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#### Question 1:

State whether the given algebraic expressions are polynomials? Justify.

(i)   (ii)   (iii)   (iv)   (v) 10

In an algebraic expression, if the powers of the variables are whole numbers then the algebraic expression is a polynomial.

(i)
$y+\frac{1}{y}=y+{y}^{-1}$

Here, one of the powers of y is −1, which is not a whole number. So,  is not a polynomial.

(ii)

Here, the power of x is $\frac{1}{2}$, which is not a whole number. So,  is not a polynomial.

(iii)

Here, the powers of the variable x are 2, 1 and 0, which are whole numbers. So,  is a polynomial.

(iv)

Here, one of the powers of m is −2, which is not a whole number. So,  is not a polynomial.

(v)
10 = 10 × 1 = 10x0

Here, the power of x is 0, which is a whole numbers. So, 10 is a polynomial (or constant polynomial).

#### Question 2:

Write the coefficient of m3 in each of the given polynomial.

(i) m (ii)  (iii)

(i)
Coefficient of m3 = 1

(ii)

Coefficient of m3 = $-\sqrt{3}$

(iii)

Coefficient of m3 = $-\frac{2}{3}$

#### Question 3:

Write the polynomial in x using the given information.
(i) Monomial with degree 7
(ii) Binomial with degree 35
(iii) Trinomial with degree 8

(i) A polynomial having only one term is called a monomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

3xis a monomial in x with degree 7.

(ii) A polynomial having only two terms is called a binomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

2x35 +  1 is a binomial in x with degree 35.

(iii) A polynomial having only three terms is called a trinomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

5x+ 6x4 + 7x is a trinomial in x with degree 8.

#### Question 4:

Write the degree of the given polynomials.

(i) $\sqrt{5}$    (ii)${x}^{°}$     (iii) ${x}^{2}$      (iv)          (v)       (vi)      (vii)      (viii)

The highest power of the variable in a polynomial of one variable is called the degreee of the polynomial. Also, the highest sum of the powers of the variables in each term of the polynomial in more than one variable is the degree of the polynomial.

(i)
$\sqrt{5}=\sqrt{5}×1=\sqrt{5}{x}^{0}$

The degree of the polynomial $\sqrt{5}$ is 0.

(ii)
The degree of the polynomial x0 is 0.

(iii)
The degree of the polynomial x2 is 2.

(iv)
The degree of the polynomial $\sqrt{2}{m}^{10}-7$ is 10.

(v)
The degree of the polynomial $2p-\sqrt{7}$ is 1.

(vi)
The degree of the polynomial $7y-{y}^{3}+{y}^{5}$ is 5.

(vii)
The sum of the powers of the variables in the polynomial $xyz+xy-z$ are 1 + 1 + 1 = 3 and 1 + 1 = 2.

The degree of the polynomial $xyz+xy-z$ is 3.

(viii)
The sum of the powers of the variables in the polynomial ${m}^{3}{n}^{7}-3{m}^{5}n+mn$ are 3 + 7 = 10, 5 + 1 = 6 and 1 + 1 = 2.

The degree of the polynomial ${m}^{3}{n}^{7}-3{m}^{5}n+mn$ is 10.

#### Question 5:

Classify the following polynomials as linear, quadratic and cubic polynomial.

(i) 2x2 + 3 x + 1 (ii) $5p$ (iii) (iv) (v) ${a}^{2}$ (vi) $3{r}^{3}$

(i)
The degree of the polynomial 2x2 + 3x + 1 is 2.

So, the polynomial 2x2 + 3x + 1 is a quadratic polynomial.

(ii)
The degree of the polynomial 5p is 1.

So, the polynomial 5p is a linear polynomial.

(iii)
The degree of the polynomial $\sqrt{2}y-\frac{1}{2}$ is 1.

So, the polynomial $\sqrt{2}y-\frac{1}{2}$ is a linear polynomial.

(iv)
The degree of the polynomial ${m}^{3}+7{m}^{2}+\frac{5}{2}m-\sqrt{7}$ is 3.

So, the polynomial ${m}^{3}+7{m}^{2}+\frac{5}{2}m-\sqrt{7}$ is a cubic polynomial.

(v)
The degree of the polynomial a2 is 2.

So, the polynomial a2 is a quadratic polynomial.

(vi)
The degree of the polynomial 3r3 is 3.

So, the polynomial 3r3 is a cubic polynomial.

#### Question 6:

Write the following polynomials in standard form.

(i)  (ii)

A polynomial written in either descending or ascending powers of its variable is called the standard form of the polynomial.

(i)
The given polynomial is m3 + 3 + 5m.

The standard form of the polynomial is 3 + 5mm3 or m3 + 5m + 3.

(ii)
The given polynomial is $-7y+{y}^{5}+3{y}^{3}-\frac{1}{2}+2{y}^{4}-{y}^{2}$.

The standard form of the polynomial is ${y}^{5}+2{y}^{4}+3{y}^{3}-{y}^{2}-7y-\frac{1}{2}$ or $-\frac{1}{2}-7y-{y}^{2}+3{y}^{3}+2{y}^{4}+{y}^{5}$.

#### Question 7:

Write the following polynomials in coefficient form.

(i)   (ii) $5y$   (iii)    (iv)

(i)
${x}^{3}-2={x}^{3}+0{x}^{2}+0x-2$

The coefficient form of the polynomial is (1, 0, 0, −2).

(ii)
5y = 5y + 0

The coefficient form of the polynomial is (5, 0).

(iii)
$2{m}^{4}-3{m}^{2}+7=2{m}^{4}+0{m}^{3}-3{m}^{2}+0m+7$

The coefficient form of the polynomial is (2, 0, −3, 0, 7).

(iv)
The coefficient form of the polynomial $-\frac{2}{3}$ is $\left(-\frac{2}{3}\right)$.

#### Question 8:

Write the polynomials in index form.

(i) (1, 2, 3) (ii) (5, 0, 0, 0, $-$1) (iii) ($-$2, 2,$-$​2, 2)

(i)
The coefficient form of the polynomial is (1, 2, 3).

Therefore, the index form of the polynomial is x+ 2x + 3.

(ii)
The coefficient form of the polynomial is (5, 0, 0, 0, −1).

Therefore, the index form of the polynomial is 5x+ 0x3 + 0x+ 0x − 1 or 5x4 − 1.

(iii)
The coefficient form of the polynomial is (−2, 2, −2, 2).

Therefore, the index form of the polynomial is −2x3 + 2x−2x + 2.

#### Question 9:

Write the appropriate polynomials in the boxes.

#### Question 1:

Use the given letters to write the answer.

(i) There are ‘a’ trees in the village Lat. If the number of trees increases every year by ‘b’, then how many trees will there be after ‘x’ years?

(ii) For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?

(iii) The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.

(i)
Initial number of trees in the village = a

Increase in the number of trees every year = b

∴ Number of trees in the village after x years

= Initial number of trees in the village + Increase in the number of trees every year × x

= abx

Thus, the number of trees after x years is abx.

(ii)
Number of students in each row = y

Number of rows = x

∴ Total number of students in the parade = Number of students in each row × Number of rows = y × xyx xy

Thus, there are in all xy students for the parade.

(iii)
Digit at the tens place = m

Digit at the units place = n

∴ Two digit number = Digit at the tens place × 10 + Digit at the units place = m × 10 + = 10mn

Thus, the polynomial which represents the two digit number is 10mn.

#### Question 2:

Add the given polynomials.

(i)   (ii)   (iii)

(i)
$\left({x}^{3}-2{x}^{2}-9\right)+\left(5{x}^{3}+2x+9\right)\phantom{\rule{0ex}{0ex}}={x}^{3}+5{x}^{3}-2{x}^{2}+2x-9+9\phantom{\rule{0ex}{0ex}}=6{x}^{3}-2{x}^{2}+2x$
(ii)
$\left(-7{m}^{4}+5{m}^{3}+\sqrt{2}\right)+\left(5{m}^{4}-3{m}^{3}+2{m}^{2}+3m-6\right)\phantom{\rule{0ex}{0ex}}=-7{m}^{4}+5{m}^{4}+5{m}^{3}-3{m}^{3}+2{m}^{2}+3m+\sqrt{2}-6\phantom{\rule{0ex}{0ex}}=-2{m}^{4}+2{m}^{3}+2{m}^{2}+3m-6+\sqrt{2}$
(iii)
$\left(2{y}^{2}+7y+5\right)+\left(3y+9\right)+\left(3{y}^{2}-4y-3\right)\phantom{\rule{0ex}{0ex}}=2{y}^{2}+3{y}^{2}+7y+3y-4y+5+9-3\phantom{\rule{0ex}{0ex}}=5{y}^{2}+6y+11$

#### Question 3:

Subtract the second polynomial from the first.

(i)          (ii)

(i)

(ii)
$\left(2a{b}^{2}+3{a}^{2}b-4ab\right)-\left(3ab-8a{b}^{2}+2{a}^{2}b\right)\phantom{\rule{0ex}{0ex}}=2a{b}^{2}+3{a}^{2}b-4ab-3ab+8a{b}^{2}-2{a}^{2}b\phantom{\rule{0ex}{0ex}}=2a{b}^{2}+8a{b}^{2}+3{a}^{2}b-2{a}^{2}b-4ab-3ab\phantom{\rule{0ex}{0ex}}=10a{b}^{2}+{a}^{2}b-7ab$

#### Question 4:

Multiply the given polynomials.

(i)       (ii)      (iii)

(i)

(ii)
$\left({x}^{5}-1\right)×\left({x}^{3}+2{x}^{2}+2\right)\phantom{\rule{0ex}{0ex}}={x}^{5}\left({x}^{3}+2{x}^{2}+2\right)-1\left({x}^{3}+2{x}^{2}+2\right)\phantom{\rule{0ex}{0ex}}={x}^{8}+2{x}^{7}+2{x}^{5}-{x}^{3}-2{x}^{2}-2$
(iii)
$\left(2y+1\right)×\left({y}^{2}-2{y}^{3}+3y\right)\phantom{\rule{0ex}{0ex}}=2y\left({y}^{2}-2{y}^{3}+3y\right)+1\left({y}^{2}-2{y}^{3}+3y\right)\phantom{\rule{0ex}{0ex}}=2{y}^{3}-4{y}^{4}+6{y}^{2}+{y}^{2}-2{y}^{3}+3y\phantom{\rule{0ex}{0ex}}=-4{y}^{4}+2{y}^{3}-2{y}^{3}+6{y}^{2}+{y}^{2}+3y\phantom{\rule{0ex}{0ex}}=-4{y}^{4}+7{y}^{2}+3y$

#### Question 5:

Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor $×$ Quotient + Remainder’.

(i)    (ii)

(i)
${x}^{3}-64={x}^{3}+0{x}^{2}+0x-64$

Using long division method,

Dividend = Divisor ×​ Quotient + Remainder

$\therefore {x}^{3}-64=\left(x-4\right)×\left({x}^{2}+4x+16\right)+0$

(ii)

Using long division method,

Dividend = Divisor ×​ Quotient + Remainder

$\therefore 5{x}^{5}+4{x}^{4}-3{x}^{3}+2{x}^{2}+2=\left({x}^{2}-x\right)×\left(5{x}^{3}+9{x}^{2}+6x+8\right)+\left(8x+2\right)$

#### Question 6:

Write down the information in the form of algebraic expression and simplify.

There is a rectangular farm with length  metre and breadth ( ) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was ( ) metre. What is the area of the remaining part of the farm ?

Lenght of the rectangular farm = (2a2 + 3b2) m

Breadth of the rectangular farm = (a2 + b2) m

Side of the square plot = (a2 − b2) m

∴ Area of the remaining part of the farm

= Total area of the farm − Area of the square plot

= Lenght of the rectangular farm × Breadth of the rectangular farm − (Side of the square plot)2

= (2a2 + 3b2) × (a2 + b2) − (a2 − b2)2

= 2a2(a2 + b2) + 3b2(a2 + b2) − (a4 + b− 2a2b2)

= 2a+ 2a2b2 + 3a2b2 + 3b4 − a4 − b+ 2a2b2

= 2a4 − a4 + 2a2b2  + 3a2b+ 2a2b+ 3b− b4

= (a4 + 7a2b+ 2b4) m2

Thus, the area of the remaining part of the farm is (a4 + 7a2b+ 2b4) m2.

#### Question 1:

Divide each of the following polynomials by synthetic division method and also by
linear division method. Write the quotient and the remainder.

(i)                       (ii)                         (iii)

(iv)          (v)           (vi)

(i)
Synthetic Division:

Dividend = $2{m}^{2}-3m+10$

Divisor = $m-5$

Opposite of −5 = 5

The coefficient form of the quotient is (2, 7).

∴ Quotient = 2m + 7 and Remainder = 45

Linear Method:

$2{m}^{2}-3m+10\phantom{\rule{0ex}{0ex}}=2m\left(m-5\right)+10m-3m+10\phantom{\rule{0ex}{0ex}}=2m\left(m-5\right)+7\left(m-5\right)+35+10\phantom{\rule{0ex}{0ex}}=\left(m-5\right)×\left(2m+7\right)+45$
(ii)
Synthetic Division:

Dividend = ${x}^{4}+2{x}^{3}+3{x}^{2}+4x+5$

Divisor = $x+2$

Opposite of 2 = −2

The coefficient form of the quotient is (1, 0, 3, −2).

∴ Quotient = x3 + 3x − 2 and Remainder = 9

Linear Method:

${x}^{4}+2{x}^{3}+3{x}^{2}+4x+5\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x+2\right)+3x\left(x+2\right)-6x+4x+5\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x+2\right)+3x\left(x+2\right)-2x+5\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x+2\right)+3x\left(x+2\right)-2\left(x+2\right)+4+5\phantom{\rule{0ex}{0ex}}=\left(x+2\right)×\left({x}^{3}+3x-2\right)+9$
(iii)
Synthetic Division:

Dividend = ${y}^{3}-216={y}^{3}+0{y}^{2}+0y-216$

Divisor = $y-6$

Opposite of −6 = 6

The coefficient form of the quotient is (1, 6, 36).

∴ Quotient = y+ 6y + 36 and Remainder = 0

Linear Method:

${y}^{3}-216\phantom{\rule{0ex}{0ex}}={y}^{2}\left(y-6\right)+6{y}^{2}-216\phantom{\rule{0ex}{0ex}}={y}^{2}\left(y-6\right)+6y\left(y-6\right)+36y-216\phantom{\rule{0ex}{0ex}}={y}^{2}\left(y-6\right)+6y\left(y-6\right)+36\left(y-6\right)+216-216\phantom{\rule{0ex}{0ex}}={y}^{2}\left(y-6\right)+6y\left(y-6\right)+36\left(y-6\right)\phantom{\rule{0ex}{0ex}}=\left(y-6\right)×\left({y}^{2}+6y+36\right)$
(iv)
Synthetic Division:

Dividend = $2{x}^{4}+3{x}^{3}+4x-2{x}^{2}=2{x}^{4}+3{x}^{3}-2{x}^{2}+4x+0$

Divisor = $x+3$

Opposite of 3 = −3

The coefficient form of the quotient is (2, −3, 7, −17).

∴ Quotient = 2x− 3x2 + 7x − 17 and Remainder = 51

Linear Method:

$2{x}^{4}+3{x}^{3}-2{x}^{2}+4x\phantom{\rule{0ex}{0ex}}=2{x}^{3}\left(x+3\right)-6{x}^{3}+3{x}^{3}-2{x}^{2}+4x\phantom{\rule{0ex}{0ex}}=2{x}^{3}\left(x+3\right)-3{x}^{2}\left(x+3\right)+9{x}^{2}-2{x}^{2}+4x\phantom{\rule{0ex}{0ex}}=2{x}^{3}\left(x+3\right)-3{x}^{2}\left(x+3\right)+7x\left(x+3\right)-21x+4x\phantom{\rule{0ex}{0ex}}=2{x}^{3}\left(x+3\right)-3{x}^{2}\left(x+3\right)+7x\left(x+3\right)-17\left(x+3\right)+51\phantom{\rule{0ex}{0ex}}=\left(x+3\right)×\left(2{x}^{3}-3{x}^{2}+7x-17\right)+51$
(v)
Synthetic Division:

Dividend = ${x}^{4}-3{x}^{2}-8={x}^{4}+0{x}^{3}-3{x}^{2}+0x-8$

Divisor = $x+4$

Opposite of 4 = −4

The coefficient form of the quotient is (1, −4, 13, −52).

∴ Quotient = x− 4x2 + 13x − 52 and Remainder = 200

Linear Method:

${x}^{4}-3{x}^{2}-8\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x+4\right)-4{x}^{3}-3{x}^{2}-8\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x+4\right)-4{x}^{2}\left(x+4\right)+16{x}^{2}-3{x}^{2}-8\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x+4\right)-4{x}^{2}\left(x+4\right)+13x\left(x+4\right)-52x-8\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x+4\right)-4{x}^{2}\left(x+4\right)+13x\left(x+4\right)-52\left(x+4\right)+208-8\phantom{\rule{0ex}{0ex}}=\left(x+4\right)×\left({x}^{3}-4{x}^{2}+13x-52\right)+200$
(vi)
Synthetic Division:

Dividend = ${y}^{3}-3{y}^{2}+5y-1$

Divisor = $y-1$

Opposite of −1 = 1

The coefficient form of the quotient is (1, −2, 3).

∴ Quotient = y− 2y + 3 and Remainder = 2

Linear Method:

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

#### Question 1:

For $x$ = 0 find the value of the polynomial  .

Let $p\left(x\right)={x}^{2}-5x+5$.

$\therefore p\left(0\right)={\left(0\right)}^{2}-5×0+5=0-0+5=5$

Hence, for x = 0 the value of the polynomial is 5.

#### Question 2:

If  then find  .

$p\left(y\right)={y}^{2}-3\sqrt{2}y+1$

$\therefore p\left(3\sqrt{2}\right)={\left(3\sqrt{2}\right)}^{2}-3\sqrt{2}×3\sqrt{2}+1=18-18+1=1$

#### Question 3:

If  then

$p\left(m\right)={m}^{3}+2{m}^{2}-m+10$

$\therefore p\left(a\right)={a}^{3}+2{a}^{2}-a+10$     .....(1)

Also,

$p\left(-a\right)={\left(-a\right)}^{3}+2{\left(-a\right)}^{2}-\left(-a\right)+10$

$⇒p\left(-a\right)=-{a}^{3}+2{a}^{2}+a+10$        .....(2)

Adding (1) and (2), we get

$p\left(a\right)+p\left(-a\right)\phantom{\rule{0ex}{0ex}}=\left({a}^{3}+2{a}^{2}-a+10\right)+\left(-{a}^{3}+2{a}^{2}+a+10\right)\phantom{\rule{0ex}{0ex}}={a}^{3}-{a}^{3}+2{a}^{2}+2{a}^{2}-a+a+10+10\phantom{\rule{0ex}{0ex}}=4{a}^{2}+20$
$\therefore p\left(a\right)+p\left(-a\right)=4{a}^{2}+20$

#### Question 4:

If  then find $p\left(2\right).$

$p\left(y\right)=2{y}^{3}-6{y}^{2}-5y+7$

$\therefore p\left(2\right)=2×{\left(2\right)}^{3}-6×{\left(2\right)}^{2}-5×2+7=16-24-10+7=-11$

#### Question 1:

Find the value of the polynomial     using given values  for $x$.

(i)     (ii)   (iii) x = 0

Let $p\left(x\right)=2x-2{x}^{3}+7$.

(i)
$p\left(3\right)=2×3-2×{\left(3\right)}^{3}+7\phantom{\rule{0ex}{0ex}}=6-2×27+7\phantom{\rule{0ex}{0ex}}=6-54+7\phantom{\rule{0ex}{0ex}}=-41$
Thus, the value of polynomial for x = 3 is −41.

(ii)
$p\left(-1\right)=2×\left(-1\right)-2×{\left(-1\right)}^{3}+7\phantom{\rule{0ex}{0ex}}=-2-2×\left(-1\right)+7\phantom{\rule{0ex}{0ex}}=-2+2+7\phantom{\rule{0ex}{0ex}}=7$
Thus, the value of polynomial for x = −1 is 7.

(iii)
$p\left(0\right)=2×0-2×{\left(0\right)}^{3}+7\phantom{\rule{0ex}{0ex}}=0-0+7\phantom{\rule{0ex}{0ex}}=7$
Thus, the value of polynomial for x = 0 is 7.

#### Question 2:

For each of the following polynomial, find  and $p\left(-2\right)$.

(i)     (ii)  (iii)

(i)
$p\left(x\right)={x}^{3}\phantom{\rule{0ex}{0ex}}\therefore p\left(1\right)={\left(1\right)}^{3}=1\phantom{\rule{0ex}{0ex}}p\left(0\right)={\left(0\right)}^{3}=0\phantom{\rule{0ex}{0ex}}p\left(-2\right)={\left(-2\right)}^{3}=-8$
(ii)
$p\left(y\right)={y}^{2}-2y+5\phantom{\rule{0ex}{0ex}}\therefore p\left(1\right)={\left(1\right)}^{2}-2×1+5=1-2+5=4\phantom{\rule{0ex}{0ex}}p\left(0\right)={\left(0\right)}^{2}-2×0+5=0-0+5=5\phantom{\rule{0ex}{0ex}}p\left(-2\right)={\left(-2\right)}^{2}-2×\left(-2\right)+5=4+4+5=13$
(iii)
$p\left(x\right)={x}^{4}-2{x}^{2}-x\phantom{\rule{0ex}{0ex}}\therefore p\left(1\right)={\left(1\right)}^{4}-2×{\left(1\right)}^{2}-1=1-2-1=-2\phantom{\rule{0ex}{0ex}}p\left(0\right)={\left(0\right)}^{4}-2×{\left(0\right)}^{2}-0=0-0-0=0\phantom{\rule{0ex}{0ex}}p\left(-2\right)={\left(-2\right)}^{4}-2×{\left(-2\right)}^{2}-\left(-2\right)=16-2×4+2=16-8+2=10$

#### Question 3:

If the value of the polynomial   is 12  for  , then  find the value of a.

Let $p\left(m\right)={m}^{3}+2m+a$

For m = 2, p(2) = 12.

$\therefore {\left(2\right)}^{3}+2×2+a=12\phantom{\rule{0ex}{0ex}}⇒8+4+a=12\phantom{\rule{0ex}{0ex}}⇒12+a=12\phantom{\rule{0ex}{0ex}}⇒a=12-12=0$
Thus, the value of a is 0.

#### Question 4:

For the polynomial   if   then find m.

Let $p\left(x\right)=m{x}^{2}-2x+3$.

$\therefore p\left(-1\right)=7\phantom{\rule{0ex}{0ex}}⇒m×{\left(-1\right)}^{2}-2×\left(-1\right)+3=7\phantom{\rule{0ex}{0ex}}⇒m+2+3=7\phantom{\rule{0ex}{0ex}}⇒m=7-5=2$
Thus, the value of m is 2.

#### Question 5:

Divide the first polynomial by the second polynomial and find the remainder using factor theorem .

(i)     (ii)    (iii)

(i)
By synthetic division:

Dividend = ${x}^{2}-7x+9$

Divisor = x + 1

Opposite of 1 = −1

The coefficient form of the quotient is (1, −8).

∴ Quotient = x − 8

Remainder = 17

By remainder theorem:

Let $p\left(x\right)={x}^{2}-7x+9$.

Divisor = x + 1

By remainder theorem,

Remainder = p(−1) = ${\left(-1\right)}^{2}-7×\left(-1\right)+9=1+7+9=19$

(ii)
By synthetic division:

Dividend = $2{x}^{3}-2{x}^{2}+ax-a$

Divisor = x − a

Opposite of −aa

The coefficient form of the quotient is (2, 2a − 2, 2a2​ − a).

∴ Quotient = 2x2 + (2a − 2)x + 2a2​ − a

Remainder = 2a3​ − a− a

By remainder theorem:

Let $p\left(x\right)=2{x}^{3}-2{x}^{2}+ax-a$.

Divisor = x − a

By remainder theorem,

Remainder = p(a) = $2×{a}^{3}-2×{a}^{2}+a×a-a=2{a}^{3}-2{a}^{2}+{a}^{2}-a=2{a}^{3}-{a}^{2}-a$

(iii)
By synthetic division:

Dividend = $54{m}^{3}+18{m}^{2}-27m+5$

Divisor = m − 3

Opposite of −3 = 3

The coefficient form of the quotient is (54, 180, 513).

∴ Quotient = 54x2 + 180x + 513

Remainder = 1544

By remainder theorem:

Let $p\left(m\right)=54{m}^{3}+18{m}^{2}-27m+5$.

Divisor = m − 3

By remainder theorem,

Remainder = p(3) = $54×{\left(3\right)}^{3}+18×{\left(3\right)}^{2}-27×3+5=54×27+18×9-27×3+5=1458+162-81+5=1544$

#### Question 6:

If the polynomial  is divided  by  y + 2  and the remainder is 50 then find the value of m.

Let $p\left(y\right)={y}^{3}-5{y}^{2}+7y+m$.

When the polynomial is divided by (y + 2), the remainder is 50. This means that the value of the polynomial when y = −2 is 50.

By remainder theorem,

Remainder = p(−2) = 50

$\therefore {\left(-2\right)}^{3}-5×{\left(-2\right)}^{2}+7×\left(-2\right)+m=50\phantom{\rule{0ex}{0ex}}⇒-8-5×4-14+m=50\phantom{\rule{0ex}{0ex}}⇒-8-20-14+m=50\phantom{\rule{0ex}{0ex}}⇒-42+m=50\phantom{\rule{0ex}{0ex}}⇒m=50+42=92$
Thus, the value of m is 92.

#### Question 7:

Use factor theorem to determine whether x + 3 is factor of x 2 + 2x − 3 or not.

Let p(x) = x2 + 2x − 3.

Divisor = x + 3

$\therefore p\left(-3\right)={\left(-3\right)}^{2}+2×\left(-3\right)-3=9-6-3=0$

So, by factor theorem, (x + 3) is a factor of x2 + 2x − 3.

#### Question 8:

If ( x  ) is a factor of  then find the value of m.

Let $p\left(x\right)={x}^{3}-m{x}^{2}+10x-20$.

It is given that (x − 2) is a factor of $p\left(x\right)={x}^{3}-m{x}^{2}+10x-20$.

$\therefore p\left(2\right)=0\phantom{\rule{0ex}{0ex}}⇒{\left(2\right)}^{3}-m×{\left(2\right)}^{2}+10×\left(2\right)-20=0\phantom{\rule{0ex}{0ex}}⇒8-4m+20-20=0\phantom{\rule{0ex}{0ex}}⇒8-4m=0\phantom{\rule{0ex}{0ex}}⇒4m=8\phantom{\rule{0ex}{0ex}}⇒m=2$
Thus, the value of m is 2.

#### Question 9:

By using factor theorem in the following examples, determine whether q ( x ) is a factor p ( x ) or not.

(i)

(ii)

(i)
$p\left(x\right)={x}^{3}-{x}^{2}-x-1$

Divisor = $q\left(x\right)=x-1$

$\therefore p\left(1\right)={\left(1\right)}^{3}-{\left(1\right)}^{2}-1-1=1-1-1-1=-2$

Since p(1) ≠ 0, so by factor theorem $q\left(x\right)=x-1$ is not a factor of polynomial $p\left(x\right)={x}^{3}-{x}^{2}-x-1$.

(ii)
$p\left(x\right)=2{x}^{3}-{x}^{2}-45$

Divisor = $q\left(x\right)=x-3$

$\therefore p\left(3\right)=2×{\left(3\right)}^{3}-{\left(3\right)}^{2}-45=2×27-9-45=54-54=0$

So, by factor theorem $q\left(x\right)=x-3$ is a factor of polynomial $p\left(x\right)=2{x}^{3}-{x}^{2}-45$.

#### Question 10:

If ( x31 + 31) is divided by (x + 1) then find the remainder.

Let p(x) = x31 + 31.

Divisor = x + 1

By remainder theorem, we have

Remainder = p(−1) = (−1)31 + 31 = −1 + 31 = 30

Thus, the remainder when (x31 + 31) is divided by (x + 1) is 30.

#### Question 11:

Show that m $-$1 is a factor of m21 $-$ 1 and m22 $-$ 1.

Let p(m) = m21 − 1 and q(m) = m22 − 1.

Divisor = m − 1

Now,

p(1) = (1)21 − 1 = 1 − 1 = 0

Therefore, by factor theorem (m − 1) is a factor of p(m) = m21 − 1.

Also,

q(1) = (1)22 − 1 = 1 − 1 = 0

Therefore, by factor theorem (m − 1) is a factor of q(m) = m22 − 1.

Hence, (m − 1) is a factor of m21 − 1 and m22 − 1.

#### Question 12:

If  and  both are the factors of the polynomial nx2 − 5x + m, then show that

Let p(x) = nx2 − 5xm.

It given that (x − 2) and $\left(x-\frac{1}{2}\right)$ are the factors of the polynomial p(x) = nx2 − 5x + m.

∴ By factor theorem, p(2) = 0 and $p\left(\frac{1}{2}\right)=0$.

Also,

From (1) and (2), we have

4nmn + 4m

⇒ 4n − n = 4m − m

⇒ 3n = 3m

⇒ n = m

Putting nm in (1), we have

4mm = 10

⇒ 5m = 10

⇒ m = 2

∴ nm = 2

#### Question 13:

(i) If  then  .

(ii)  then $p\left(5\sqrt{3}\right)$.

(i)
$p\left(x\right)=2+5x\phantom{\rule{0ex}{0ex}}\therefore p\left(2\right)=2+5×2=2+10=12\phantom{\rule{0ex}{0ex}}p\left(-2\right)=2+5×\left(-2\right)=2-10=-8\phantom{\rule{0ex}{0ex}}p\left(1\right)=2+5×1=2+5=7$
$\therefore p\left(2\right)+p\left(-2\right)-p\left(1\right)=12+\left(-8\right)-7=12-8-7=12-15=-3$

(ii)
$p\left(x\right)=2{x}^{2}-5\sqrt{3}x+5$
$\therefore p\left(5\sqrt{3}\right)=2×{\left(5\sqrt{3}\right)}^{2}-5\sqrt{3}×5\sqrt{3}+5\phantom{\rule{0ex}{0ex}}=2×75-25×3+5\phantom{\rule{0ex}{0ex}}=150-75+5\phantom{\rule{0ex}{0ex}}=80$

#### Question 1:

Find the factors of the polynomials given below.
(i) 2x2 + x – 1

(ii) 2m2 + 5m – 3

(iii) 12x2 + 61x + 77

(iv) 3y2 – 2y – 1

(v) $\sqrt{3}{x}^{2}+4x+\sqrt{3}$

(vi) $\frac{1}{2}{x}^{2}-3x+4$

(i)
$2{x}^{2}+x-1\phantom{\rule{0ex}{0ex}}=2{x}^{2}+2x-x-1\phantom{\rule{0ex}{0ex}}=2x\left(x+1\right)-1\left(x+1\right)\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left(2x-1\right)$
(ii)
$2{m}^{2}+5m-3\phantom{\rule{0ex}{0ex}}=2{m}^{2}+6m-m-3\phantom{\rule{0ex}{0ex}}=2m\left(m+3\right)-1\left(m+3\right)\phantom{\rule{0ex}{0ex}}=\left(m+3\right)\left(2m-1\right)$
(iii)
$12{x}^{2}+61x+77\phantom{\rule{0ex}{0ex}}=12{x}^{2}+28x+33x+77\phantom{\rule{0ex}{0ex}}=4x\left(3x+7\right)+11\left(3x+7\right)\phantom{\rule{0ex}{0ex}}=\left(3x+7\right)\left(4x+11\right)$
(iv)
$3{y}^{2}-2y-1\phantom{\rule{0ex}{0ex}}=3{y}^{2}-3y+y-1\phantom{\rule{0ex}{0ex}}=3y\left(y-1\right)+1\left(y-1\right)\phantom{\rule{0ex}{0ex}}=\left(y-1\right)\left(3y+1\right)$
(v)
$\sqrt{3}{x}^{2}+4x+\sqrt{3}\phantom{\rule{0ex}{0ex}}=\sqrt{3}{x}^{2}+3x+x+\sqrt{3}\phantom{\rule{0ex}{0ex}}=\sqrt{3}x\left(x+\sqrt{3}\right)+1\left(x+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x+\sqrt{3}\right)\left(\sqrt{3}x+1\right)$
(vi)
$\frac{1}{2}{x}^{2}-3x+4\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{x}^{2}-2x-x+4\phantom{\rule{0ex}{0ex}}=\frac{1}{2}x\left(x-4\right)-1\left(x-4\right)\phantom{\rule{0ex}{0ex}}=\left(x-4\right)\left(\frac{1}{2}x-1\right)$

#### Question 2:

Factorize the following polynomials.
(i) (x2x)2 – 8 (x2x) + 12
(ii) (x – 5)2 – (5x – 25) – 24
(iii) (x2 – 6x)2 – 8 (x2 – 6x + 8) – 64
(iv) (x2 – 2x + 3) (x2 – 2x + 5) – 35
(v) (y + 2) (y – 3) (y + 8) (y + 3) + 56
(vi) (y2 + 5y) (y2 + 5y – 2) – 24
(vii) (x – 3) (x – 4)2 (x – 5) – 6

(i)
(x2 – x)2 – 8(x2 – x) + 12
Let x2 – xz.
$\therefore {\left({x}^{2}-x\right)}^{2}-8\left({x}^{2}-x\right)+12\phantom{\rule{0ex}{0ex}}={z}^{2}-8z+12\phantom{\rule{0ex}{0ex}}={z}^{2}-6z-2z+12\phantom{\rule{0ex}{0ex}}=z\left(z-6\right)-2\left(z-6\right)\phantom{\rule{0ex}{0ex}}=\left(z-6\right)\left(z-2\right)$

(ii)
(x – 5)2 – (5x – 25) – 24
= (x – 5)2 – 5(x – 5) – 24
Let x – 5 = z.
$\therefore {\left(x-5\right)}^{2}-5\left(x-5\right)-24\phantom{\rule{0ex}{0ex}}={z}^{2}-5z-24\phantom{\rule{0ex}{0ex}}={z}^{2}-8z+3z-24\phantom{\rule{0ex}{0ex}}=z\left(z-8\right)+3\left(z-8\right)\phantom{\rule{0ex}{0ex}}=\left(z-8\right)\left(z+3\right)$

(iii)
(x2 – 6x)2 – 8(x2 – 6x + 8) – 64
= (x2 – 6x)2 – 8(x2 – 6x) – 64 – 64
= (x2 – 6x)2 – 8(x2 – 6x) – 128
Let x2 – 6x z.
$\therefore {\left({x}^{2}-6x\right)}^{2}-8\left({x}^{2}-6x\right)-128\phantom{\rule{0ex}{0ex}}={z}^{2}-8z-128\phantom{\rule{0ex}{0ex}}={z}^{2}-16z+8z-128\phantom{\rule{0ex}{0ex}}=z\left(z-16\right)+8\left(z-16\right)\phantom{\rule{0ex}{0ex}}=\left(z-16\right)\left(z+8\right)$

(iv)
(x2 – 2x + 3)(x2 – 2x + 5) – 35
Let x2 – 2xz.
$\therefore \left({x}^{2}-2x+3\right)\left({x}^{2}-2x+5\right)-35\phantom{\rule{0ex}{0ex}}=\left(z+3\right)\left(z+5\right)-35\phantom{\rule{0ex}{0ex}}={z}^{2}+5z+3z+15-35\phantom{\rule{0ex}{0ex}}={z}^{2}+8z-20\phantom{\rule{0ex}{0ex}}={z}^{2}+10z-2z-20$

(v)
(y + 2)(y – 3)(y + 8)(y + 3) + 56
= (y + 2)(y + 3)(y + 8)(y – 3) + 56
= (y2 + 5y + 6)(y2 + 5– 24) + 56
Let y2 + 5yz.
$\therefore \left({y}^{2}+5y+6\right)\left({y}^{2}+5y-24\right)+56\phantom{\rule{0ex}{0ex}}=\left(z+6\right)\left(z-24\right)+56\phantom{\rule{0ex}{0ex}}={z}^{2}-18z-144+56\phantom{\rule{0ex}{0ex}}={z}^{2}-18z-88$

$=\left({y}^{2}+5y-22\right)\left({y}^{2}+4y+y+4\right)\phantom{\rule{0ex}{0ex}}=\left({y}^{2}+5y-22\right)\left[y\left(y+4\right)+1\left(y+4\right)\right]\phantom{\rule{0ex}{0ex}}=\left({y}^{2}+5y-22\right)\left(y+4\right)\left(y+1\right)$
(vi)
(y2 + 5y)(y2 + 5y – 2) – 24
Let y2 + 5yz.
$\therefore \left({y}^{2}+5y\right)\left({y}^{2}+5y-2\right)-24\phantom{\rule{0ex}{0ex}}=z\left(z-2\right)-24\phantom{\rule{0ex}{0ex}}={z}^{2}-2z-24\phantom{\rule{0ex}{0ex}}={z}^{2}-6z+4z-24\phantom{\rule{0ex}{0ex}}=z\left(z-6\right)+4\left(z-6\right)\phantom{\rule{0ex}{0ex}}=\left(z-6\right)\left(z+4\right)$

(vii)
(x – 3)(x – 4)2(x – 5) – 6
= (x – 3)(x – 5)(x – 4)2 – 6
= (x2 – 8x + 15)(x2 – 8x + 16) – 6
Let x2 – 8xz.
$\therefore \left({x}^{2}-8x+15\right)\left({x}^{2}-8x+16\right)-6\phantom{\rule{0ex}{0ex}}=\left(z+15\right)\left(z+16\right)-6\phantom{\rule{0ex}{0ex}}={z}^{2}+31z+240-6\phantom{\rule{0ex}{0ex}}={z}^{2}+31z+234$

#### Question 1:

Write the correct alternative answer for each of the following questions.

(i) Which of the following is a polynomial ?
(A)  $\frac{x}{y}$  (B)    (C)    (D)

(ii) What is the degree of the polynomial $\sqrt{7}$?
(A) $\frac{1}{2}$     (B)   5    (C)  2   (D)  0

(iii) What is the degree of the 0 polynomial ?
(A) 0 (B)   1   (C)   undefined   (D) any real number

(iv) What is the degree of the polynomial 2x2 + 5x3 + 7?
(A)   3    (B)   2   (C)     5   (D) 7

(v) What is the coefficient form of  ?
(A) (1, $-$ 1)    (B) (3,$-$1)    (C) (1, 0, 0, $-$1)      (D) (1, 3, $-$ 1)

(vi)    then
(A) 3      (B)   $7\sqrt{7}$   (C)      (D)  $49\sqrt{7}$

(vii) When    , what is the value of the polynomial 2x3 + 2x ?
(A) 4    (B) 2  (C) $-2$     (D)$-4$

(viii) If , what is a factor of the polynomial  then find the value of m.
(A) 2   (B) $-$2  (C) $-3$     (D)  3

(ix) Multiply  ( x2$-$3) (2x $-$ 7x 3 + 4) and write the degree of the product.
(A) 5  (B) 3  (C) 2   (D)  0

(x) Which of the following is a linear polynomial ?
(A) x + 5
(B) x2 + 5
(C) x3 + 5
(D) x4 + 5

(i)
In an algebraic expression, if the powers of the variables are whole numbers then the algebraic expression is a polynomial.

In the expression , the power of the variable x is 2, which is a whole number. So, the expression  is a polynomial.

Hence, the correct answer is option (D).

(ii)
$\sqrt{7}=\sqrt{7}×1=\sqrt{7}{x}^{0}$

The degree of the polynomial $\sqrt{7}$ is 0.

Hence, the correct answer is option (D).

(iii)
The degree of 0 polynomial is not defined.

Hence, the correct answer is option (C).

(iv)
The highest power of the variable in a polynomial is called the degree of the polynomial.

The degree of the polynomial 2x+ 5x+ 7 is 3.

Hence, the correct answer is option (A).

(v)

The coefficient form of the polynomial is (1, 0, 0, −1).

Hence, the correct answer is option (C).

(vi)

Hence, the correct answer is option (A).

(vii)
Let p(x) = 2x+ 2x

∴ p(−1) = 2 × (−1)3 + 2 × (−1) = 2 × (−1) − 2 = −2 − 2 = −4

Thus, the value of the polynomial when x = −1 is −4.

Hence, the correct answer is option (D).

(viii)
Let p(x) = 3x+ mx.

(x − 1) is a factor of p(x).

∴ p(1) = 0

⇒ 3 × (1)2 + m × 1 = 0

⇒ 3 + m = 0

⇒ m = −3

Hence, the correct answer is option (C).

(ix)
$\left({x}^{2}-3\right)\left(2x-7{x}^{3}+4\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(2x-7{x}^{3}+4\right)-3\left(2x-7{x}^{3}+4\right)\phantom{\rule{0ex}{0ex}}=2{x}^{3}-7{x}^{5}+4{x}^{2}-6x+21{x}^{3}-12\phantom{\rule{0ex}{0ex}}=-7{x}^{5}+2{x}^{3}+21{x}^{3}+4{x}^{2}-6x-12\phantom{\rule{0ex}{0ex}}=-7{x}^{5}+23{x}^{3}+4{x}^{2}-6x-12$
Thus, the degree of the resultant polynomial is 5.

Hence, the correct answer is option (A).

(x)
A polynomial with degree one is called a linear polynomial.

Thus, the polynomial x + 5 is a linear polynomial.

Hence, the correct answer is option (A).

#### Question 2:

Write the degree of the polynomial for each of the following.

(i) 5 + 3x4 (ii) 7 (iii) ax7 + bx9 ( a, b are constants.)

The highest power of the variable in a polynomial in one variable is called the degree of the polynomial.

(i)
The degree of the polynomial 5 + 3xis 4.

(ii)
7 = 7 × 1 = 7x0

The degree of the constant polynomial 7 is 0.

(iii)
The degree of the polynomial axbx9 is 9.

#### Question 3:

Write the following polynomials in standard form.

(i) 4x2 + 7x4 $-$x3$-$x + 9
(ii) p + 2 p3 + 10 p2 + 5 p4 $-$8

A polynomial written in either descending or ascending power of its variable is called the standard form of the polynomial.

(i)
The given polynomial is 4x2 + 7x− x− x + 9.
The standard form of the polynomial is 7x4
− x3 + 4x2 − x + 9 or 9 − + 4x2 − x37x4.

(ii)
The given polynomial is + 2p3 + 10p2 + 5p4 − 8.

The standard form of the polynomial is 5p4 + 2p3 + 10p2p − 8 or −8 + + 10p2 + 2p3 + 5p4.

#### Question 4:

Write the following polynomial in coefficient form.
(i) x4 + 16 (ii) m5 + 2m2 + 3m + 15

(i)
${x}^{4}+16={x}^{4}+0{x}^{3}+0{x}^{2}+0x+16$

Therefore, the given polynomial in coefficient form is (1, 0, 0, 0, 16).

(ii)
${m}^{5}+2{m}^{2}+3m+15={m}^{5}+0{m}^{4}+0{m}^{3}+2{m}^{2}+3m+15$

Therefore, the given polynomial in coefficient form is (1, 0, 0, 2, 3, 15).

#### Question 5:

Write the index form of the polynomial using variable x from its coefficient form.

(i) (3, $-$2, 0, 7, 18)
(ii) (6, 1, 0, 7)
(iii) (4, 5, $-$3, 0)

(i)
The coefficient form of the polynomial is (3, −2, 0, 7, 18).

Therefore, the index form the polynomial is 3x4 − 2x+ 0x2 + 7x + 18 or 3x4 − 2x+ 7x + 18.

(ii)
The coefficient form of the polynomial is (6, 1, 0, 7).

Therefore, the index form the polynomial is 6xx2 + 0x + 7 or 6xx2 + 7.

(iii)
The coefficient form of the polynomial is (4, 5, −3, 0).

Therefore, the index form the polynomial is 4x+ 5x2 − 3+ 0 or 4x+ 5x2 − 3x.

#### Question 6:

Add the following polynomials.
(i) 7x4 $-$2x3  +  x  +  10 ; 3x4 + 15x3 + 9x2 $-$8x + 2
(ii) 3p3q+ 2p2q + 7 ; 2p2q + 4pq$-$2p3q

(i)
$\left(7{x}^{4}-2{x}^{3}+x+10\right)+\left(3{x}^{4}+15{x}^{3}+9{x}^{2}-8x+2\right)\phantom{\rule{0ex}{0ex}}=7{x}^{4}+3{x}^{4}-2{x}^{3}+15{x}^{3}+9{x}^{2}+x-8x+10+2\phantom{\rule{0ex}{0ex}}=10{x}^{4}+13{x}^{3}+9{x}^{2}-7x+12$
(ii)
$\left(3{p}^{3}q+2{p}^{2}q+7\right)+\left(2{p}^{2}q+4pq-2{p}^{3}q\right)\phantom{\rule{0ex}{0ex}}=3{p}^{3}q-2{p}^{3}q+2{p}^{2}q+2{p}^{2}q+4pq+7\phantom{\rule{0ex}{0ex}}={p}^{3}q+4{p}^{2}q+4pq+7$

#### Question 7:

Subtract the second polynomial from the first.

(i) 5x2$-$2y + 9 ; 3x2 + 5y $-$7 (ii) 2x2 + 3x + 5 ; x 2 $-$2x + 3

(i)
$\left(5{x}^{2}-2y+9\right)-\left(3{x}^{2}+5y-7\right)\phantom{\rule{0ex}{0ex}}=5{x}^{2}-2y+9-3{x}^{2}-5y+7\phantom{\rule{0ex}{0ex}}=5{x}^{2}-3{x}^{2}-2y-5y+9+7\phantom{\rule{0ex}{0ex}}=2{x}^{2}-7y+16$
(ii)
$\left(2{x}^{2}+3x+5\right)-\left({x}^{2}-2x+3\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}+3x+5-{x}^{2}+2x-3\phantom{\rule{0ex}{0ex}}=2{x}^{2}-{x}^{2}+3x+2x+5-3\phantom{\rule{0ex}{0ex}}={x}^{2}+5x+2$

#### Question 8:

Multiply the following polynomials.

(i) (m3 $-$2m + 3)(m4$-$2m2 + 3+ 2)

(ii) (5m3 $-$2)(m2 $-$ + 3)

(i)
$\left({m}^{3}-2m+3\right)\left({m}^{4}-2{m}^{2}+3m+2\right)\phantom{\rule{0ex}{0ex}}={m}^{3}\left({m}^{4}-2{m}^{2}+3m+2\right)-2m\left({m}^{4}-2{m}^{2}+3m+2\right)+3\left({m}^{4}-2{m}^{2}+3m+2\right)\phantom{\rule{0ex}{0ex}}={m}^{7}-2{m}^{5}+3{m}^{4}+2{m}^{3}-2{m}^{5}+4{m}^{3}-6{m}^{2}-4m+3{m}^{4}-6{m}^{2}+9m+6\phantom{\rule{0ex}{0ex}}={m}^{7}-2{m}^{5}-2{m}^{5}+3{m}^{4}+3{m}^{4}+2{m}^{3}+4{m}^{3}-6{m}^{2}-6{m}^{2}-4m+9m+6\phantom{\rule{0ex}{0ex}}={m}^{7}-4{m}^{5}+6{m}^{4}+6{m}^{3}-12{m}^{2}+5m+6$
(ii)
$\left(5{m}^{3}-2\right)\left({m}^{2}-m+3\right)\phantom{\rule{0ex}{0ex}}=5{m}^{3}\left({m}^{2}-m+3\right)-2\left({m}^{2}-m+3\right)\phantom{\rule{0ex}{0ex}}=5{m}^{5}-5{m}^{4}+15{m}^{3}-2{m}^{2}+2m-6$

#### Question 9:

Divide polynomial 3x3 $-$ 8x2 + x + 7 by x $-$ 3 using synthetic method and write the quotient and remainder.

Dividend = $3{x}^{3}-8{x}^{2}+x+7$

Divisor = $x-3$

Opposite of −3 = 3

The coefficient form of the quotient is (3, 1, 4).

∴ Quotient = 3x+ x + 4 and Remainder = 19

#### Question 10:

For which the value of m, x + 3 is the factor of the polynomial x3 $-$ 2mx + 21?

Let $p\left(x\right)={x}^{3}-2mx+21$.

(x + 3) is the factor of the polynomial $p\left(x\right)={x}^{3}-2mx+21$.

$\therefore p\left(-3\right)=0\phantom{\rule{0ex}{0ex}}⇒{\left(-3\right)}^{3}-2m×\left(-3\right)+21=0\phantom{\rule{0ex}{0ex}}⇒-27+6m+21=0\phantom{\rule{0ex}{0ex}}⇒6m-6=0\phantom{\rule{0ex}{0ex}}⇒6m=6\phantom{\rule{0ex}{0ex}}⇒m=1$
Thus, the value of m is 1.

#### Question 11:

At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x2 $-$3 y2 , 7 y2 + 2 xy and 9 x2 + 4 xy respectively. At the beginning of the year 2017, x2 + xy $-$ y2 , 5 xy and 3 x2 + xy persons from each of the three villages respectively went to another village for education then what is the remaining total population of these three villages ?

Total population of the three villages
$=\left(5{x}^{2}-3{y}^{2}\right)+\left(7{y}^{2}+2xy\right)+\left(9{x}^{2}+4xy\right)\phantom{\rule{0ex}{0ex}}=5{x}^{2}+9{x}^{2}-3{y}^{2}+7{y}^{2}+2xy+4xy\phantom{\rule{0ex}{0ex}}=14{x}^{2}+4{y}^{2}+6xy$
Total number of persons who went to another village for education
$=\left({x}^{2}+xy-{y}^{2}\right)+5xy+\left(3{x}^{2}+xy\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+3{x}^{2}-{y}^{2}+xy+5xy+xy\phantom{\rule{0ex}{0ex}}=4{x}^{2}-{y}^{2}+7xy$
∴ Remaining total population of the three villages = Total population of the three villages − Total number of persons who went to another village for education
$=\left(14{x}^{2}+4{y}^{2}+6xy\right)-\left(4{x}^{2}-{y}^{2}+7xy\right)\phantom{\rule{0ex}{0ex}}=14{x}^{2}+4{y}^{2}+6xy-4{x}^{2}+{y}^{2}-7xy\phantom{\rule{0ex}{0ex}}=14{x}^{2}-4{x}^{2}+4{y}^{2}+{y}^{2}+6xy-7xy\phantom{\rule{0ex}{0ex}}=10{x}^{2}+5{y}^{2}-xy$
Thus, the remaining total population of these three villages is 10x2 + 5y2 − xy.

#### Question 12:

Polynomials bx2 + x + 5 and bx3 $-$2x + 5 are divided by polynomial x$-$3 and the remainders are m and n respectively. If m$-$ n = 0 then find the value of b.

Let $p\left(x\right)=b{x}^{2}+x+5$ and $q\left(x\right)=b{x}^{3}-2x+5$.

The remainder when $p\left(x\right)=b{x}^{2}+x+5$ is divided by (x − 3) is m.

By remainder theorem,

Remainder = $p\left(3\right)=m$

The remainder when $q\left(x\right)=b{x}^{3}-2x+5$ is divided by (x − 3) is n.

By remainder theorem,

Remainder = $q\left(3\right)=n$

Now,

Thus, the value of b is $\frac{1}{2}$.

#### Question 13:

Simplify

(8m2 + 3m$-$6)$-$ (9m $-$ 7) + (3m2$-$ 2+ 4)

$\left(8{m}^{2}+3m-6\right)-\left(9m-7\right)+\left(3{m}^{2}-2m+4\right)\phantom{\rule{0ex}{0ex}}=8{m}^{2}+3m-6-9m+7+3{m}^{2}-2m+4\phantom{\rule{0ex}{0ex}}=8{m}^{2}+3{m}^{2}+3m-9m-2m-6+7+4\phantom{\rule{0ex}{0ex}}=11{m}^{2}-8m+5$

#### Question 14:

Which polynomial is to be subtracted from x2 +  13+ 7 to get the polynomial  3x2 + 5x - 4?

Let p(x) be the polynomial which is to be subtracted from x2 + 13+ 7 to get the polynomial  3x+ 5x − 4.

∴ (x2 + 13+ 7) − p(x) = 3x+ 5x − 4

⇒ p(x) = (x2 + 13+ 7) − (3x+ 5x − 4)

⇒ p(x) = x2 + 13+ 7 − 3x− 5x + 4

⇒ p(x) = x2 − 3x+ 13− 5x + 7 + 4

⇒ p(x) = −2x+ 811

Thus, the required polynomial is −2x+ 811.

#### Question 15:

Which polynomial is to be added to 4+ 2+ 3 to get the polynomial 6m + 3+ 10?

The required polynomial can be obtained by subtracting the polynomial 4+ 2+ 3 from 6m + 3+ 10.

∴ Required polynomial

= (6m + 3+ 10) − (4+ 2+ 3)

= 6m + 3+ 10 − 4− 2− 3

= 6− 4m + 3− 2n + 10 − 3

= 2mn + 7

Thus, the polynomial 2m + n + 7 is to be added to 4+ 2+ 3 to get the polynomial 6m + 3+ 10.

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