| 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) =-6Â \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 \\
\therefore \quad x=-\frac{b}{a}(Ans)\\
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.
12. Short answer type question:
(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:Â
\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 \\
\quad=54+63-72-45=0 \\
\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:Â
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 \quad x+2=0 \\
\quad 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) } \\
\therefore \quad a=1 \\
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:Â
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 \\
\therefore(c) is correct answer.
(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 + 1Â
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)
\therefore(b) is correct answer
(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)
Solution: (a) is correct answer.
(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
Solution: (a) is correct answer
(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\\
which is not given answer
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
\therefore(a) is correct answer.
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 \\
\therefore \mathrm{x}=2 \quad \text { and } \quad x=\frac{1}{2}\\
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)\\
