2} + 4 – 2*x*^{3} is called a ___________ polynomial.

#### Answer:

Polynomial with degree 3 is known as a cubic polynomial.

â€‹Hence, 12*x* – 7*x*^{2} + 4 – 2*x*^{3} is called a __cubic__ polynomial.

#### Page No 6.36:

#### Question 3:

13*x*^{2} – 88*x*^{3} + 9*x*^{4} is called a _________ polynomial.

#### Answer:

Polynomial with degree 4 is known as a biquadratic polynomial.

â€‹Hence, 12*x* – 7*x*^{2} + 4 – 2*x*^{3} is called a __biquadratic__ polynomial.

#### Page No 6.36:

#### Question 4:

If *x* + 1 is a factor of the polynomial *ax*^{3} +* x*^{2} – 2*x* + 4*a* – 9, then *a* = __________.

#### Answer:

Let *f*(*x*) = *ax*^{3} +* x*^{2} – 2*x* + 4*a* – 9

It is given that one factor of *f*(*x*) is (*x* + 1).

Therefore, $f\left(x\right)=0\mathrm{when}x=-1$.

On putting *x* = –1 in *f*(*x*) = 0, we get

$a{\left(-1\right)}^{3}+{\left(-1\right)}^{2}-2\left(-1\right)+4a-9=0\phantom{\rule{0ex}{0ex}}\Rightarrow -a+1+2+4a-9=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3a-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3a=6\phantom{\rule{0ex}{0ex}}\Rightarrow a=2$

Hence, if *x* + 1 is a factor of the polynomial *ax*^{3} +* x*^{2} – 2*x* + 4*a* – 9, then *a* = __2__.

#### Page No 6.36:

#### Question 5:

If 3*x* – 1 is a factor of the polynomial 81*x*^{3} – 45*x*^{2} +3*a* – 6, then the value of *a* is ___________.

#### Answer:

Let *f*(*x*) = 81*x*^{3} – 45*x*^{2} +3*a* – 6

It is given that one factor of *f*(*x*) is (3*x* – 1).

Therefore, $f\left(x\right)=0\mathrm{when}x=\frac{1}{3}$.

On putting *x* = $\frac{1}{3}$ in *f*(*x*) = 0, we get

$81{\left(\frac{1}{3}\right)}^{3}-45{\left(\frac{1}{3}\right)}^{2}+3a-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 81\left(\frac{1}{27}\right)-45\left(\frac{1}{9}\right)+3a-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3-5+3a-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3a-8=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3a=8\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{8}{3}$

Hence, if 3*x* – 1 is a factor of the polynomial 81*x*^{3} – 45*x*^{2} +3*a* – 6, then the value of *a* is $\overline{)\frac{8}{3}}.$

#### Page No 6.36:

#### Question 6:

The remainders obtained when *x*^{3} + *x*^{2} – 9*x *– 9 is divided by *x, x *+ 1 and *x* + 2 respectively are _________.

#### Answer:

Let *f*(*x*) = *x*^{3} + *x*^{2} – 9*x *– 9

To find the remainder obtained when *x*^{3} + *x*^{2} – 9*x *– 9 is divided by *x,*

we use remainder theorem, put *x *= 0.

*f*(0) is the remainder.

Now,

*f*(0) = (0)^{3} + (0)^{2} – 9(0) – 9

= –9

Hence, the remainder obtained when *x*^{3} + *x*^{2} – 9*x *– 9 is divided by *x* is –9.

To find the remainder obtained when *x*^{3} + *x*^{2} – 9*x *– 9 is divided by *x* + 1,

put *x *+1 = 0.

*f*(–1) is the remainder.

Now,

*f*(–1) = (–1)^{3} + (–1)^{2} – 9(–1) – 9

= –1 + 1 + 9 – 9

= 0

Hence, the remainder obtained when *x*^{3} + *x*^{2} – 9*x *– 9 is divided by *x* + 1 is 0.

To find the remainder obtained when *x*^{3} + *x*^{2} – 9*x *– 9 is divided by *x* + 2,

put *x *+2 = 0.

*f*(–2) is the remainder.

Now,

*f*(–2) = (–2)^{3} + (–2)^{2} – 9(–2) – 9

= –8 + 4 + 18 – 9

= 5

Hence, the remainder obtained when *x*^{3} + *x*^{2} – 9*x *– 9 is divided by *x* + 2 is 5.

Hence, the remainders obtained when *x*^{3} + *x*^{2} – 9*x *– 9 is divided by *x, x *+ 1 and *x* + 2 respectively are __–9, 0 and 5__.

#### Page No 6.36:

#### Question 7:

The remainder when *f*(*x*) = 4*x*^{3 }– 3*x*^{2} + 2*x* – 1 is divided by 2*x* + 1 is __________.

#### Answer:

Let *f*(*x*) = 4*x*^{3 }– 3*x*^{2} + 2*x* – 1

To find the remainder obtained when 4*x*^{3 }– 3*x*^{2} + 2*x* – 1 is divided by 2*x* + 1,

we use remainder theorem, put 2*x* + 1* *= 0.

*f*($-\frac{1}{2}$) is the remainder.

Now,

$f\left(-\frac{1}{2}\right)=4{\left(-\frac{1}{2}\right)}^{3}-3{\left(-\frac{1}{2}\right)}^{2}+2\left(-\frac{1}{2}\right)-1\phantom{\rule{0ex}{0ex}}=4\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)-1-1\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}-\frac{3}{4}-2\phantom{\rule{0ex}{0ex}}=\frac{-2-3-8}{4}\phantom{\rule{0ex}{0ex}}=-\frac{13}{4}$

Hence, the remainder when *f*(*x*) = 4*x*^{3 }– 3*x*^{2} + 2*x* – 1 is divided by 2*x* + 1 is $\overline{)-\frac{13}{4}}.$

#### Page No 6.36:

#### Question 8:

The degree of a polynomial *f*(*x*) is 7 and that of polynomial *f*(*x*) *g*(*x*) is 56, then degree of *g*(*x*) is ________.

#### Answer:

Given:

Degree of a polynomial *f*(*x*) = 7

Degree of polynomial *f*(*x*) *g*(*x*) = 56

Degree of polynomial *f*(*x*) *g*(*x*) = Degree of â€‹polynomial *f*(*x*) × Degree of â€‹polynomial *g*(*x*)

⇒ 56 = 7 × Degree of â€‹polynomial *g*(*x*)

⇒ Degree of â€‹polynomial *g*(*x*) = $\frac{56}{7}$

⇒ Degree of â€‹polynomial *g*(*x*) = 8

Hence, degree of *g*(*x*) is __8__.

#### Page No 6.36:

#### Question 9:

The remainder when *x*^{15 }is divided by *x *+ 1 is __________.

#### Answer:

Let *f*(*x*) = *x*^{15}

To find the remainder obtained when *x*^{15} is divided by *x* + 1,

we use remainder theorem, put *x* + 1* *= 0.

*f*(−1) is the remainder.

Now,

$f\left(-1\right)={\left(-1\right)}^{15}\phantom{\rule{0ex}{0ex}}=-1$

Hence, the remainder when *x*^{15 }is divided by *x *+ 1 is __−1__.

#### Page No 6.36:

#### Question 10:

If $p\left(x\right)={x}^{2}-4x+3,\mathrm{then}p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=\_\_\_\_\_\_\_\_\_\_\_\_\_.$

#### Answer:

Let *p*(*x*) = *x*^{2} – 4*x* + 3

$p\left(2\right)={\left(2\right)}^{2}-4\left(2\right)+3\phantom{\rule{0ex}{0ex}}=4-8+3\phantom{\rule{0ex}{0ex}}=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}p\left(-1\right)={\left(-1\right)}^{2}-4\left(-1\right)+3\phantom{\rule{0ex}{0ex}}=1+4+3\phantom{\rule{0ex}{0ex}}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^{2}-4\left(\frac{1}{2}\right)+3\phantom{\rule{0ex}{0ex}}=\frac{1}{4}-2+3\phantom{\rule{0ex}{0ex}}=\frac{1}{4}+1\phantom{\rule{0ex}{0ex}}=\frac{1+4}{4}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=-1-8+\frac{5}{4}\phantom{\rule{0ex}{0ex}}=-9+\frac{5}{4}\phantom{\rule{0ex}{0ex}}=\frac{-36+5}{4}\phantom{\rule{0ex}{0ex}}=-\frac{31}{4}\phantom{\rule{0ex}{0ex}}$

Hence, if $p\left(x\right)={x}^{2}-4x+3,\mathrm{then}p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=\overline{)-\frac{31}{4}}.$

#### Page No 6.36:

#### Question 11:

If the polynomial* f*(*x*) = 5*x*^{5} – 3*x*^{3} + 2*x*^{2} – *k* gives remainder 1 when divided by *x* + 1, then *k* = __________.

#### Answer:

Let *f*(*x*) = 5*x*^{5} – 3*x*^{3} + 2*x*^{2} – *k*

To find the remainder obtained when 5*x*^{5} – 3*x*^{3} + 2*x*^{2} – *k* is divided by *x* + 1,

we use remainder theorem, put *x* + 1* *= 0.

*f*(−1) is the remainder.

Now,

$f\left(-1\right)=5{\left(-1\right)}^{5}-3{\left(-1\right)}^{3}+2{\left(-1\right)}^{2}-k\phantom{\rule{0ex}{0ex}}\Rightarrow 1=5\left(-1\right)-3\left(-1\right)+2\left(1\right)-k\phantom{\rule{0ex}{0ex}}\Rightarrow 1=-5+3+2-k\phantom{\rule{0ex}{0ex}}\Rightarrow 1=-k\phantom{\rule{0ex}{0ex}}\Rightarrow k=-1$

Hence, *k* = __–1__.

#### Page No 6.36:

#### Question 12:

The remainder when *f*(*x*) = *x*^{45} is divided by *x*^{2} – 1 is ____________.

#### Answer:

Let *f*(*x*) = *x*^{45}

To find the remainder obtained when *x*^{45} is divided by *x*^{2} – 1,

Let the remainder (*r*) be *ax + b*.

Then,

*f*(*x*) = (*x*^{2} – 1) *q* + *r*

$\Rightarrow {x}^{45}=\left({x}^{2}-1\right)q+\left(ax+b\right)...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Putting}x=1\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{\left(1\right)}^{45}=\left({\left(1\right)}^{2}-1\right)q+\left(a\left(1\right)+b\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 1=0+a+b\phantom{\rule{0ex}{0ex}}\Rightarrow a+b=1...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Putting}x=-1\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{\left(-1\right)}^{45}=\left({\left(-1\right)}^{2}-1\right)q+\left(a\left(-1\right)+b\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -1=0-a+b\phantom{\rule{0ex}{0ex}}\Rightarrow -a+b=-1...\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Solving}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}b=0\mathrm{and}a=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore, the remainder (*r*) = 1(*x*) + 0 =* x.*

Hence, the remainder when *f*(*x*) = *x*^{45} is divided by *x*^{2} – 1 is __ x__.

#### Page No 6.36:

#### Question 1:

Define zero or root of a polynomial.

#### Answer:

A real number *a* is *a* zero (or root) of the polynomial *f*(*x*), if *f *(*a*) = 0

#### Page No 6.36:

#### Question 2:

If $x=\frac{1}{2}$ is a zero of the polynomial f(x) = 8*x*^{3} + *ax*^{2} − 4*x* + 2, find the value of *a*.

#### Answer:

Since is *a* zero of polynomial *f*(*x*).

Therefore

The value of *a* is .

#### Page No 6.37:

#### Question 3:

Write the remainder when the polynomial* f*(*x*) = *x*^{3} + *x*^{2} − 3*x* + 2 is divided by *x *+ 1.

#### Answer:

When the polynomial *f*(*x*), divided by the remainder will be

Thus, the reminder = 5

#### Page No 6.37:

#### Question 4:

Find the remainder when* **x*^{3} + 4*x*^{2} + 4*x* − 3 is divided by *x*.

#### Answer:

When the polynomial *f*(*x*) = *x*^{3} + 4*x*^{2} + 4*x* − 3 is divided by *x* the remainder will be.

Thus, the reminder = −3

#### Page No 6.37:

#### Question 5:

If *x* + 1 is a factor of *x*^{3} + *a*, then write the value of *a*.

#### Answer:

As is a factor of polynomial

i.e.

${\left(-1\right)}^{3}+a=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=1$

Thus, the value of *a* = 1

#### Page No 6.37:

#### Question 6:

If *f*(*x*) = *x*^{4} − 2*x*^{3} + 3*x*^{2} − *ax* − *b* when divided by* x* − 1, the remainder is 6, then find the value of *a* + *b*

#### Answer:

When polynomial divided by

The remainder is 6.

i.e.

Thus, the value of

#### Page No 6.8:

#### Question 1:

If f(x) = 2*x*^{3} − 13*x*^{2} + 17*x* + 12, find (i)* f*(2) (ii) *f*(−3) (iii) *f*(0)

#### Answer:

Let be the given polynomial

(i) The value of *f* (2) can be found by putting *x* = 2

= 10

(ii) The value of *f* (–3) can be found by putting *x* = –3

(iii) The value of *f* (0) can be found by putting *x* = 0

#### Page No 6.8:

#### Question 2:

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:

(i) $f\left(x\right)=3x+1,x=-\frac{1}{3}$

(ii) $f\left(x\right)={x}^{2}-1,x=1,-1$

(iii) $g\left(x\right)=3{x}^{2}-2,x=\frac{2}{\sqrt{3}},-\frac{2}{\sqrt{3}}$

(iv) $p\left(x\right)={x}^{3}-6{x}^{2}+11x-6,x=1,2,3$

(v) $f\left(x\right)=5x-\mathrm{\pi},\mathrm{x}=\frac{4}{5}$

(vi) $f\left(x\right)={x}^{2},x=0$

(vii) $f\left(x\right)=lx+m,x=-\frac{m}{1}$

(viii) $f\left(x\right)=2x+1,x=\frac{1}{2}$

#### Answer:

(i) To check whether the given number is the zero of the polynomial or not we have to find

Hence, is the zeros of the given polynomial.

(ii) To check whether the given number is the zero of the polynomial or not we have to find and

Now,

And ,

Hence, and *x* = 1 are the zeros of the polynomial.

(iii) To check whether the given number is the zero of the polynomial or not we have to find

Now,

And

Here, both the value of , are not satisfied the polynomial. Therefore, they are not the zeros of the polynomial.

(iv) To check whether the given number is the zero of the polynomial or not we have to find

And

Hence, *x* = 1, 2, 3 are the zeros of the polynomial *p*(*x*).

(v)To check whether the given number is the zero of the polynomial or not we have to find

Here, , does not satisfy therefore, is not a zero of the polynomial.

(vi)To check whether the given number is the zero of the polynomial or not we have to find

Hence, *x* = 0 is the zeros of the polynomial.

(vii) To check whether the given number is the zero of the polynomial or not we have to find

Hence, is zeros of the polynomial.

(viii)To check whether the given number is the zero of the polynomial or not we have to find

Hence, does not satisyf the polynomial, so, is not zero of the polynomial.

#### Page No 6.8:

#### Question 3:

If *x* = 2 is a root of the polynomial *f*(*x*) = 2*x*^{2} − 3*x* + 7*a*, find the value of *a*.

#### Answer:

The given polynomial is

If *x* = 2 is the root of the polynomial.

Then

#### Page No 6.8:

#### Question 4:

If $x=-\frac{1}{2}$ is a zero of the polynomial *p*(*x*) = 8*x*^{3} − *ax*^{2}−*x* + 2, find the value of *a*.

#### Answer:

The given polynomial is

If is a zeros of the polynomial *p*(*x*).

then

Therefore,

Hence the value of

#### Page No 6.8:

#### Question 5:

If x = 0 and x = −1 are the roots of the polynomial f(x) =2*x*^{3} − 3*x*^{2}^{ }+ *ax* + *b*, find the value of *a* and *b*.

#### Answer:

The given polynomial is

f(x) =2*x*^{3} − 3*x*^{2}^{ }+ *ax* + *b*

If is zeros of the polynomial *f*(*x*), then f(0) = 0

Similarly, if *x* = − 1 is the zeros of the polynomial of,

Then,

Putting the value of *b* from equation (1)

Thus,

#### Page No 6.8:

#### Question 6:

Find the integral roots of the polynomial f(x) = *x*^{3} + 6*x*^{2} + 11*x* + 6.

#### Answer:

The given polynomial is

Here, *f*(*x*) is a polynomial with integer coefficient and the coefficient of highest degree term is 1. So, the integer roots of *f*(*x*) are factors of 6. Which are by observing.

= 0

Also,

And similarly,

*f*(−3) = 0

Therefore, the integer roots of the polynomial *f*(*x*) are −1, −2, − 3

#### Page No 6.8:

#### Question 7:

Find rational roots of the polynomial f(x) = 2*x*^{3} + *x*^{2} − 7*x* − 6.

#### Answer:

The given polynomial is

f(x) is a cubic polynomial with integer coefficients. If $\frac{b}{c}$ is rational root in lowest terms, then the values of b are limited

to the factors of 6 which are $\pm 1,\pm 2,\pm 3,\pm 6$ and the values of c are limited to the factor of 2 as $\pm 1,\pm 2$. Hence, the possible

rational roots are $\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}$.

Since,

So, 2 is a root of the polynomial

Now, the polynomial can be written as,

Also,

Therefore,

Hence, the rational roots of the polynomialare 2, – 3/2 and – 1.

View NCERT Solutions for all chapters of Class 9