Math Ncert Exemplar 2019 Solutions for Class 9 Maths Chapter 2 - Polynomials

Answer:

For an expression to be a polynomial, any variable in the expression must have the power of a whole number such as $x^2, y^{100}$,etc.
a variable term with negative power or fractional power such as $\frac{1}{x^2}, x^{-3}$, etc. are not a polynomial.
Option A, is not a polynomial because of $-\frac{2}{x^2}$.
Option B, is not a polynomial because of $\sqrt{2x}=\sqrt{2}{x}^{\frac{1}{2}}$
Option C, $x^2+\frac{3x^{{\displaystyle \frac{3}{2}}}}{\sqrt{x}}$ 
$$\begin{gathered} =x^2+\frac{3x^{{\displaystyle \frac{3}{2}}}}{x^{{\displaystyle \frac{1}{2}}}}=x^2+3x^{\frac{3}{2}-\frac{1}{2}} \\ =x^2+3x^1 \end{gathered}$$
Hence it is a polynomial.
In option D, $\frac{x-1}{x+1}$
$$\begin{gathered} =\frac{x-1+1-1}{x+1} \\ =\frac{x+1}{x+1}-\frac{2}{x+1} \end{gathered}$$
$=1-\frac{2}{x+1}$
Hence, it is not a polynomial.

We can conclude that option C is correct.

Answer:

$\sqrt{2}$ is a constant polynomial.
It can be represented as $\sqrt{2}{x}^0$.
Thus, the degree of the non-zero constant polynomial is 0.

Hence, the correct answer is option B.

Answer:

Given: 4x⁴ + 0x³ + 0x⁵ + 5x + 7.
The highest power of the variable in a polynomial is called degree of the polynomial.

Thus, the term with the highest power of x is 4x⁴. So, the degree of polynomial is 4.
Hence, the correct answer is option A.

Answer:

Degree of the zero polynomial (0) is not defined.
Hence, the correct answer is option D.

Answer:

Given: $p\left(x\right)=x^2-2\sqrt{2}x+1$
$\begin{array}{rcl}p\left(2\sqrt{2}\right)& =& {\left(2\sqrt{2}\right)}^2-2\sqrt{2}\left(2\sqrt{2}\right)+1\\ & =& \left(8\right)-\left(8\right)+1\\ & =& 1\end{array}$

Hence, the correct answer is option B.

Answer:

Given: p(x) = 5x – 4x² + 3
p(–1) – 5(–1) – 4(–1)² + 3
= –5 – 4 + 3
= – 6
So, value of polynomial at x = –1 is –6.
Hence, the correct answer is option A.

Answer:

Given: p(x) = x + 3
∴ p(–x) = –x + 3
Now,
p(x) + p(–x) = (x + 3) + (–x + 3)
= 6
Hence, the correct answer is option D.

Answer:

Zero of a polynomial is the value of a variable for which polynomial becomes zero.
In zero polynomial, all coefficients are equal to zero.
Therefore, zero of a zero polynomial is not defined.

Hence, the correct answer is option D.

Answer:

Given: p(x) = 2x + 5
For zero of polynomial, p(x) = 0
$$\begin{gathered} \therefore 2x+5=0 \\ \Rightarrow x=-\frac{5}{2}. \end{gathered}$$
Hence, zero of polynomial p(x) is $-\frac{5}{2}.$
Hence, the correct answer is option B.

Answer:

Given: p(x) = 2x² + 7x – 4
For zero of polynomial, p(x) = 0
$$\begin{gathered} \Rightarrow 2x^2+7x-4=0 \\ \Rightarrow 2x^2+8x-x-4=0 \\ \Rightarrow 2x(x+4)-1(x+4)=0 \\ \Rightarrow (2x-1) (x+4)=0 \\ \Rightarrow 2x-1=0 \mathrm{or} x+4=0 \\ \Rightarrow x=-4, \frac{1}{2} \end{gathered}$$
Hence, the correct answer is option B.

Answer:

Let p(x) = x⁵¹ + 51
By Remainder Theorem, we have
Remainder = p(–1), since divisor is (x + 1)
p(–1) = (1)⁵¹ + 51

​= –1 + 51

= 50 

Hence, the correct answer is option D.

Answer:

Let p(x) = 2x² + kx
By Remainder Theorem
Remainder = p(–1) = 0    ((x + 1) is a factor of p(x))
∴ p(–1) = 2(–1)² + k(–1) = 0
⇒ 2 – k = 0
⇒ k = 2
Hence, the correct answer is option C.

Answer:

Apply Remainder Theorem.
x + 1 = 0
⇒ x = –1
Putting the value of x = –1 in all equations
(A) x³ + x² – x + 1
= (–1)³ + (–1)² – (–1) + 1
= –1 + 1 + 1 + 1
= 2
Therefore, (x + 1) is not a factor of equation.

(B) x³ + x² + x + 1
= (–1)³ + (–1)² + (–1) + 1
= –1 + 1 – 1 + 1
= 0
Therefore, (x + 1) is a factor of equation.

(C) x⁴ + x³ + x² + 1
= (–1)⁴ + (–1)³ + (–1)² + 1
= 1 – 1 + 1 + 1
= 2
Therefore, (x + 1) is not a factor of equation.

(D) x⁴ + 3x³ + 3x² + x + 1
= (–1)⁴ + 3(–1)³ + 3(–1)² + (–1) + 1
= 1 – 3 + 3 – 1 +  1
Therefore, (x + 1) is not a factor of equation.

Hence, the correct answer is option B.

Answer:

Let Q(x) = (25x² – 1) + (1 + 5x)²
= (25x² – 1) + (25x² + 10x + 1)
= 50x² + 10x
= 10x (5x + 1)
Thus, factors of Q(x) are 10x and (5x + 1).

Hence, the correct answer is option D.

Answer:

Given: (249)² – (248)²
we know,
a² – b² = (a – b) (a + b)
∴ (249)² – (248)²
= (249 – 248)(249 + 248)
= (1)(497)
= 497

Hence, the correct answer is option D.

Answer:

Let p(x) = 4x² + 8x + 3
= 4x² + 6x + 2x + 3
= 2x(2x + 3) + 1(2x + 3)
= (2x + 1) (2x + 3)
Hence, the correct answer is option B.

Answer:

(x + y)³ – (x³ + y³)
= x³ + y³ + 3xy (x + y) – x³  – y³
= 3xy (x + y)
Thus, factors are 3xy and (x + y).

Hence, the correct answer is option D.

Answer:

We know,
(x + y)³ = x³ + y³ + 3x²y + 3xy²

∴ (x + 3)³ = x³ + 27 + 9x² + 27x
So, coefficient of x is 27.

Hence, the correct answer is option D.

Answer:

Given: $\frac{x}{y}+\frac{y}{x}=-1 \left(x, y \ne 0\right)$
⇒ x² + y² = –xy
⇒ x² + y² + xy = 0      ...(1)
Now,
 x³ – y³ = (x – y) (x² + y² + xy)
= (x – y) (0)        (from (1))
= 0
Hence, the correct answer is option C.

Answer:

From RHS, we have
$49x^2-b=\left(7x+\frac{1}{2}\right) \left(7x-\frac{1}{2}\right)=\left(49x^2-\frac{1}{4}\right) \left(\because \left(a^2-b^2\right)=(a-b)(a+b)\right)$
Now, in LHS, we have
⇒ 49x² – b
Comparing LHS with RHS, we get
$b=\frac{1}{4}$

Hence, the correct answer is option C.
 

Answer:

Given: a + b + c = 0
we know,
a³ + b³ + c³ – 3abc = (a + b + c ) (a² + b² + c² – ab – bc – ca)
∴ a³ + b³ + c³ – 3abc = 0
⇒ a³ + b³ + c³ = 3abc
Hence, the correct answer is option C.
 

Answer:

(i) Polynomial, because the exponent of the variable of 8 or $8x^0$ is 0, which is a whole number.
(ii) Polynomial, because the exponent of  variable $\sqrt{3}{x}^2-2x$ is a whole number.
(iii) Not polynomial, because the exponent of the variable of $\left(1-\sqrt{5x}\right)$ is $\frac{1}{2},$ whole number.
(iv) Polynomial, because the exponent of the variable $\frac{1}{5x^{-2}}+5x+7=\frac{1}{5}{x}^2+5x+7$ is a whole number.
(v) Not polynomial, because the exponent of $\frac{\left(x-2\right)\left(x-4\right)}{x}=\frac{x^2-6x+8}{x}=x-6+8x^{-1}$ is –1, which is not a whole number.
(vi) Not polynomial, $\frac{1}{x+1}$ is a rational expression $\left(f\left(x\right)=\frac{q\left(x\right)}{r\left(x\right)} \mathrm{is} \mathrm{a} \mathrm{rational} \mathrm{expression}, r\left(x\right)\ne 0, q\left(x\right) \mathrm{and} r\left(x\right) \mathrm{are} \mathrm{polynomial}\right)$
(vii) Polynomial. because exponent of  $\frac{1}{7}{a}^3-\frac{2}{\sqrt{3}}{a}^2+4a-7$ is whole number.
(viii) Not polynomial, because exponent of  $\frac{1}{2x}=\frac{1}{2}{x}^{-1}$ is –1, which is not a whole number.

Answer:

(i) False, because a binomial has exactly two terms.
(ii) False, because every polynomial is not binomial 
e.g., 5x is a polynomial but not binomial.
(iii) True, because a binomial is a polynomial whose degree is a whole number greater than or equal to one.
(iv) False, because zero of polynomial can be any real number.
e.g., Q(x) = x – 6, then 6 is zero of polynomial Q(x).
(v) False, because a polynomial can have any number of zeros, it depends upon degree of polynomial.
(vi) False, because the sum of any two polynomial of same degree is not always the same.
e.g.,  f(x) = x⁵ + 3x² ; g(x) = –x⁵ + 2
∴ f(x) + g(x) = (x⁵ + 3x²) + (–x⁵ + 2)
= 3x² + 2
Which is not a polynomial of degree 5.

 

Answer:

(i) x² + x + 1, Polynomial in one variable i.e. x.
(ii) y³ – 5y, Polynomial in one variable because it contains only one variable i.e., y.
(iii) xy + yz + zx , polynomial in three variable, i.e., x, y and z.
(iv)  x² – 2xy + y² + 1, polynomial in two variable, i.e., x and y.

Answer:

(i) 2x – 1, degree is 1, because the maximum exponent of x is 1.
(ii) –10, degree is 0, because the exponent of x is 0.
(iii) x³ – 9x + 3x⁵, degree is 5, because the maximum exponent of x is 5.
(iv) y³(1 – y⁴) = y³ – y⁷, degree is 7, because the maximum exponent of y is 7. 

Answer:

Given: $\frac{x^3+2x+1}{5}-\frac{7}{2}{x}^2-x^6$
$=\frac{x^3}{5}+\frac{2x}{5} +\frac{1}{5}-\frac{7}{2}{x}^2-x^6$
(i) Degree of polynomial is the highest degree of its terms.
Here, degree of $\frac{x^3}{5}+\frac{2x}{5} +\frac{1}{5}-\frac{7}{2}{x}^2-x^6$  is 6.
(ii) Coefficient of x³ is $\frac{1}{5}$.
(iii) Coefficient of x⁶ is –1.
(iv) The constant term in polynomial is $\frac{1}{5}$.

Answer:

(i) Coefficient of x² in $\frac{\pi }{6}x+x^2-1$ is 1.
(ii) Coefficient of  x² in 3x – 5 is 0.
(iii)  (x – 1) (3x – 4) = 3x² – 7x + 4
 Coefficient of x² in 3x² – 7x + 4 is 3.
(iv) (2x – 5) (2x² – 3x + 1) = 4x³ – 6x² + 2x – 10x² + 15x – 5
= 4x³ – 16x² + 17x – 5
Coefficient of x² in 4x³ – 16x² + 17x – 5 is –16.

Answer:

(i) 2 – x² + x³ is a cubic polynomial.
(ii) 3x³ is a cubic polynomial because its highest degree is 3.
(iii) $5t-\sqrt{7}$ is a linear polynomial.
(iv) 4 – 5y² is a quadratic polynomial.
(v) 3 is a constant polynomial.
(vi) 2 + x is a linear polynomial.
(vii) y³ – y is a cubic polynomial.
(viii) 1 + x + x² is a quadratic polynomial.
(ix) t² is a quadratic polynomial.
(x) $\sqrt{2}x-1$ is a linear polynomial.

 

Answer:

(i) Monomial of degree 1  :  5x
(ii) Binomial of degree 20  :  6y²⁰ + 5
(iii) Tri-nomial of degree 2  :  x² + x + 1

Answer:

Given: p(x) = 3x³ – 4x² + 7x – 5
∴ p(x) = 3(3)³ – 4(3)² + 7(3) – 5
           = 81 – 36 + 21 – 5

Answer:

Given: p(x) = x² – 4x + 3
Now, p(2) = (2)² – 4(2) + 3

Answer:

(i)  Given p(x) = 10x – 4x² – 3
Now, 
p(0) = 10(0) – 4(0)² – 3

Answer:

(i) False, because 
x – 3 = 0
⇒ x = 3
Thus, zero of x – 3 is 3 not –3.
(ii) True, because
3x + 1 =0
⇒ x = $\frac{-1}{3}$
Thus, $\frac{-1}{3}$ is zero of 3x + 1.
(iii) False, because
4 – 5y = 0
⇒ y = $\frac{4}{5}$
Thus, $\frac{4}{5}$ is zero of 4 – 5y.
(iv) True, because
t² – 2t = 0
⇒ t(t – 2) = 0
⇒ t = 0, 2
Thus, 0 and 2 are zeros of t² – 2t.
(v) True, because
y² + y – 6 = 0
⇒ y² + 3y – 2y – 6 = 0
⇒ (y + 3)(y – 2) = 0
⇒ y = –3, 2
Thus, –3 is a zero of y² + y – 6.

Answer:

Given: (i) p(x) = x – 4
For zero of polynomial, p(x) = 0
⇒ x – 4 = 0
⇒ x = 4
Hence, zero of polynomial, p(x) is 4.
(ii) Given: g(x) = 3 – 6x
For zero of polynomial, g(x) = 0
⇒ 3 – 6x = 0
⇒ x = $\frac{1}{2}$
Hence, zero of polynomial, g(x) is $\frac{1}{2}.$
(iii) Given: q(x) = 2x – 7
For zero of polynomial, q(x) = 0
⇒ 2x – 7 = 0
⇒ x = $\frac{7}{2}.$
Hence, zero of polynomial q(x) is $\frac{7}{2}.$
(iv) Given: h(y) = 2y
For zero of polynomial, h(y) = 0
⇒ 2y = 0
⇒ y = 0
Hence, zero of polynomial h(y) is 0.

Answer:

Given: p(x) = (x – 2)² – (x + 2)²
For zero of polynomial,
p(x) = 0
⇒ (x – 2)² – (x + 2)² = 0
⇒ (x² – 4x + 4) – (x² + 4x + 4) = 0
⇒ –8x = 0
⇒ x = 0
Hence, zero of polynomial p(x) is 0.

Answer:

From Long Division Method, we have


Hence, Quotient : x³ + x² + x + 1 and Remainder : 2

Answer:

(i) Given: p(x) = x³ – 2x² – 4x – 1, g(x) = x + 1
Now, g(x), we get
x + 1 = 0
⇒ x = –1
By remainder theorem, when p(x) divided by g(x), remainder is p(–1).
∴ p(–1) = (–1)³ – 2(–1)² – 4(–1) – 1
= –1 – 2 + 4 – 1
= 0
Hence, value of remainder is 0.

(ii) Given: p(x) = x³ – 3x² + 4x + 50, g(x) = x – 3 
Now, p(x) = 0, we get
x – 3 = 0
⇒ x = 3
By remainder theorem, when p(x) is divided by g(x), remainder is p(3).
∴ p(3) = (3)³ – 3(3)² + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
Hence, value of remainder is 62.
(iii)  Given:  p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1
Now, g(x) = 0 we get
2x – 1 = 0
⇒ x = $\frac{1}{2}$
By Remainder Theorem, when p(x) is divided by g(x) remainder is $p\left(\frac{1}{2}\right)$.
$\begin{array}{rcl}\therefore p\left(\frac{1}{2}\right)& =& 4{\left(\frac{1}{2}\right)}^3-12{\left(\frac{1}{2}\right)}^2+14\left(\frac{1}{2}\right)-3\\ & =& \frac{1}{2}-3+7-3\\ & =& \frac{3}{2}\end{array}$
Hence, value of remainder is $\frac{3}{2}$.
(iv) Given: p(x) = x³ – 6x² + 2x – 4, g(x) = $1-\frac{3}{2}x$
Now, g(x) = 0, we get
$$\begin{gathered} 1-\frac{3x}{2}=0 \\ \Rightarrow x=\frac{2}{3} \end{gathered}$$ 
By remainder theorem, when p(x) is divided by g(x) remainder is $p\left(\frac{2}{3}\right)$.
$\begin{array}{rcl}\therefore p\left(\frac{2}{3}\right)& =& {\left(\frac{2}{3}\right)}^3-6{\left(\frac{2}{3}\right)}^2+2\left(\frac{2}{3}\right)-4\\ & =& \frac{8}{27}-6\left(\frac{4}{9}\right)+\frac{5}{3}-4\\ & =& -\frac{136}{27}\end{array}$
Hence, value of remainder is $-\frac{136}{27}$.

Answer:

(i) Given: p(x) = x³ – 5x² + 4x – 3, g(x) = x – 2
zero of polynomial g(x)
⇒ g(x) = 0
⇒ x – 2 = 0
⇒ x – 2
For p(x) to be a multiple of g(x), remainder must be equal to zero 
​From remainder theorem, we have
Remainder = p(2) = (2)³ – 5(2)² + 4(2) – 3
= 8 – 20 + 8 – 3
= –7
Remainder ≠ 0
Here, p(x) is not a multiple of g(x).
(ii) Given: p(x) = 2x³ – 11x² – 4x + 5, g(x) = 2x + 1
zero of polynomial g(x),
⇒ g(x) = 0
⇒ 2x + 1 = 0
⇒ x = $\frac{-1}{2}$
For p(x) to be a multiple of g(x), remainder must be equal to zero.
From remainder theorem, we have
Remainder = $\begin{array}{rcl}p\left(\frac{-1}{2}\right)& =& 2{\left(\frac{-1}{2}\right)}^3-11{\left(\frac{-1}{2}\right)}^2-4\left(\frac{-1}{2}\right)+5\end{array}$

Answer:

(i) Let p(x) = 69 + 11x – x² + x³ and g(x) = x + 3
Now, g(x) = 0
⇒ x + 3 = 0
⇒ x = –3
Now, for g(x) to be a factor of p(x).
p(–3) = 0
∴ p(–3) = 69 + 11(–3) – (–3)² + (–3)³
= 69 – 33 – 9 – 27
= 0
Hence, (x + 3) is a factor of 69 + 11x – x² + x³.

(ii) Let p(x) = x + 2x³ – 9x² + 12 and g(x) = 2x – 3
Now, g(x) = 0
⇒ 2x – 3 = 0
⇒ x = $\frac{3}{2}$
Now, for g(x) to be a factor of g(x).
$p\left(\frac{3}{2}\right)=0$
$\begin{array}{rcl}\therefore p\left(\frac{3}{2}\right)& =& \left(\frac{3}{2}\right)+2{\left(\frac{3}{2}\right)}^3-9{\left(\frac{3}{2}\right)}^2+12\\ & =& \frac{3}{2}+\frac{27}{4}-\frac{81}{4}+12\\ & =& \frac{3}{2}-\frac{27}{2}+12\\ & =& -12+12\\ & =& 0\end{array}$
Hence, (2x – 3) is a factor of x + 2x³ – 9x² + 12.

Answer:

(i) Let p(x) = 3x² + 6x – 24 and g(x) = x – 2
Now, g(x) = 0
⇒ x – 2 = 0
⇒ x = 0
For g(x) to be a factor of p(x).
p(2) = 0
∴ p(2) = 3(2)² + 6(2) – 24
= 12 + 12 – 24
= 0
Hence, (x – 2) is a factor of 3x² + 6x – 24.

(ii) Let p(x) =  4x² + x – 2 and g(x) = x – 2
Now, g(x) = 0
⇒ x – 2 = 0
⇒ x = 2
For g(x) to be a factor of p(x).
p(2) = 0
∴ p(2) = 4(2)² + 2 – 2
= 16
Hence, (x – 2) is not a factor of 4x² + x – 2.

Answer:

(i) Let p(x) = p¹⁰ – 1;  q(x) = p¹¹ – 1 and g(x) = p – 1
⇒ p – 1 = 0
⇒ â€‹p = 1
For (p – 1) to be a factor of p(x) and q(x),
p(1) = 0 and q(1) = 0
∴ p(1) = (1)¹⁰ – 1 = 0
q(1) = (1)¹¹ – 1 = 0

Hence, (p – 1) is a factor of both (p¹⁰ – 1) and (p¹¹ – 1).

Answer:

Let p(x) = x³ – 2mx² + 16
Since, p(x) is divisible by (x + 2).
∴ p(–2) = 0 (By remainder theorem)
⇒ (–2)³ – 2m(–2)² + 16 = 0
⇒ –8 – 8m + 16 = 0
⇒  8m = 8
⇒ m = 1
Hence, value of m is 1.

Answer:

Let p(x) = x⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a
Since, g(x) is a factor of p(x).
∴ By remainder theorem.
p(–2a) = 0
⇒ (–2a)⁵ – 4a²(–2a)³ + 2(–2a) + 2a + 3 = 0
⇒ –32a⁵ + 32a⁵ – 4a + 2a + 3 = 0
⇒ –2a + 3 = 0
⇒ a = $\frac{3}{2}$
Hence, value of a is $\frac{3}{2}$.

Answer:

Let p(x) =  8x⁴ + 4x³ – 16x² + 10x + m and g(x) =  2x – 1
g(x) = 0
⇒ x = $\frac{1}{2}$
For g(x) to be a factor of p(x),
$p\left(\frac{1}{2}\right)=0$
$$\begin{gathered} \Rightarrow 8{\left(\frac{1}{2}\right)}^4+4{\left(\frac{1}{2}\right)}^3-16{\left(\frac{1}{2}\right)}^2+10\left(\frac{1}{2}\right)+m=0 \\ \Rightarrow \frac{1}{2}+\frac{1}{2}-4+5+m=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 1+1+m=0 \\ \Rightarrow m=-2 \end{gathered}$$

Hence, value of m is –2.

Answer:

Given: p(x) = ax³ + x² – 2x + 4a – 9
g(x) = x + 1
⇒ x + 1 = 0
⇒ x = –1
If g(x) is a factor of p(x)
∴ p(–1) = 0
⇒ a(–1)³ + (–1)² – 2(–1) + 4a – 9 = 0
⇒ –a + 1 + 2 + 4a – 9 = 0
⇒ 3a – 6 = 0
⇒ a = 2
Hence, value of a is 2.

Answer:

(i) x² + 9x + 18
⇒ x² + 3x +  6x + 18
⇒ x(x + 3) + 6(x + 3)
⇒ (x + 3) (x + 6)

(ii) 6x² + 7x – 3
⇒ 6x² + 9x – 2x – 3
⇒ 3x(2x + 3) – 1(2x + 3)
⇒ (2x + 3) (3x – 1)

(iii) 2x² – 7x – 15
⇒ 2x² – 10x + 3x – 15
⇒ 2x(x – 5) + 3(x – 5)
⇒ (2x + 3) (x – 5)

(iv) 84 – 2r – 2r²
⇒ –2(r² + r – 42)
⇒ –2(r² + 7r – 6r – 42)
⇒ –2(r(r + 7) – 6(r + 7))
⇒ –2(r + 7) (r – 6)

Answer:

(i) 2x³ – 3x² – 17x + 30
⇒  2x³ + 2x² – 12x – 5x² – 5x + 30
⇒ 2x(x² + x – 6) – 5(x² + x – 6) 
⇒ (2x – 5) (x² + x –  6)
⇒  (2x – 5) (x² + 3x – 2x – 6)
⇒  (2x – 5) (x(x + 3) – 2(x + 3))
⇒  (2x – 5) (x –  2) (x + 3)

(ii) x³ – 6x² + 11x – 6
⇒ x³ – 5x² – x² + 5x + 6x + 30
⇒ x³ – 5x² + 6x – x² + 5x – 6
⇒ x( x² – 5x + 6) – 1(x² – 5x + 6)
⇒  (x – 1) (x² – 5x + 6)
⇒   (x – 1) (x(x – 2) – 3 (x – 2))
⇒   (x – 1) (x – 2) (x – 3) 

(iii) x³ + x² – 4x – 4
⇒  x²(x + 1) – 4(x + 1)
⇒ (x + 1) (x² – 4)
⇒ (x + 1) (x – 2) (x + 2)

(iv) 3x³ – x² – 3x + 1
⇒  3x³ – 3x – x² + 1
⇒  3x(x² – 1) – 1(x² – 1)  
⇒  (3x – 1) (x² – 1)
⇒  (3x – 1) (x – 1) â€‹(x + 1)

Answer:

(i) (103)³
⇒ (100 + 3)³
⇒ (100)³ + (3)³ + 3 × 100 × 3 (100 + 3)
⇒ 1000000 + 27 + 92700
⇒ 1092727

(ii) 101 × 102
⇒(100 + 1) (100 + 2)
⇒ (100)² + 100(1 + 2) + 1 × 2
⇒ 10000 + 300 + 2
⇒ 10302

(iii) 999²
⇒ (1000 – 1)²
⇒ (1000)² + (1)² – 2 × 1000 × 1
⇒ 1000000 + 1 – 2000
⇒ 998001

Answer:

(i) 4x² + 20x + 25
= (2x)² + 2 × (2x) × (5) + (5)²
= (2x + 5)²                (using (a + b)² = a² + 2ab + b²)

(ii) 9y² – 66yz + 121z²
= (3y)² – 2 × (3y) × (11z) + (11z)²
=(3y – 11z)²            (using (a – b)² = a² – 2ab + b²)

(iii) ${\left(2x+\frac{1}{3}\right)}^2-{\left(x-\frac{1}{2}\right)}^2$
$$\begin{gathered} =\left\{\left(2x+\frac{1}{3}\right)+\left(x-\frac{1}{2}\right)\right\} \left\{\left(2x+\frac{1}{3}\right)-\left(x-\frac{1}{2}\right)\right\} \\ =\left\{2x+\frac{1}{3}+x-\frac{1}{2}\right\} \left\{2x+\frac{1}{3}-x+\frac{1}{2}\right\} \\ =\left(3x-\frac{1}{6}\right)\left(x+\frac{5}{6}\right) \end{gathered}$$

Answer:

(i) 9x² – 12x + 3
= 9x² – 9x – 3x + 3
= 9x(x – 1) – 3(x – 1)
= (9x – 3) (x – 1)
= 3(3x – 1) (x – 1)

(ii) 9x² – 12x + 4
= (3x)² – 2 × (3x) × (2) + (2)²
=  (3x – 2)²                           (∵ (a – b)² = a² – 2ab + b²)

Answer:

(i) (4a – b + 2c)²

Answer:

(i) 9x² + 4y² + 16z² + 12xy – 16yz – 24xz
= (3x)² + (2y)² + (–4z)² + 2(3x)(2y) + 2(2y)(–4z) + 2(–4z)(3x)
= (3x + 2y – 4z)²
= (3x + 2y – 4z) (3x + 2y – 4z) 

(ii) 25x² + 16y² + 4z² – 40xy + 16yz – 20xz
= (–5x)² + (4y)² + (2z)² + 2(–5x)(4y) + 2(4y)(2z) + 2(2z)(–5x)
= (–5x + 4y + 2z)²
= (–5x + 4y + 2z) (–5x + 4y + 2z)

(iii) 16x² + 4y² + 9z² – 16xy – 12yz + 24xz
= (–4x)² + (2y)² + (–3z)² + 2(–4x)(2y) + 2(2y)(–3z) + 2(–3z)(–4x)
= (–4x + 2y – 3z)² 
= (–4x + 2y – 3z)  (–4x + 2y – 3z)​

Answer:

Given: a + b + c = 9 and ab + bc + ca = 26
we know,
(a + b + c)² = (a² + b² + c²) + 2(ab + bc + ca)
⇒ (9)² = (a² + b² + c²) + 2(26)
⇒ 81 = (a² + b² + c²) + 52
⇒ (a² + b² + c²) = 29

Hence, value of (a² + b² + c²) is 29.

Answer:

(i) (3a – 2b)³ 
we know,
(a – b)³ = a³ – b³ – 3ab(a – b)
∴ (3a – 2b)³ = (3a)³ – (2b)³ – 3(3a)(2b)(3a – 2b)
= 27a³ – 8b³ – 18ab (3a – 2b)
= 27a³ – 8b³ – 54a²b + 36ab²

(ii) ${\left(\frac{1}{x}+\frac{y}{3}\right)}^3$
we know,
(a + b)³ = a³ + b³ + 3a²b + 3ab²
$\begin{array}{rcl}{\left(\frac{1}{x}+\frac{y}{3}\right)}^3& =& {\left(\frac{1}{x}\right)}^3+{\left(\frac{y}{3}\right)}^3+3{\left(\frac{1}{x}\right)}^2\left(\frac{y}{3}\right)+3\left(\frac{1}{x}\right){\left(\frac{y}{3}\right)}^2\\ & =& \frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x^2}+\frac{y^2}{3x}\end{array}$

(iii) ${\left(4-\frac{1}{3x}\right)}^3$
we know,
(a – b)³ = a³ – b³ – 3a²b + 3ab²
$\begin{array}{rcl}\therefore {\left(4-\frac{1}{3x}\right)}^3& =& {\left(4\right)}^3-{\left(\frac{1}{3x}\right)}^3-3{\left(4\right)}^2\left(\frac{1}{3x}\right)+3\left(4\right){\left(\frac{1}{3x}\right)}^2\\ & =& 64-\frac{1}{27x^3}-\frac{16}{x}+\frac{4}{3x^2}\end{array}$




 

Answer:

(i) 1 – 64a³ – 12a + 48a²
= (1)³ – (4a)³ – 3(1)²(4a) + 3(1)(4a)² 
= (1 – 4a)³   (∵ Using (a – b)³ = a³ – b³ – 3a²b + 3ab²)

(ii) $8p^3+\frac{12}{5}{p}^2+\frac{6}{25}p+\frac{1}{125}$
$$\begin{gathered} ={\left(2p\right)}^3+3{\left(2p\right)}^2\left(\frac{1}{5}\right)+3\left(2p\right){\left(\frac{1}{5}\right)}^2+{\left(\frac{1}{5}\right)}^3 \\ ={\left(2p+\frac{1}{5}\right)}^3 \left[\because {(a + b)}^3=a^3+b^3+3a^{2}b+3a{b}^2\right] \end{gathered}$$

Answer:

(i) $\left(\frac{x}{2}+2y\right) \left(\frac{x^2}{4}-xy+4y^2\right)$
$=\left(\frac{x}{2}+2y\right) \left[{\left(\frac{x}{2}\right)}^2- \left(\frac{x}{2}\right)\left(2y\right) + {\left(2y\right)}^2\right]$
Using Identity,
a³ + b³ = (a + b) (a² – ab + b²)
we get,
$$\begin{gathered} ={\left(\frac{x}{2}\right)}^3+{\left(2y\right)}^3 \\ =\frac{x^3}{8}+8y^3 \end{gathered}$$

(ii) (x² – 1) (x⁴ + x² + 1)
= (x² – 1) ((x²)² + (x)²(1) + (1)²)
Using Identity,
a³ – b³ = (a – b) (a² + ab + b²)
we get,
= (x²)³ – (1)³
= x⁶ – 1

Answer:

(i) 1 + 64x³ 
we know,
a³ + b³ = (a + b) (a² – ab + b²)
Now,
1 + 64x³ = (1)³ + (4x)³
= (1 + 4x) (1 – 4x + 16x²)

(ii) $a^3-2\sqrt{2}{b}^3$ 
we know,
a³ – b³ = (a – b) (a² + ab + b²)
Now,
$\begin{array}{rcl}{a}^{3 }- 2\sqrt{2} & =& {\left(a\right)}^3 - {\left(\sqrt{2}b\right)}^3\\ & =& (a-\sqrt{2}b) (a^2+\sqrt{2}ab+2b^2)\end{array}$

Answer:

we know,
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ – 3abc
Now,
(2x – y + 3z) (4x² + y² + 9z² + 2xy + 3yz – 6xz)
= (2x + (–y) + 3z)((2x)² + (–y)² + (3z)² – (2x)(–y) – (–y)(3z) – (3z)(2x))
= (2x)³ + (–y)³ + (3z)³ – 3(2x) (–y) (3z)
= 8x³ – y³ + 27z³ + 18xyz

Answer:

(i) a³ – 8b³ – 64c³ – 24abc 
we know,
x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)
Now,
(a)³ + (–2b)³ + (–4c)³ – 3(a)(–2b)(–4c)
= {(a – 2b – 4c) ((a)² + (–2b)² + (–4c)² –(a)(–2b) – (–2b)(–4c) – (–4c)(a)}
= {(a – 2b –4c) (a² + 4b² + 16c² + 2ab – 8bc + 4ca)}

(ii) $2\sqrt{2}{a}^3+8b^3-27c^3+18\sqrt{2}abc$.
we know,
x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)
Now,
$$\begin{gathered} {\left(\sqrt{2}a\right)}^3+{\left(2b\right)}^3+{\left(-3c\right)}^3 -3\left(\sqrt{2}a\right)\left(2b\right)\left(-3c\right) \\ =\left(\sqrt{2}a+2b-3c\right)\left[{\left(\sqrt{2}a\right)}^2+{\left(2b\right)}^2+{\left(–3c\right)}^2-\left(\sqrt{2}a\right)\left(2b\right)-\left(2b\right)\left(-3c\right)-\left(-3c\right)\left(\sqrt{2}a\right)\right] \\ =\left(\sqrt{2}a+2b-3c\right)\left(2a^2+4b^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}ac\right) \end{gathered}$$

Answer:

(i) ${\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{3}\right)}^3-{\left(\frac{5}{6}\right)}^3$
Let $x=\frac{1}{2}, y=\frac{1}{3} \mathrm{and} z=\frac{-5}{6}$
$\therefore x+y+z=0$
We know,
if  x + y + z = 0
then, x³ + y³ + z³ = 3xyz
$\begin{array}{rcl}\therefore {\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{3}\right)}^3+ {\left(\frac{-5}{6}\right)}^3& =& 3\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\left(\frac{-5}{6}\right)\\ & =& \frac{-5}{12}\end{array}$

(ii) (0.2)³ – (0.3)³ + (0.1)³
Let x = 0.2, y = –0.3, z = 0.1
$\therefore x+y+z=0$
we know,
if  x + y + z = 0
then, x³ + y³ + z³ = 3xyz

$\begin{array}{rcl}\therefore {(0.2)}^3+{(-0.3)}^3+{(0.1)}^3& =& 3(0.2)(-0.3)(0.1)\\ & =& -0.018\\ & =& \frac{-9}{500}\end{array}$

Answer:

(x – 2y)³ + (2y – 3z)³ + (3z – x)³ 
Let a = x – 2y , b = 2y – 3z, c = 3z – x
If a + b + c = 0  (∴ x – 2y + 2y + 2y – 3z + 3z + 3z + 3z – x = 0)
then,  a³ + b³ + c³ = 3abc
∴ (x – 2y)³ +  (2y – 3z)³ + (3z – x)³
= 3(x – 2y) (2y – 3z) (3z – x)
 

Answer:

(i) we know
a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)
∴ x³ + y³ – 12xy + 64
= (x)³ + (y)³ + (4)³ – 3(x)(y)(4)
= 0(x² + y² + 16 – xy – 4y – 4x)                        (∵ x + y = –4)
= 0

(ii) we know,
a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)
∴ x³ – 8y³ – 36xy – 216
= (x)³ + (–2y)³ + (–6)³ – 3(x)(–2y)(–6)
= (x – 2y – 6) (x² + 4y² + 36 + 2xy – 12y + 6x)
Now,  x – 2y = 6   (given)
= (6 – 6) (x² + 4y² + 36 + 2xy – 2xy – 12y + 6x)
= 0

Answer:

Given: Area = 4a² + 4a – 3
we know,
Area of rectangle = length × breadth
Using the Method of splitting middle term,
4a² + 4a – 3
= 4a² + 6a – 2a – 3
= 2a(2a + 3) – 1(2a + 3)
= (2a – 1) (2a + 3)
∴ length × breadth = (2a – 1) (2a + 3)
Hence, Possible expression for length and breadth
length = (2a – 1), breadth = 2a + 3 or
length = (2a + 3), breadth = 2a – 1

Answer:

Let p(z) = az³ + 4z² + 3z – 4, q(z) = z³ – 4z + a, r(z) = z – 3
Now, zero of r(z),
r(z) = 0
⇒ z – 3 = 0
⇒ z = 3

By remainder theorem, we have
p(3) = q(3)
⇒ p(3) = q(3)
⇒ a(3)³ + 4(3)² + 3(3) – 4 = (3)³ – 4(3) + a
⇒ 27a + 36 + 9 – 4 = 27 – 12 + a
⇒ 27a – a = 15 – 41
⇒ 26a = –26
⇒ a = –1

Answer:

Given:  p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7 
when p(x) is divided by (x + 1), remainder is 19.
By Remainder Theorem, we have
p(–1) = 19
⇒ (–1)⁴ – 2(–1)³ + 3(–1)² – a(–1) + 3a – 7 = 19
⇒ 1 + 2 + 3 + a + 3a – 7 = 19
⇒ 4a – 1 = 19
⇒4a = 20
⇒a = 5
Thus, the value of a is 5.

Therefore,  p(x) = x⁴ – 2x³ + 3x² – 5x + 8
Now, when p(x) is divided by (x + 2), remainder will be p(–2)
∴ p(–2) = (–2)⁴ – 2(–2)³ + 3(–2)² –5(–2) + 8
= 16 + 16 + 12 + 10 + 8
= 62

Hence, the remainder of polynomial p(x) when divided by (x + 2) is 62.

Answer:

 Let Q(x) = px² + 5x + r
(x – 2) and $\left(x-\frac{1}{2}\right)$ are factors of Q(x).
∴ Q(2) = 0
⇒ p(2)² + 5(2) + r = 0
​⇒ 4p + r + 10 = 0          .....(1)

Also,
    $Q\left(\frac{1}{2}\right)=0$
$$\begin{gathered} \Rightarrow p{\left(\frac{1}{2}\right)}^2+5\left(\frac{1}{2}\right)+r = 0 \\ \Rightarrow \frac{p}{4}+r+\frac{5}{2}=0 \\ \Rightarrow p+4r+10=0 .....\left(2\right) \end{gathered}$$

Now, subtracting the equations (1) and (2),
(4p – p) + (r – 4r) + (10 – 10) = 0
⇒ 3p – 3r = 0
⇒ p = r

Hence, proved.

Answer:

Let p(x) = 2x⁴ – 5x³ + 2x² – x + 2
Now,
x² –3x + 2
= x² – 2x – x + 2        (By splitting Middle term)
= x(x – 2) – 1(x – 2)
= (x – 1) (x – 2)

Therefore, p(x) is divisible by x² –3x + 2, if p(1) = 0 and p(2) = 0
∴ p(1) = 2(1)⁴ – 5(1)³ + 2(1)² – (1) + 2
= 2 – 5 + 2 – 1 + 2
= 0
Also,
p(2) = 2(1)⁴ – 5(2)³ + 2(2)² – (2) + 2
= 32 – 40 + 8 – 2 + 2
= 0
Hence, p(x) is divisible by x² – 3x + 2.

Answer:

(2x – 5y)³ – (2x + 5y)³
we have,
(a – b)³ = a³ – b³ – 3ab(a – b)
(a + b)³ = a³ + b³ + 3ab(a + b)

∴ (2x – 5y)³ – (2x + 5y)³
= (8x³ – 125y³ – 30xy(2x – 5y)) – (8x³ + 125y³ + 30xy(2x + 5y))
= 8x³ – 125y³ – 60x²y + 150xy² – 8x³ – 125y³ – 60x²y – 150xy² 
= – 250y³ – 120x²y

Answer:

Now, (x² + 4y² + z² + 2xy + xz – 2yz) (–z + x – 2y)
= x²(–z + x – 2y) + 4y²(–z + x – 2y) + z²(–z + x – 2y) + 2xy(–z + x – 2y) + xz(–z + x – 2y) – 2yz(–z + x – 2y)
= –x²z + x³ – 2x²y – 4y²z + 4xy² – 8y³ – z³ + xz² – 2yz² – 2xyz + 2x²y – 4xy² – xz² + x²z – 2xyz + 2yz² – 2xyz + 4y²z
= (–x²z + x²z) + x³ + (–2yz² –2x²y) + (–4y²z + 4y²z) + (4xy² – 4xy²) – 3y³ – z³ + (xz² – xz²⁾ + (2yz² – 2yz²) + (–2xyz – 2xyz – 2xyz)
= x³ – 3y³ – z³ – 6xyz

Answer:

Given: a + b + c = 0
Now,
$\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3$
$$\begin{gathered} =\frac{a^2+b^2+c^2}{abc} \\ =\frac{(a+b+c) (a^2+b^2+c^2-ab-bc-ca)+3abc}{abc} \end{gathered}$$
$$\begin{gathered} =\frac{3abc}{abc} \left( \because a+b+c=0\right) \\ = 3 \end{gathered}$$

Hence, the required value is 3.

Answer:

Given: a + b + c = 5, ab + bc + ca = 10
Now,
(a + b + c)² = 5²
⇒ a² + b² + c² + 2(ab + bc + ca) = 25
⇒ a² + b² + c² + 2 × 10 = 25
⇒ a² + b² + c² = 5

We know,
 a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – (ab + bc + ca))

Answer:

[(a + b) + c]³ = (a + b)³ + 3(a + b)²c + 3(a + b)c² + c³
= (a³ + 3a²b + 3ab² + b³) + 3(a² + 2ab + b²)c +3(a + b)c² + c³
= a³ + 3a²b + 3ab² + b³ + 3a²c + 6abc + b²c + 3ac² + 3bc² + c³
= a³ + b³ + c³ + 3a²b + 3a²c + 3ab² + 3b²c + 3ac² + 3bc² + 6abc
= a³ + b³ + c³ + 3a(ab + ac + b² + bc) + 3c(ab + ac + b² + bc)
= a³ + b³ + c³ + 3(a + c) (ab + ac + b² + bc)
= a³ + b³ + c³ + 3(a + c) (a + b) (b + c)
(a + b + c)³ – a³ – b³ – c³ = 3(a + c) (b + c) (a + b)

Hence, proved.