# Chapter – 7 : Polynomials | Chapter Solution Class 9

 Book Name : Ganit Prakash Subject : Mathematics (Maths) Class : 9 (Madhyamik/WB) Publisher : Prof. Nabanita Chatterjee Chapter Name : Polynomials (7th Chapter)

## Let us work out – 7.1

1. If f(x)=x^5+3 x^3-7 x^2+6 \quad h(x)=3 x^3-8 x^2+7 \quad g(x)=x+1 \\p(x)=x^4-x^2+2 \text { and } q(y)=7 y^3-y+10 \\then let us calculate and write what would be the following polynomials.

(i) f(x)+g(x)\\

(ii) f(x)-h(x)\\

(iii) f(x)-p(x)\\

(iv) f(x)+p(x)\\

(v) p(x)+g(x)+f(x)\\

(vi) p(x)-q(x)\\

(vii) f(x) \cdot g(x)\\

(viii) p(x) \cdot g(x)\\

Solution:

\text { (i) } f(x)+g(x) \\

=x^5+3 x^3-7 x^2+6+x+1 \\

=x^5+3 x^3-7 x^2+x+7 \text { (Ans) } \\

(ii) f(x)-h(x)\\

=x^5+3 x^3-7 x^2+6-\left(3 x^3-8 x^2+7\right) \\

=x^5+7 x^3-7 x^2+6-3 x^3+8 x^2-7 \\

=x^5+x^2-1 \text { (Ans) }\\

\text { (iii) } f(x)-p(x) \\

=x^5+3 x^3-7 x^2+6-\left(x^4-x^2+2\right) \\

=x^5+3 x^3-7 x^2+6-x^2-2 \\

=x^5-x^4+3 x^3-6 x^2+4(\text { Ans })\\

\text { (iv) } f(x)+p(x)=x^{5}+3 x^{3}-7 x^{2}+6+x^{4}-x^{2}+2 \\

=x^{5}+x^{4}+3 x^{2}-8 x^{2}+8 \quad(\text { Ans }) \\

\text { (v) } p(x)+g(x)+f(x)

= x^{4}-x^{2}+2+x+1+x^{5}+3 x^{3}-7 x^{2}+6 \\

= x^{5}+x^{4}+3 x^{3}-8 x^{2}+x+9(\text { Ans) } \\

\text { (vi) } p(x)-q(y) \\

=x^{4}-x^{2}+2-7 y^{3}

y+10 =x^{4}-x^{2}+2-7 y^{3}+y-10 \\

=x^{4}-x^{2}-7 y^{3}+y-8(\text { Ans) }\\

\text {(vii)} f(x) \cdot g(x) =\left(x^{5}+3 x^{3}-7 x^{2}+6\right)(x+1) \\

=x^{6}+x^{5}+3 x^{4}+3 x^{3}-7 x^{3}-7 x^{2}+6 x+6 \\

=x^{6}+x^{5}+3 x^{4}-4

x^{3}-7 x^{2}+6 x+6(\text { Ans) }\\

\text{(viii)} p(x) \cdot g(x) =\left(x^{4}-x^{2}+2\right) \cdot(x+1) \\

-x^{5}+x^{4}-x^{3}-x^{2}+2 x+2(\text { Ans })

## Let us work out – 7.1

1. Let us write which are the polynomial in the following algebraic expression. Let us write the degree of each of the polynomials.

(i) 2 x^{6}-4 x^{5}+7 x^{2}+3 \\

(ii) x^{-2}+2 x^{-1}+4 \\

(iii) y^{3}-\frac{3}{4} y+\sqrt{7} \\

(iv) \frac{1}{x}-x+2 \\

(v) x^{51}-1 \\

(vi) \sqrt[3]{t}+\frac{t}{27} \\

(vii) 15

(viii) 0

(ix) z+\frac{3}{z}+2 \\

(x) y^{3}+4 (xi) \frac{1}{\sqrt{2}} x^{2}-\sqrt{2} x+2 \\

Solution:

(i) Power of a polynomial is 6.

(ii) It is not a polynomial.

(iii) Power of polynomial is 3.

(iv) It is not a polynomial.

(v) Power of polynomial is 51.

(vi) It is not a polynomial.

(vii) Power of polynomial is 0.

(viii) power of polynomial is undefined

(ix) It is not a polynomial.

(x) Power of polynomial is 3.

(xi) Power of polynomial is 2.

4. I write the degree of each of the following polynomials:

(i) x^{4}+2 x^{3}+x^{2}+x

(ii) 7 x-5

(iii) 16

(iv) 2-y^{2} y^{3}

(vi) 5 x^{2}+x^{19}

Solution:

(i) Degree = 4

(ii) Degree = 1

(iii) Degree = 0

(iv) Degree = 3

(v) Degree = 1

(vi) Degree = 19

5. I write two separate biomonial in one variable whose degree are 17

Solution: x^{17}+5,6 x^{17}-13 \\

6. I write two separate monomials in one variable whose degrees are 4

Solution: x^{4}, 5 y^{4} \\

7. I write two separate nomials in one variable whose degrees are 3

Solutions: x^{3}+x^{2}+1,7 y^{3}-9 y^{2}-10 \\

8. In the following algebraic expression, which are polynomial in one variable which are polynomial in two variables and which are not polynomial-Let us with them

(i) x^{2}+3 x+2 \\

(ii) x^{2}+y^{2}+a^{2} \\

(iii) y^{2}-4 a x

(iv) x+y+2\\

(v) x^{8}+y^{4}+x^{5} y^{9} \\

(vi) x+\frac{5}{x} \\

Solution. (i),(ii),(iii),(iv),(v) are polynomials

(i) within variable

(ii), (iii),(iv), and (v) withtwo variadles.(Ans)-1

## Let us work out – 7.2

1. If f(x)=x^{2}+9 x-6 ,then let us write by calculating the values of f(0), f(1) and f(3).\\

Solution: f(x) =x^{2}+9 x-6 \\

\therefore f(0) =(0)^{2}+9(0)-6 \\

=0+0-6 \\

=-6.\\

\therefore f(0) =-6 \quad(\text { Ans }) \\

f(1) =(1)^{2}+9(1)-6 \\

=1+9-6 \\

=10-6=4\\

\therefore f(1)=4.(Ans)\\

and f(3)=(3)^{2}+9(3)-6 \\

= 9+27-6 \\

=36-6=30 \\

\therefore f(3) =30 .(\text { Ans ) }\\

2. By calculating the following polynominals f(x) let us write the values of f(1) and f(-1).

(i) f(x)=2 x^{3}+x^{2}+x+4\\

(ii) f(x)=3 x^{4}-5 x^{3}+x^{2}+8 \\

(iii) f(x)=4+3 x-x^{3}+5 x^{6} \\

(iv) f(x)=6+10 x-7 x^{2} \\

Solution:

(i)\text { (i) } f(x) =2 x^{3}+x^{2}+x+4 \\

f(x) =2(1)^{3}+(1)^{2}+1+4 \\

=2+1+1+4=8 \quad(\text { Ans }) \\

\text { and } f(-1) =2(-1)^{3}+(-1)^{2}+(-1)+4 \\

=2 \times(-1)+1-1+4 \\

=-2+4=2 . \quad(\text { Ans })\\

(ii) f(x) =3 x^{4}-5 x^{3}+x^{2}+8 \\

f(1) =3 \cdot(1)^{4}-5(1)^{3}+(1)^{2}+8 \\

=3-5+1+8 \\ =12-5=7 \text { (Ans })\\

andf(-1) =3(-1)^{4}-5(-1)^{3}+(-1)^{2}+8 \\

=3 \times 1-5 \times(-1)+1+8 \\

=3+5+1+8=17\\

\text { (iii) } f(x) =4+3 x-x^{3}+5 x^{6} \\

f(1) =4+3(1)-(1)^{3}+5(1)^{6} \\

=4+3-1+5 \Rightarrow 12-1=11 \\

and f(-1) =4+3(-1)-(-1)^{3}+5(-1)^{6} \\

=1+1+5=7\\

(iv) f(x)=6+10 x-7 x^{2} \\

f(1)=6+10(1)-7(1)^{2} \\

=6+10-7=9\\

andf(-1) =6+10(-1)-7(-1)^{2} \\

=6-10-7=-11\\

3. Let us check the following statements-

(i) The zero of the polynomial P(x)=x-1 is 1.

solution: p(x)=0

\text { or, } x-1=0 \\

\therefore x=1\\

The statement is-true.

(ii) The zero of the polynomial P(x)=3-x is 3.

solution: If p(x)=0 \\

or, 3-x=0 \\

\therefore \mathrm{x}=0\\

The statement is true.

(iii) The zero of the polynomial P(x)=5 x+1 is -\frac{1}{5} \\

Solution: If p(x) = 0 \\

or, 5 x+1=0 \\

or, 5 x=-1.\\

\therefore x=-\frac{1}{5} \\

The statement is true.

(iv) The two zeros of the polynomial P(x)=x^{2}-9 \\ are 3 and -3.

solution: \text { If } p(x)=0 \\

\text { or, } x^{2}-9=0 \\

\text { or, } x^{2}=9 \\

\text { or, } x= \pm \sqrt{9} \\

\text { or, } x= \pm 3 \\

\therefore x=3,-3\\

The statement is true.

(v) The two zeros of the polynomial P\left(x=x^{2}\right)-5 x \\are 0 and 5

solution: If p(x)=0 \\

or, x^{2}-5 x=0 \\

or, x(x-5)=0 \\

either x=0 \quad, x-5=0 \Rightarrow x=5 \\

\therefore x=0,5.\\

The statement is true.

(vi) The two zeros of the polynomial \mathrm{P}(\mathrm{x})=\mathrm{x}^{2}-2 x-8 are 4 and (-2) \\

solution: If p(x)=0 \\

or, x^{2}-2 x-8=0 \\

or, x^{2}-4 x+2 x-8=0 \\

or, x(x-4)+2(x-4)=0 \\

or, (x-4)(x-2) \\

eitherx-4=0 \Rightarrow x=4 \\

or, x+2=0 \Rightarrow x=-2 \\

The statement is true.

4. Let us determine the zeros of the following polynomial –

(i) \mathrm{f}(\mathrm{x})=2-\mathrm{x} \\

(ii) f(x)=7 x+2 \\

(iii) f(x)=x+9 \\

(iv) f(x)=6-2 x \\

(v) f(x)=2 x \\

(vi) f(x)=a x+b(a \neq 0) \\

Solution: (i) f(x)=2-x \\

\therefore f(x)=0 gives\\

2-x=0 \text { or, } x=2 \text { (Ans) }\\

(ii) f(x)=7 x+2 \\

\therefore f(x)=0 gives\\

or, 7 x+2=0 \\

\text { or, } 7 x=-2 \\

\therefore x=-\frac{2}{7}(\text { Ans })\\

(iii) f(x)=x+9 \\

\therefore f(x)=0 \\

\text { gives } \\

x+9=0 \\

\therefore x=-9(\text { Ans) }\\

(iv)f(x)=6-2 x \\

\therefore f(x)=0 \text { gives } \\

\therefore 6-2 x=0 \\

\text { or, } 2 x=6 \\

\text { or, } x=\frac{6}{2} \\

\therefore x=3 \text { (Ans) }\\

(v)f(x)=2 x \\

\therefore f(x)=0 \text { gives } \\

\text { or, } 2 x=0 \\

\therefore x=0(Ans).\\

(vi) f(x)=ax + b\\

\therefore a x+b=0 \text { gives } \\

\text { or }, a x=-b \\

## Let us work out – 7.3

1. By applying the reminder theorem let us calculate and write the reminder that I shall get in every cases, when x^{3}-3 x^{2}+x+5 \\is divided by

(i) \mathrm{x}-2

(ii) \mathrm{x}+2

(iii) 2 \mathrm{x}-1

(iv) 2 \mathrm{x}+1.\\

Solution:

(i) At first we find out the zero of the linear polynomial (x – 2)

\therefore x-2=0 \\

\therefore x=2\\

From the remainder theorem we know that the division of f(x)

=x^{3}-3 x^{2}+x+5 by (x-2)gives the remainder f(2)

\therefore The required remainder = f(2).

=(2)^{3}-3 \cdot(2)^{2}+2+5 \\

=8-3 \times 4+2+5 \\

=8-12+2+5 \\

=15-12 \\

=3 \text { (Ans) }

(ii) At first we find out the zero of the linear polynimial

(x + 2)

\therefore x+2=0 \\

x=-2\\

From the remiander theorem, we know that the division of f(x)

= x^{3}-3 x^{2}+x+5 by (x+2) gives the remainder f(-2)

\therefore The required remainder =f(-2) \\

=(-2)^{3}-3(-2)^{2}+(-2)+5 \\

=-8-3 \times 4-2+5 \\

=-8-12-2+5 \\

=-22+5=-17(\text { Ans })\\

(iii) At first we find the zeros of linear polynomial (2x – 1).

\therefore 2 x-1=0 \\

\text { or, } 2 x=1 \\

\therefore x=\frac{1}{2} \\

\therefore From the remainder theorem we know that the division of f (x)

=x^{3}-3 x^{2}+x+5 by (2 \mathrm{x}-1) gives the remainderf\left(\frac{1}{2}\right) \\

\therefore The required remainder =f\left(\frac{1}{2}\right) \\

=\left(\frac{1}{2}\right)^{3}-3\left(\frac{1}{2}\right)^{2}+\frac{1}{2}+5 \\

=\frac{1}{8}-3 \times \frac{1}{4}+\frac{1}{2}+5 \\

=\frac{1}{8}-\frac{3}{4}+\frac{1}{2}+5 \\

=\frac{1-6+4+40}{8} \\

=\frac{45-6}{8}\\

=\frac{39}{8}\\

=4 \frac{7}{8}(\text { Ans })\\

(iv) first we find the zero of linear polynomial (2x + 1)

\therefore 2 x+1=0 \\

2 x=-1 \\

x=-\frac{1}{2}\\

From the remainder theorem, we know that the the division

of f(x)=x^{3}-3 x^{2}+x+5 by (2 x+1) gives the remainder f\left(-\frac{1}{2}\right)\\

\therefore The required remainder =f\left(-\frac{1}{2}\right) \\

=\left(-\frac{1}{2}\right)^{3}-3\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)+5\\

= -\frac{1}{8}-3 \times \frac{1}{4}-\frac{1}{2}+5 \\

=\frac{-1-6-4+40}{8} \\

=\frac{40-11}{8}=\frac{29}{8}=3 \frac{5}{8}(\text { Ans })\\

2. By applying Remainder theorem, let us calculate and write that I shall get when the following polynomial is divided by (x-1).

(i) x^{3}-6 x^{2}+13 x+60 \\

(ii) x^{3}-3 x^{2}+4 x+50 \\

(iii) 4 x^{3}+4 x^{2}-x-1 \\

(iv) 11 x^{3}-12 x^{2}-x+7 \\

Solution: (i) At the first we find the zero of the linear polynomial (x – 1)

\therefore x-1=0 \\

x=1\\

Let us suppose f(x)=x^{3}-6 x^{2}+13 x+60 \\

From remainder theorem, we know that the division of f(x)

= x^{3}-6 x^{2}+13 x+60 \text{by} (x-1) gives the remainder f(1).

\therefore The required remainder =f(1)

=(1)^{3}-6(1)^{2}+13(1)+60 \\

=1-6+13+60 \\

=74-6 \\

= 68 (Ans)

(ii) Let f(x)=x^{3}-3 x^{2} 4 x+50 \\

From the remainder theorem, we know that the division of

f(x)=x^{3}-3 x^{2} 4 x+50 by (x-1) gives the remainder =f(1)

\therefore The required remainder =f(1)

=(1)^{3}-3(1)^{2}+4(1)+50 \\

=1-3+4+50 \\

=55-3 \\

=52 \text { (Ans) }\\

(iii) Let f(x)=4 x^{3}+4 x^{2}-x-1 \\

From remainder theorem, we know that the division of

f(x) =4 x^{3}+4 x^{2}-x-1 \text{by} (x-1) gives the remainder f(1)

\therefore The required remainder = f(1)

=4(1)^{3}+4(1)^{2}-1-1 \\

=4+4-2 \\

=8-2 \\

=6(\text { Ans })\\

(iv) Let f(x)=11 x^{3}-12 x^{2}-x+7 \\

From the remainder theorem, we know that the division of

f(x)=11 x^{3}-12 x^{2}-x+7 \text{by} (x-1) give the remainder f(1)

\therefore The required remainder = f(1)

=11 .(1)^{3}-12(1)^{2}-1+7 \\

=11-12-1+7 \\

=18-13 \\

=5(\text { Ans })\\

3. Applying Remainder Theorem let us write the remainder, when-

(i) The polynomial x^{3}-6 x^{2}+9 x-8 \text { is divided by (x-3)} \\

Solution:

At the first, we find the zero of the linear polynomial (x-3)

\therefore x-3=0 \\

x = 3.

Let us f(x)=\mathrm{x}^{3}-6 x^{2}+9 x-8 \\

From the remainder theorem, we know that the division of f(x)

=\mathrm{x}^{3}-6 x^{2}+9 x-8 \ by \ (x-3) gives the remainder f(3)

\therefore The required remainder = f(3)

=(3)^{3}-6(3)^{2}+9(3)-8 \\

=27-6 \times 9+27-8 \\

=54-54-8 \\

=-8(\text { Ans })\\

(ii) The polynomial x^{3}-a x^{2}+2 x-a is divided by (x-a)

Solution:

At first, we find the zero of the linear polynomial (x – a)

\therefore x-a=0.\\

x=aLet f(x)=x^{3}-a x^{2}+2 x-a \\

From remainder Theoram, we know that the division of f(x)

= x^{3}-a x^{2}+2 x-a b y(x-a) gives the remainder f(a)

The required remainder = f(a)

=(a)^{3}-a(a)^{2}+2 .(a)-a \\

=a^{3}-a^{3}+2 a-a \\

= a

4. Applying Remainder Theorem, let us calculate whether the polynomial p(x)=4 x^{3}+4 x^{2}-x-1 is a multiple of (2 x+1) or not

Solution:

At the first we find the zero of the linear polynomial (2x + 1)

\therefore 2 x+1=0 \\

\text { or } 2 x=-1 \\

\therefore x=-\frac{1}{2} \\

\therefore The zero of the linear polynomial (2 x+1) is -\frac{1}{2} \\

Let p(x)=4 x^{3}+4 x^{2}-x-1 \\

\therefore p(x) will be multiple of (2 x+1) if p\left(-\frac{1}{2}\right)=0 \\

\therefore p\left(-\frac{1}{2}\right) =4\left(-\frac{1}{2}\right)^{3}+4\left(-\frac{1}{2}\right)^{2}-\left(-\frac{1}{2}\right)-1 \\

=-4 \times \frac{1}{8}+4 \times \frac{1}{4}+\frac{1}{2}-1 \\

=-\frac{1}{2}+\frac{1}{1}+\frac{1}{2}-1 \\

= 0

\therefore p(x) is a multiple of (2 x+1)\\

5. For what value of the division of the two polynomial \left(a x^{3}+3 x^{2}-3\right) and \left(2 x^{3}-5 x+a\right) \ by \ (x-4) give the same remainder-let us calculate and write it.

Solution:

Let us suppose,

f(x)=a x^{3}+3 x^{2}-3 \\

and g(x)=2 x^{3}-5 x+a \\

If f(x) is divided by (x – 4), then the remainder is

f(4) =a(4)^{3}+3(4)^{2}-3 \\

=a .64+3 \times 16-3 \\

=64 a+48-3=64 a+45\\

If g(x) is divided by (x – 4), then the remainder is

g(4)=2 .(4)^{3}-5(4)+a \\

=2 \times 64-20+a \\

=128-20+a \\

=108+a \\

\therefore f(4)=g(4)\\

or, 64 a+45=108+a \\

or, 64 a-a=108-45 \\

or, 63 a=63 \\

or, a=\frac{63}{63}=1 \\

\therefore a=1(\text { Ans })\\

6. The two polynomial x^{3}+2 x^{2}-p x-7 \ and \ x^{3}+p x^{2}+6 are divided by (x+1) \ and \ (x-2) respectively and if the respectively and if the remainder R_{1} \ and \ R_{2} are obtained and if 2 R_{1}+R_{2}=6, then let us calculate the the value of p.

Solution:

We find the zero of the linear polynomial (x+1)

\therefore x+1=0 \\

\text { or, } x=-1\\

Let f(x)=x^{3}+2 x^{2}-p x-7 \\

\therefore R_{1} =f(-1) \\

=(-1)^{3}+2(-1)^{2}-p(-1)-7 \\

=2+p-8 \\

=p-6\\

Again, we find the zero of the linear polynomial (x – 2)

\therefore x-2=0 \\

\therefore x=2\\

Let g(x)=x^{3}+p x^{2}-12 x+6 \\

R_{2} =g(2) \\

=(2)^{3}+p(2)^{2}-12(2)+6 \\

=8+4 p-24+6 \\

=4 p+14-24 \\

=4 p-10 \\

\therefore 2R_{1}+R_{2}=6\\

or, 2(p-6)+4 p-10=6 \\

or, 2 p-12+4 p-10=6\\

or, 6 p-22=6 \\

or, 6 p=6+22 \\

\therefore p=\frac{14}{3}=4 \frac{2}{3}(Ans)\\

7. If the polynomial x^{4}-2 x^{3}+3 x^{2}-a x+b is divided by (x-1) \ and \ (x+1) and the remainder are 5 and 19 respectively. But if that polynomial is divided by (x + 2), then what will be the remainder -let us calculate.

Solution:

We find the zero of the linear polynomial (x – 1)

\therefore x-1=0 \\

x = 1.

Let f(x)=x^{4}-2 x^{3}+3 x^{2}-a x+b \\

f(1) =(1)^{4}-2(1)^{3}+3(1)^{2}-a \cdot 1+b \\

=1-2+3-a+b \\

=4-2-a+b \\

=2-a+b \\

\therefore Given, f(1)=5 \\

\therefore 2-a+b=5\\

or, -a+b=5-2 \\

or, -a+b=3 \rightarrow(i)\\

Again, we find the zero of the linear

\text { polynomial }(x+1) \\

\therefore x+1=0 \\

x=-1 \\

\therefore f(-1)=(-1)^{4}-2(-1)^{3}+3(-1)^{2}-a(-1)+b \\

=1+2+3+a+b \\

=a+b+6 \\

Given, f(-1)=19 \\

\text { or, } a+b+6=19 \\

\text { or, } a+b=19-6 \\

\therefore a+b=13 \rightarrow(ii)\\

We get,-a+b=3----(i) \\ \underline{a+b=13----(ii)} \\ \text { or, } 2 b=16 \quad \text { (By adding) } \\

\text { or, } b=\frac{16}{2}=8 \\

\therefore b=8\\

putting the value of b in equation (i),

-a+b=3 \\

\text { or, }-a+8=3 \\

\text { or, }-a=3-8 \\

\text { or, }-a=-5 \\

\therefore a=5\\

We find the zero of the polynomial (x + 2)

\therefore x+2=0 \\

x=-2 \\

\therefore f(-2)=(-2)^{4}-2(-2)^{3}+3(-2)^{2}-a x+b \\

=16-2(-8)+3 \cdot(4)-5(-2)-8 \\

=16+16+12+10+8 \\

=62(Ans) \\

8. If f(x)=\frac{a(x-b)}{a-b}+\frac{b(x-a)}{b-a}, then let us show that f(a)+f(b)=f(a+b)\\

Solution:

f(x)= \frac{a(x-b)}{a-b}+\frac{b(x-a)}{b-a} \\

\therefore f(a) =\frac{a(a-b)}{a-b}+\frac{b(a-a)}{b-a} \\

=a+0 \\

=a \\

\therefore f(a)=a \\

\text { Again, } f(b) =\frac{a(b-b)}{a-b}+\frac{b(b-a)}{b-a}=0+b=b \\

\therefore f(a+b) =\frac{a(a+b-b)}{a-b}+\frac{b(a+b-a)}{b-a} \\

=\frac{a^{2}}{a-b}-\frac{b^{2}}{a-b} \\

=\frac{a^{2}-b^{2}}{a-b} \\

=\frac{(a-b)(a+b)}{(a-b)}=(a+b)\\

\therefore We see that,

f(a)+f(b)=f(a+b)(\text { Proved })\\

9. If f(x)=a x+b and f(0)=3, f(2)=5, let us determine the value of a and b.

Solution:

f(x)=a x+b \\

\therefore f(0)=a .0+b \\

\text { or, } 3=0+b \\

\therefore b=3\\

and f(2)=a .2+b \\

or, 5=2 a+3 \\

or, 2 a=5-3 \\

or, 2 a=2 \\

or, a=\frac{2}{2}=1 \\

\left\{\begin{array}{l}\therefore a=1 \\\therefore b=3\end{array}\right\}

10. If f(x)=a x^{2}+b x+c \ and \ f(0)=2, f(1)=1 \ and \ f(4)=6 , let y us write and calculate the values of a, b and c.

Solution:

f(x)=a x^{2}+b x+c \\

f(0)=a(0)^{2}+b .0+c \\

2=0+0+c \\

\therefore c=2 \\

f(1)=a(1)^{2}+b .1+c \\

1=a+b+2 \\

\text { or, } a+b=1-2 \\

\text { or, } a+b=-1 \rightarrow (i) \\

\text { Again, } f(4)=a(4)^{2}+b .4+c \\

\text { or, } 6=16 a+4 b+2 \\

\text { or, } 16 a+4 b=6-2 \\

\text { or }, 16 a+4 b=4 \\

\text { or, } 4 a+b=1 \rightarrow(\text { (ii })\\

We get a+b=-1 \rightarrow(i) \\

4 a+b=1 \rightarrow(ii)\\ \underline{(-)(-) \quad(-)}\\ \quad -3 a=-2 \ (by \ subtracting) \\

\therefore a=\frac{2}{3}\\

Putting the value of a in equation(i),

a+b=-1\\

or, \frac{2}{3}+b=-1\\

\text { or, } b=-1-\frac{2}{3} \\

\therefore b=-\frac{5}{3} \\

\therefore a=\frac{2}{3}, b=-\frac{5}{3}, c=2(\mathrm{Ans})\\

## 11. M.C.Q

Which of the following is polynomial in one variable?

(a) x+\frac{2}{x}+3 \\

(b) 3 \sqrt{x}+\frac{2}{\sqrt{x}}+5 \\

(c) \sqrt{2} x^{2}-\sqrt{3} x+6 \\

(d) x^{10}+y^{5}+8 \\

Solution : (c) is a poly nomial in one variable

(ii) which of the following is the polynomial?

(a) x-1 \\

(b) \frac{x-1}{x+1} \\

(c) x^{2}-\frac{2}{x^{2}}+5 \\

(d) x^{2}+\frac{2 x^{\frac{2}{3}}}{\sqrt{x^{2}}}+6 \\

Solution : (a) is a linear polynomial

(iii) which of the following is a linear polynomial?

(a) x+x^{2} \\

(b) x+1 \\

(c) 5 x^{2}-x+3 \\

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

Solution: (b) is a linear polynomial.

(iv) Which of the following is a second degree polynomial?

(a) \sqrt{x}-4

(b) x^{3}+x

(c) x^{3}+2 x+6

(d) x^{2}+5 x+6

Solution: (d) is a second degree polynomial

(v) The degree of the polynomial \sqrt{3} is

(a) \frac{1}{2}

(b) 2

(c) 1

(d) 0

Solution : (d) is correct option.

(i) Let us write the zero of the polynomial p(x)=2 x-3

Solution:

\therefore 2 x-3=0 \\

\text { or, } 2 x=3 \\

\therefore x=\frac{3}{2}\\

(ii) If p(x)=x+4 , let us w rite the value of p(x)+p(-x).\\

Solution:

\therefore p(x)=x+4 \\

\text { or, } p(-x)=-x+4 \\

\therefore p(x)+p(-x) =x+4-x-4 \\

= 8

(iii) Let us write the remainder, if the polynomial x^{3}+4 x^{2}+4 x-3 is divided by x.\\

Solution:

We find that zero of the linear polynomial x,

\therefore x=0\\

Let f(x)=x^{3}+4 x^{2}+4 x-3\\

f(0) =(0)^{3}+4 \cdot(0)^{2}+4 \cdot 0-3 \\

=0+0+0-3 \\

=-3(Ans)\\

(iv) If (3 x-1)^{7}=a_{7} x^{7}+a_{6} x^{6}+a_{5} x^{5}+\ldots \ldots \ldots \ldots+a_{1} x+a_{0}, then let us write the value of a_{7}+a_{6}+a_{5}+ +a_{0} ( \ where a_{7}+a_{6}+a_{5}+\ldots \ldots .+a_{0} \ are \ constants).\\

Solution:

\therefore (3 x-1)^{7}=a_{7} x^{7}+a_{6} x^{6}+a_{5} x^{5}+\ldots \ldots+a_{1}x+a_{0} \\

\text { putting } x=1 \\

a_{7}+a_{6}+a_{5}+\ldots \ldots \ldots+a_{0}=(3.1-1)^{7} \\

=(3-1)^{7} \\

=2^{7} \\

=128 \text { (Ans) }\\

## Let us work out – 7.4

1. Let us calculate and write, which of the following will have a factor (x + 1)?

(i) 2 x^{3}+3 x^{2}-1 \\

(ii) x^{4}+x^{3}-x^{2}+4 x+5 \\

(iii) 7 x^{3}+x^{2}+7 x+1 \\

(iv) 3+3 x-5 x^{3}-5 x^{4}(v) x^{4}+x^{2}+x+1 (vi) x^{3}+x^{2}+x+1 \\

Solution:

(i) At the first, we find out the zero of the linear polynomial (x + 1)

\therefore x+1=0 \\

x=-1\\

Let us suppose, f(x)=2 x^{3}+3 x^{2}-1 \\

f(-1) =2 .(-1)^{3}+3 \cdot(-1)^{2}-1 \\

=2 \times(-1)+3 \times 1-1 \\

=-2+3-1=0 \\

\therefore(x+1) is factor of 2 x^{3}+3 x^{2}-1 (Ans)\\

(ii)\text { Let } f(x)=x^{4}+x^{3}-x^{2}+4 x+5 \\

f(-1)=(-1)^{4}+(-1)^{3}-(-1)^{2}+4 \cdot(-1)+5 \\

=1-1-1-4+5 \\

=6-6 = 0 \\

\therefore(x+1) is a factor of x^{4}+x^{3}-x^{2}+4 x+5( Ans ) \\

(iii) Let f(x)= 7 x^{3}+x^{2}+7 x+1 \\

f(-1) =7 \cdot(-1)^{3}+(-1)^{2}+7 \cdot(-1)+1 \\

=-7+1-7+1=-12 \\

\therefore(x+1) is a factor of 7 x^{3}+x^{2}+7 x+1 . (Ans)\\

(iv) Let f(x)= 3+3 x-5 x^{3}-5 x^{4} \\

f(-1) =3+3(-1)-5(-1)^{3}-5(-1)^{4} \\

=3-3+5-5=0 \\

\therefore(x+1)is a factor of 3+3 x-5 x^{3}-5 x^{4} (Ans)\\

(v) Let f(x)=x^{4}+x^{2}+x+1 \\

f(-1) =(-1)^{4}+(-1)^{2}+(-1)+1 \\

=1+1-1+1=2 \\

\therefore(x+1) is not a factor of x^{4}+x^{2}+x+1 (Ans)\\

(vi) \text { 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) \text { is a factor of } x^{3}+x^{2}+x+1(\text { Ans })\\

2. By applying the Factor theorem, let us write g(x) is factor of the following polynomial f(x).

(i) f(x)=x^{4}-x^{2}-12 \ and \ g(x)=x+2 \\

(ii) f(x)=2 x^{3}+9 x^{2}-11 x-30 \ and \ g(x)=x+5 \\

(iii) f(x)=2 x^{3}+7 x^{2}-24 x-45 \ and \ g(x)=x-3 \\

(iv) f(x)=3 x^{3}+x^{2}-20 x+12 \ and \ g(x)=3 x-2 \\

Solution:

(i) Given f(x)=x^{4}-x^{2}-12 \ and \ g(x)=x+2 \\

At first, we find the zero of the linear polynomial (x + 2)

\therefore x+2=0 \\

x=-2 \\

\therefore f(-2)=(-2)^{4}-(-2)^{2}-12 \\

=16-4-12 \\

=16-16 \\

=0 \\

\therefore, g(x) is a factor of f(x)( Ans)\\

(ii) Given f(x)=2 x^{3}+9 x^{2}-11 x-30 \ and \ g(x)=x+5 \\

At first, we find the zero of the linear polynomial (x + 5)

\therefore x+5=0 \\

\text { or, } x=-5\\

\therefore f(-5) =2(-5)^{3}+9(-5)^{2}-11(-5)-30 \\

=2 \times(-125)+9(25)-11(-5)-30 \\

=-250+225+55-30 \\

=-280+280=0 \\

\therefore g(x) is a factor of f(x)( Ans ) \\

(iii) Given f(x)=2 x^{3}+7 x^{2}-24 x-45 \ and \ g(x)=x-3 \\

At first, we find the zero of the linear polynomial (x – 3)

\therefore x-3=0 \\

\text { or, } x=-3 \\

\therefore f(3)=2(3)^{3}+7(3)^{2}-24(3)-45 \\

\quad=2 \times 27+7 \times 9-24 \times 3-45 \\

\therefore g(x) \text { is a factor of } f(x)(\text { Ans })\\

(iv) Given,f(x)=3 x^{3}+x^{2}-20 x+12 and g(x)=3 x-2 \\

At first, we find the zero of the linear polynomial (3x – 2)

\therefore 3 x-2=0 \\

\text { or, } x=\frac{2}{3} \\

\therefore f\left(\frac{2}{3}\right)=3 \cdot\left(\frac{2}{3}\right)^{3}+\left(\frac{2}{3}\right)^{2}-20\left(\frac{2}{3}\right)+12 \\

=3 \times \frac{8}{27}+\frac{4}{9}-20 \times \frac{2}{3}+12 \\

=\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12 \\

=\frac{8+4-120+108}{9}=\frac{120-120}{9}=\frac{0}{9}=0 \\

\therefore g(x) is a factor of f(x) (Ans)\\

3. Let us calculate the value of K for which the polynomial 2 x^{4}+3 x^{3}+2 k x^{2}+3 x+6 is divided by x + 2

Solution:

At the first we find the zero of the linear polynomial (x + 2)

\therefore x+2=0\\

or, x=-2 \\

\therefore \text { Let } f(x)=2 x^{4}+3 x^{3}+2 k x^{2}+3 x+6 \\

\because f(x) \text { is divided by }(x+2) \\

\therefore f(-2)=0 \\

\therefore f(-2)=2(-2)^{4}+3(-2)^{3}+2 k(-2)^{2}+3(-2)+6 \\

=2 \times 16+3(-8)+2 k(4)-6+6 \\

=32-24+8 k+0=8+8 k \\

\therefore f(-2)=0 \\

\text { or, } 8+8 k=0 \\

or, 8 k=-8 \\

or, k=-\frac{8}{8} \\

\therefore k=-1(Ans)\\

4. Let us calculate the value of k for which g(x) will be the factor of the following polynomial f(x).

(i) f(x)=2 x^{3}+9 x^{2}+x+k \ and \ g(x)=x-1 \\

(ii) f(x)=k x^{2}-3 x+k \ and \ g(x)=x-1 \\

(iii) f(x)=2 x^{4}+x^{3}-k x^{2}-x+6 \ and \ g(x)=2 x-3 \\

(iv) f(x)=2 x^{3}+k x^{2}+11 x+k+3 \ and \ g(x)=2 x-1\\

Solution:

(i) Given f(x)=2 x^{3}-9 x^{2}+x+k \ and \ g(x)=x-1 \\

At first we find the zero of the linear polynomial (x – 1).

\therefore x-1=0 \\

\text { or, } x=1 \\

\therefore g(x) is a factor of f(x) \\

\therefore f(1)=0\\

\text { or, } 2 . (1)^{3}+9(1)^{2}+1+k=0\\

\text { or, } 2 \times 1+9+1+k=0 \\

\text { or, } 2+9+1+k=0 \\

\text { or }, 12+k=0\\

\text { or, } k=-12(Ans)\\

(ii) Given f(x)=k x^{2}-3 x+k \ and \ g(x)=(x-1) \\

At first we find the zero of the linear polynomial (x – 1)

\therefore x-1=0 \\

\text { or, } x=1 \\

\therefore g(x) is a factor of f(x)

\therefore f(1)=0\\

\text { or, } k \cdot(1)^{2}-3.1+k=0\\

\text { or, } k-3+k=0 \\

\text { or, } 2 k=3\\

\text { or, } k=\frac{3}{2}(Ans)\\

(iii) Given f(x)=2 x^{4}+x^{3}-k x^{2}-x+6 \ and \ g(x)=2 x-3 \\

At first we find the zero of the linear polynomial (2x – 3)

\therefore 2 x-3=0\\

\text { or, } x=\frac{3}{2} \\

\therefore g(x) is a factor of f(x)

\therefore f\left(\frac{3}{2}\right)=0 \\

\text { or, } 2\left(\frac{2}{3}\right)^{4}+\left(\frac{2}{3}\right)^{3}-k\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)+6=0 \\

\text { or, } 2 \times \frac{81}{16}+\frac{27}{8}-k \times \frac{9}{4}-\frac{3}{2}+6=0 \\

\text { or, } \frac{81}{8}+\frac{27}{8}-\frac{9 k}{4}-\frac{3}{2}+6=0 \\

\text { or, } \frac{81+27-18 k-12+48}{8}=0 \\

\text { or, } \frac{156-12-18 k}{8}=0 \\

\text { or, } \frac{144-18 k}{8}=0 \\

\text { or, } 144-18 k=0 \\

\text { or, } 144=18 k \\

\text { or, } \frac{144}{18}=k \\

\therefore k=8(\text { Ans })\\

(iv) Given f(x)=f(x)=2 x^{3}+k x^{2}+11 x+k+3 \ and \ g(x)=2 x-1 \\

At first, we find the zero of the linear polynomial (2x – 1)

\therefore 2 x-1=0 \\

\therefore x=\frac{1}{2} \\

\therefore g(x) is a factor of f(x) \\

\therefore f\left(\frac{1}{2}\right)=0\\

or, 2 \cdot\left(\frac{1}{2}\right)^{3}+k \cdot\left(\frac{1}{2}\right)^{2}+11\left(\frac{1}{2}\right)+k+3=0 \\

or, 2 \times \frac{1}{8}+k \times \frac{1}{4}+\frac{11}{2}+k+3=0 \\

or, \frac{1}{4}+k \times \frac{1}{4}+\frac{11}{2}+k+3=0 \\

or, \frac{1}{4}+\frac{k}{4}+\frac{11}{2}+k+3=0 \\

or, \frac{1+k+22+4 k+12}{4}=0 \\

or, \frac{5 k+35}{4}=0 \\

or, 5 k+35=0 \\

or, 5 k=-35 \\

or, k=-7( Ans )\\

5. Let us calculate and write the value of a and b if x^{2}-4 is a factor of the polynomial a x^{2}+2 x^{3}-3 x^{2}+b x-4 \\

Solution:

Let f(x)=a x^{2}+2 x^{3}-3 x^{2}+b x-4 \ and \ g(x)=x^{2}-4 \\

We find the zero of the polynomial \left(x^{2}-4\right),\\

\therefore x^{2}-4=0 \\

\text { or, } x^{2}=4 \\

\text { or, } x=\sqrt{4} \\

\therefore x= \pm 2 \\

\because g(x) is a factor of f(x)

\therefore f(2)=0 \text { and } f(-2)=0\\

\therefore f(2)=0 \text { gives }\\

\text { or, } a(2)^{4}+2(2)^{3}-3(2)^{2}+b(2)-4=0 \\

\text { or, } a \times 16+2 \times 8-3 \times 4+2 b-4=0\\

\text { or, } 16 a+16-12+2 b-4=0 \\

\text { or, } 2(8 a+b)=0 \\

\text { or, } 8 a+b=\frac{0}{2}=0 \\

\therefore 8 \mathrm{a}+\mathrm{b}=0 \rightarrow(\mathrm{i})\\

again, f(-2)=0 \text { gives }\\

a(-2)^{4}+2 \cdot(-2)^{3}-3 \cdot(-2)^{2}+b(-2)-4=0 \\

\text { or, } a \times 16+2 \times(-8)-3 \times 4-2 b-4=0 \\

\text { or, } 16 a-16-12-2 b-4=0 \\

\text { or, } 16 a-2 b-32=0 \\

\text { or, } 16 a-2 b=32 \\

\text { or, } 8 a-b=16 \\

we get,

8 a+b=0----(i) \\ \underline{8 a-b=16----(ii)} \\ \text { or, } 16 a=16 \quad \text (By \ adding)\\

\text { or, } a=\frac{16}{16}=1 \\

\therefore a=1\\

Now putting the value of a in equation (i)

8 a+b=0\\

\text { or, } 8 \times 1+b=0 \\

\text { or, } 8+b=0 \\

\text { or, } b=-8 \\

\left\{\therefore a=1\\ \therefore b=-8\right\}(\text { Ans })\\

6. If (x+1) and (x+2) are two factors of the polynomial x^{3}+3 x^{2}+2 ax+b then let us calculate and write the values of a and b.

Solution:

Let\ f(x)=x^{3}+3 x^{2}+2 a x+b \\

g(x)=x+1 \\

h(x)=x+2\\

At first we find the zero of the polynomial (x + 1)

\therefore x+1=0 \\

\therefore x=-1\\

\because g(x) \text { is a factor of } f(x) \\

\therefore f(-1)=0 \\

\text { or, }(-1)^{3}+3 \cdot(-1)^{2}+2 a(-1)+b=0 \\

\text { or, }-1+3 \times 1-2 a+b=0 \\

\text { or, }-1+3-2 a+b=0 \\

\text { or, } 2-2 a+b=0 \\

\text { or, }-2 a+b=-2----(i)\\

Again,

We find the zero of the polynomial (x + 2)

\therefore f(-2)=0 \\

\text { or, }(-2)^{3}+3(-2)^{2}+2 a(-2)+b=0 \\

\text { or, }-8+3 \times 4-4 a+b=0 \\

\text { or, }-8+12-4 a+b=0 \\

\text { or, } 4-4 a+b=0 \\

\therefore-4 a+b=-4---(ii)\\

We get,

-2 a+b=-2----(i) \\ \underline{_{(+)}-4 a+_{(-)}b=_{(+)}-4----(ii)} \\ {or, \quad 2 a=2} \quad \text { (By subtracting) } \\

Now putting the value of a in equation (i)

-2 a+b=-2 \\

\text { or, }-2 \times 1+b=-2 \\

\text { or, }-2+b=-2 \\

\text { or, } b=-2+2 \\

\therefore b=0 \\

\left\{\therefore a=1 \\\therefore b=0\right\}(Ans) \\

7. If the polynomial a x^{3}+b x^{2}+x-6 is divided by x – 2 and remainder is 4, then let us calculate the values of a and b when x + 2 is a factor of this polynomial.

Solution:

\text { Let } f(x) =a x^{3}+b x^{2}+x-6 \\

g(x) =x-2 \\

h(x) =x+2\\

At the first we find the zero of the polynomial (x – 2)

\therefore x-2=0 \\

\therefore x=2\\

If f(x) is divided by g(x) f(2)=4\\

\text { or, } a(2)^{3}+b(2)^{2}+2-6=4\\

\text { or, } a \times 8+b \times 4-4=4 \\

\text { or, } 8 a+4 b=4+4 \\

\text { or, } 8 \mathrm{a}+4 b=8 \\

\text { or, } 2 a+b=2 -----(i)\\

Again, we find the zero of the polynomial (x + 2)

\therefore x+2=0 \\

x=-2 \\

\because h(x) \text { is a factor of } f(x) \\

\therefore f(-2)=0\\

\text { or, } a(-2)^{3}+b(-2)^{2}+(-2)-6=0 \\

\text { or, } a(-8)+b \times 4-2-6=0 \\

\text { or, }-8 a+4 b-8=0 \\

\text { or, }-2 b+b=2-----(ii)\\

We get,

2 a+b=2\\

-2 a+b=2\\

\text { or, } b=\frac{4}{2}\\

\therefore b=2\\

Now putting the value of b in equation (i)

\therefore 2 a+b=2 \\

\text { or, } 2 a+2=2 \\

\text { or, } 2 a=2-2 \\

\text { or, } 2 a=0 \\

\text { or, } a=\frac{0}{2} \\

\therefore a=0 \\

\left\{\therefore a=0 \\ \therefore b=2\right\}(Ans) \\

8. Let us show that if n is any positive integer (even or odd) x – y is a factor of the polynomial x^{n}-y^{n}.\\

Solution:

Let us suppose if x^{n}-y^{n} is divided by x – y the Quotient is Q and remainder without x is R.

Dividend = divisor \times Quotient +\mathrm{Remainder} \\

\therefore x^{n}-y^{n}=(x-y) \times Q+R\\

Since x does not belong to the remainder R. the value of R will not change for any value of x. So, in the abave identiy

Putting (y) for x, we get

y^{n}-y^{n}=(y-y) \times Q+R \\

0=0 \times Q+R[\therefore \times \text { is positive even or odd integer}] \\

\therefore R=0 \\

\therefore(x-y) is a factor of the polynomial x^{n}-y^{n}, wgenn is any positive even or odd integer. (proved)

9. Let us show that if n is any positive odd integer, then x + y is a factor of x^{n}+y.\\

Solution:

Let us suppose if x^{n}+y^{n} is divided by x+y the Quotient is Q and remainder without x is R.

Dividend = Divisor \times Quotient + Re mainder\\

\therefore x^{n}+y^{n}=(x+y) \times \mathrm{Q}+\mathrm{R}\\

Since x does not belong to the remainder R, the value of R will not change for any value of x. So, in the above identity, Putting (-y) for x, we get

(-y)^{n}+y^{n}=(-y+y) \times Q+R \\

-y^{n}+y^{n}=0 \times Q+R \quad[\therefore n \text { is positive odd integer}] \\

\therefore R=0 \\

\therefore(x+y) is a factor of the polynomial x^{n}+y^{n}, when is any positive add integer: (proved)

10. Let us show that if n be any positive integer(even or odd) then x – y never be a factor of the polynomial x^{n}+y^{n} \\

Solution:

Let us suppose if x^{n}+y^{n} is divided x – y, the Quotient is Q and remainder: without x is R

Dividend = Divisor \times Quotient +\mathrm{Remainder}

\therefore x^{n}+y^{n}=(x-y) \times Q+\mathrm{R}\\

Since x does not belong to the remainder of R, the value of R will not change for any value of x. So, in the above identity, Putting (y) for x, we get

(y)^{n}+y^{n}=(y-y) \times Q+R \\

2 y^{n}=0 \times Q+R \\

\therefore R=2 y^{n} \\

\therefore(x-y) never be a factor of the polynomial x^{n}+y^{n}, when n is any positive odd integer. (proved)

## 11. M.C.Q

(i) If the polynomial x^{3}+6 x^{2}+4 x+k is divisible by (x+2), then the value of k is

(a) -6

(b) -7

(c) -8

(d) (-10)

Solution:

We find the zero of the polynomial (x + 2)

\therefore x+2=0 \\

\therefore x=-2\\

Let f(x)=x^{3}+6 x^{2}+4 x+k \ and \ g(x)=(x+2) \\

\therefore f(x) \text { is divisible by } g(x) \\

\therefore f(-2)=0\\

\text { or, } (-2)^{3}+6(-2)^{2}+4(-2)+k=0 \\

\text { or, }-8+6 \times 4-4 \times 2+k=0\\

\text { or, }-8+24-8+k=0 \\

\text { or, } 16-8+k=0\\

\text { or, } 8+k=0 \\

\therefore k=-8 \\

(ii) In the polynomial f(x) f-\frac{1}{2}, then the factor of f(x)

will be

(a) 2x – 1

(b) 2x + 1

(c) x – 1

(d) x + 1

Solution:

f\left(-\frac{1}{2}\right)=0 \\

\text { or, } x=-\frac{1}{2} \\

\text { or, } 2 x=-1 \\

\text { or, } 2 x+1=0 \\

\therefore the factor of f(x) will be (2x + 1)

(iii) (\mathrm{x}-1) is a factor of the polynomial \mathrm{f}(\mathrm{x}) but it is not the factor of \mathrm{g}(\mathrm{x}). So (x – 1) will be a factor of

(a) f(x) g(x)

(b) -f(x)+g(x)

(c) f(x)-g(x)

(d) \{f(x)+g(x)\} g(x)

(iv) (x + 1) is the factor of the polynomial x^{n}+1 when

(a) n is a positive odd integer

(b) n is a negative even integer

(c) n is negative integer

(d) n is a positiveinteger

(v) If n^{2}-1 is a factor of the polynomial an^{4}+b n^{3}+c n^{2}+d n+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 \\

Solution:

\therefore n^{2}-1 is a factor of the polynomial,

an^{4}+b n^{3}+c n^{2}+d n+e\\

We find the zeroof the polynomial \mathrm{n}^{2}-1 \\

\therefore n^{2}-1=0\\

or, n^{2} \\

or, n^{2}=1 \\

or, n= \pm 1 \\

\therefore \text { Let } f(n)=a n^{4}+b n^{3}+c n^{2}+d n+e \\

f(1)=0, f(-1)=0 \\

\therefore f(1)=0 \text { gives } \\

\text { or, } a(1)^{4}+b(1)^{3}+c(1)^{2}+d(1)+e=0 \\

\text { or, } a+b+c+d+e=0\\

Again, f( -1) = 0 gives

or, a(-1)^{4}+b(-1)^{3}+c(-1)^{2}+d(-1)+e=0 \\

or, a-b+c-d+e=0

\therefore a+c+e=b+d

## 12. Short answer type question:

(i) Let us calculate and write the value of a for which x + a will be a factor of the polynomial x^{3}+a x^{2}-2 x+a-12.

Solution:

Let f(x)=x^{3}+a x^{2}-2 x+a-12\\

We find the zero of the polynomial (x + a)

\therefore x+a=0 \\

\therefore x=-a \\

\thereforee g(x) is a factor of f(x)

2+a=b+a+d(b) \\

\therefore f( -a) = 0

or, (-a)^{3}+a(-a)^{2}-2(-a)+a-12=0\\

or, -a^{3}+a \cdot a^{2}+2 a+a-12=0\\

or, -a^{3}+a^{3}-a-12=0 \\

or, -a=12 \\

\therefore a=-12(\mathrm{Ans}) \\

(ii) Let us calculate and write the value of K for which \mathrm{x}-3 will be a factor of the polynomial k^{2} x^{3}-k x^{2}+3 k x-k \\

Solution:

Let f(x)=k^{2} x^{3}-k x^{2}+3 k x-k \\

and g(x)=x-3 \\

We find the zero of the polynomial (x – 3)

\therefore x-3=0 \\

\therefore x=3 \\

\therefore g(x) is a factor of f(x)

\text { or, } f(3)=0 \\

\therefore k^{2}(3)^{3}-k(3)^{2}+3 k(3)-k=0\\

\text { or, } k^{2} .27-9 k+9 k-k=0 \\

\text { or, } 27 k^{2}-k=0\\

\text { or, } k(27 k-1)=0\\

either, k=0 \quad \text { or, }\\

27 k-1=0\\

\text { or, } 27 k=1 \\

\therefore k=\frac{1}{27} \\

\therefore k=0, \frac{1}{27} \text { (Ans) }\\

(iii) Let us write the value of f(x)+f(-x) when f(x) = 2x + 5

Solution:

Given, f x(x)=2 x+5 \\

and f(-x)=2(-x)+5=-2 x+5 \\

\therefore f(x)+f(-x) \\

=2 x+5+(-2 x+5) \\

=2 x+5-2 x+5 \\

=10 \text { (Ans) }\\

(iv) Both (x – 2) and \left(x-\frac{1}{2}\right) are factors of the polynomial p x^{2}+5 x+r and write the relation between p and r.

Solution:

We find the zero of the polynomial (x - 2) \text { and } \left(x-\frac{1}{2}\right) \\

\therefore x-2=0 \text { and } x-\frac{1}{2}=0 \\

Let f(x)=p x^{2}+5 x+r \\

Both (x – 2) and \left(x-\frac{1}{2}\right) are factors of the polynomial f(x)

\therefore f(2)=0 \ f\left(\frac{1}{2}\right)=0 \\

\therefore f(2)=0 gives\\

p(2)^{2}+5.2+r=0\\

\text { or, } 4 p+10+r=0 \\

\text { or, } 4 p+r=-10----(\mathrm{i}) \\

\operatorname{Again} f\left(\frac{1}{2}\right)=0 gives\\

p\left(\frac{1}{2}\right)^{2}+5\left(\frac{1}{2}\right)+r=0 \\

\text { or, } p \times \frac{1}{4}+\frac{5}{2}+r=0\\

\text { or, } \frac{p}{4}+r=-\frac{5}{2} \\

\text { or, } \frac{p+4 r}{4}=\frac{-5}{2} \\

\therefore p+4 r=-10----(ii)\\

We get

4 p+r=-10---(i) \\

p+4 r=-10---(ii) \\

\therefore 4 p+r=p+4 r \\

\text { or, } 4 p-p=4 r-r \\

\text { or, } 3 p=3 r \\

\therefore p=r \text { (Ans) }\\

(v) Let us write the roots of the linear polynomial f(x)=2 x+3 \\

Solution:

We find the zero of the polynomial (2x + 3)

\therefore 2 x+3=0\\

or, 2 x=-3 \\

\therefore x=-\frac{3}{2} \\

\therefore Roots of the linear polynomial is –\frac{3}{2} (Ans)\\

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