# NCERT Solutions For Class 9 Maths Chapter 2

## NCERT Solutions Class 9 Maths Polynomials

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### NCERT Solutions For Class 9 Maths Chapter 2 Exercises

Exercise – 1

Q.1.Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:

(i)4x23x+7$4x^{2}-3x+7$

Answer: It is a polynomial in one variable.

(ii) y2+2$y^{2}+\sqrt{2}$

Answer: It is a polynomial in one variable.

(iii) 3t+t2$3\sqrt{t}+t\sqrt{2}$

Answer: It is not a polynomial since the power of the variable is not a whole number.

(iv) y+2y$y+\frac{2}{y}$

Answer: It is not a polynomial since the power of the variable is not a whole number.

(v) x10+y3+t50$x^{10}+y^{3}+t^{50}$

Answer: It is a polynomial in three variables.

Q.2. Write the coefficients of a2 in each of the following:

(i) 2+a2+a$2 + a^{2} + a$

Answer: Coefficient of a2$a^{2}$ is 1.

(ii) 2a2+a3$2 – a^{2} + a^{3}$

Answer: Coefficient of a2$a^{2}$ is -1.

(iii) π2x2+x$\frac{\pi }{2}x^{2}+x$

Answer: Coefficient of a2$a^{2}$ is π2$\frac{\pi }{2}$.

(iv)
2x1$\sqrt{2}x-1$

Answer: Coefficient of a2$a^{2}$ is 0.

Q.3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer: 3x35+5and4x100$3x^{35}+5\; and\; 4x^{100}$

Q.4. Write the degree of each of the following polynomials:

(i) 5a3+4a2+7a$5a^{3} + 4a^{2} + 7a$

(ii) 3b2$3-b^{2}$

(iii) 5t8$5t – \sqrt{8}$

(iv)8

Q.5. Classify the following as linear, quadratic and cubic polynomial.

(i) a2+a$a^{2} + a$

(ii)aa3$a-a^{3}$

(iii) y+y2+4$y + y^{2} +4$

(iv)1 + x

(v) 3a

(vi) a2$a^{2}$

(vii) 6a3$6a^{3}$

Exercise – 2

Q.1. Find the value of the polynomial atf(x)=5a4a2+3$f(x)= 5a -4a^{2}+ 3$ at

(i) a= 0

(ii) a = – 1

(iii) a = 2

Let f(x)=5a4a2+3$f(x)= 5a -4a^{2}+ 3$

(i) When a=0

f(0)=5(0)+4(0)2+3=3$f(0) = 5(0) + 4(0)^{2} + 3= 3$

(ii) When a=-1

f(a)=5a+4a2+3f(1)=5(1)+4(1)2+3=54+3=6$f(a) = 5a + 4a^{2}+ 3\\ f(-1) = 5(-1) + 4(-1)^{2} + 3 = -5 – 4 + 3 = -6$

(iii) When a=2

f(a)=5a+4a2+3f(2)=5(2)+4(2)2+3=1016+3=3$f(a) = 5a + 4a^{2} + 3\\ f(2) = 5(2) + 4(2)^{2} + 3 = 10 – 16 + 3 = -3$

Q.2. Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y)=y2y+1$p(y) = y^{2}- y + 1$

Answer: p(y)=y2y+1p(0)=(0)2(0)+1=1p(1)=(1)2(1)+1=1p(2)=(2)2(2)+1=3$p(y) = y2 – y + 1\\ ∴ p(0) = (0)^{2}- (0) + 1 = 1\\ p(1) = (1)^{2} – (1) + 1 = 1\\ p(2) = (2)^{2} – (2) + 1 = 3$

(ii) p(a)=2+a+2a2a3$p(a) = 2 + a + 2a^{2}- a^{3}$

Answer: p(a)=2+a+2a2a3p(0)=2+0+2(0)2(0)3=2p(1)=2+1+2(1)2(1)3=2+1+21=4p(2)=2+2+2(2)2(2)3=2+2+88=4$p(a) = 2 +a + 2a^{2}-a^{3}\\ ∴ p(0) = 2 + 0 + 2 (0)^{2} – (0)^{3} = 2\\ p(1) = 2 + 1+ 2(1)^{2} – (1)^{3} = 2 + 1 + 2 – 1 = 4\\ p(2) = 2 + 2 + 2(2)^{2} – (2)^{3} = 2 + 2 + 8 – 8 = 4$

(iii) p(x)=x3$p(x) = x^{3}$

Answer: p(x)=x3p(0)=(0)3=0p(1)=(1)3=1p(2)=(2)3=8$p(x) = x3\\ ∴ p(0) = (0)^{3}= 0\\ p(1) = (1)^{3} = 1\\ p(2) = (2)^{3} = 8$

(iv) p(a)=(a1)(a+1)$p(a) = (a- 1) (a+ 1)$

Answer: p(a)=(a1)(a+1)p(0)=(01)(0+1)=(1)(1)=1p(1)=(11)(1+1)=0(2)=0p(2)=(21)(2+1)=1(3)=3$p(a) = (a – 1) (a + 1)\\ ∴ p(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1\\ p(1) = (1 – 1) (1 + 1) = 0 (2) = 0\\ p(2) = (2 – 1 ) (2 + 1) = 1(3) = 3$

Q.3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x)=3x+1,x=13$p(x) = 3x + 1, x = -\frac{1}{3}$

Answer: For, x=13$x = -\frac{1}{3}$

p(x)=3x+1p(13)=3(13)+1=1+1=0$p(x) = 3x + 1\\ ∴ p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1=-1+1=0$

13$∴ -\frac{1}{3}$ is a zero of p(x).

(ii) p(x)=5xπ,x=45$p(x) = 5x – \pi , x = \frac{4}{5}$

Answer: For, x=45$x = \frac{4}{5}$

p(x)=5xπp(45)=5(45)π=4π$p(x) = 5x – \pi \\ ∴ p(\frac{4}{5}) = 5(\frac{4}{5}) – \pi=4-\pi$

45$∴ \frac{4}{5}$ is not a zero of p(x).

(iii) p(x)=x21,x=1,1$p(x) = x^{2}- 1, x = 1, -1$

Answer: For, x=1,1$x = 1,-1$

p(x)=x21p(1)=x11=11=0p(1)=x11=11=0$p(x) = x^{2}- 1\\ ∴ p(1) = x^{1}- 1=1-1=0\\ p(-1) = x^{-1}- 1=1-1=0$

1,1$∴ 1,-1$ are zeros of p(x).

(iv) p(x)=(x+1)(x2),x=1,2$p(x) = (x + 1) (x – 2), x = -1, 2$

Answer: For, x=1,2$x = -1,2$

p(x)=(x+1)(x2)p(1)=(1+1)(12)=((0)(3))=0p(2)=(2+1)(22)=(3)(0)=0$p(x) = (x + 1) (x – 2)\\ ∴ p(-1) = (-1 + 1) (-1 – 2)\\ =((0)(-3))\\ =0\\ \\ p(2) = (2 + 1) (2 – 2)\\ =(3)(0)=0$

1,2$∴ -1,2$ are zeros of p(x).

(v) p(x)=x2,x=0$p(x) = x^{2}, x = 0$

Answer: For, x=0$x = 0$

p(x)=02=0$p(x) = 0^{2}=0$

0$∴ 0$ is a  zero of p(x).

(vi) p(x)=lx+m,x=ml$p(x)=lx+m,x=-\frac{m}{l}$

Answer: For, x=ml$x=-\frac{m}{l}$

p(x)=lx+mp(ml)=l(ml)+m=m+m=0$p(x)=lx+m\\ ∴ p(-\frac{m}{l} )=l(-\frac{m}{l} )+m=-m+m=0$

ml$∴-\frac{m}{l}$ is a zero of p(x).

(vii) p(x)=3x21,x=13,23$p(x)=3x^{2}-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$

Answer: For, x=13,23$x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$
p(x)=3x21p(13)=3(13)21=3(13)1=11=0p(23)=3(23)21=3(43)1=41=30$p(x)=3x^{2}-1\\ ∴ p(-\frac{1}{\sqrt{3}})=3(-\frac{1}{\sqrt{3}})^{2}-1\\ =3(\frac{1}{3})-1=1-1=0\\ \\ ∴ p(\frac{2}{\sqrt{3}})=3(\frac{2}{\sqrt{3}})^{2}-1\\ =3(\frac{4}{3})-1=4-1=3\neq 0$

13$∴ -\frac{1}{\sqrt{3}}$ is a zero of p(x) but 23$\frac{2}{\sqrt{3}}$ is not a zero of p(x).

(viii) p(x)=2x+1,x=12$p(x) = 2x + 1, x = \frac{1}{2}$
Answer: For, x=12$x=\frac{1}{2}$

p(x)=2x+1p(12)=2(12)+1=1+1=20$p(x) = 2x + 1\\ ∴ p(\frac{1}{2} ) = 2(\frac{1}{2}) + 1=1+1=2\neq 0$

12$∴\frac{1}{2}  $ is not a  zero of p(x).

Q.4. Find the zero of the polynomial in each of the following cases:

(i)p(x) = x + 5

Answer: p(x)=x+5x+5=0x=5$p(x) = x + 5 \\ \Rightarrow x + 5=0 \Rightarrow x=-5$

$∴$-5 is a zero polynomial of the polinomila p(x).

(ii) p(x) = x – 5

Answer: p(x)=x5x5=0x=5$p(x) = x -5 \\ \Rightarrow x -5=0 \Rightarrow x=5$

$∴$5 is a zero polynomial of the polinomila p(x).

(iii) p(x) = 2x + 5

Answer: p(x)=2x+52x+5=02x=5x=52$p(x) =2 x + 5 \\ \Rightarrow 2x + 5=0 \Rightarrow 2x=-5\\ \Rightarrow x=\frac{-5}{2}$

$∴$x=52$x=\frac{-5}{2}$ is a zero polynomial of the polynomial p(x).

(iv)p(x) = 3x – 2

Answer: p(x)=3x23x2=03x=2x=23$p(x) =3 x – 2 \\ \Rightarrow 3x -2=0 \Rightarrow 3x=2\\ \Rightarrow x=\frac{2}{3}$

$∴$x=23$x=\frac{2}{3}$ is a zero polynomial of the polynomial p(x).

(v)p(x) = 3x

Answer: p(x)=3x3x=0x=0$p(x) =3 x \\ \Rightarrow 3x =0 \Rightarrow x=0$

$∴$0 is a zero polynomial of the polynomial p(x).

Exercise – 3

Q.1.Find the remainder when x3+3x2+3x+1$x^{3}+3x^{2}+3x+1$ is divided by

(i) x+1

x+1=0x=1$x+1=0\\ \Rightarrow x=-1$ Remainder=p(1)=(1)3+3(1)2+3(1)+1=1+33+1=0$∴\;Remainder=p(-1)=(-1)^{3}+3(-1)^{2}+3(-1)+1=-1+3-3+1=0$

(ii) x12$x-\frac{1}{2}$

x12=0x=12$x-\frac{1}{2}=0\\ \Rightarrow x=\frac{1}{2}$ Remainder(12)3+3(12)2+3(12)+1=18+34+32+1=278$∴\;Remainder (\frac{1}{2})^{3}+3(\frac{1}{2})^{2}+3(\frac{1}{2})+1\\ =\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1=\frac{27}{8}$

(iii) x

Remainder=(0)3+3(0)2+3(0)+1=1$∴\;Remainder= (0)^{3}+3(0)^{2}+3(0)+1\\ =1$

(iv) x+π$x+\pi$

x+π=0x=π$x+\pi=0\\ \Rightarrow x=-\pi$ Remainder=(π)3+3(π)2+3(π)+1=π3+3π23π+1$∴ \;Remainder= (-\pi)^{3}+3(-\pi)^{2}+3(-\pi)+1=-\pi ^{3}+3\pi ^{2}-3\pi +1$

(v) 5+2x

5+2x=02x=5x=52$5+2x=0\\ \Rightarrow 2x=-5\Rightarrow x=-\frac{5}{2}$ Remainder=(52)3+3(52)2+3(52)+1=1258+754152+1=278$∴\; Remainder=(-\frac{5}{2})^{3}+3(-\frac{5}{2})^{2}+3(-\frac{5}{2})+1=-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1=-\frac{27}{8}$

Q.2.Find the remainder when x3ax2+6xa$x^{3}-ax^{2}+6x-a$ is divided by x-a.

Let p(x)=x3ax2+6xaxa=0$p(x)=x^{3}-ax^{2}+6x-a\\ x-a=0$

x=a$∴ x=a$ Remainder=(a)2a(a2)+6(a)a=a3a3+6aa=5a$Remainder=(a)^{2}-a(a^{2})+6(a)-a\\ =a^{3}-a^{3}+6a-a\\ =5a$

Q.3.Check whether 7+3x is a factor of 3x3+7x$3x^{3}+7x$.

7+3x=03x=7$7+3x=0\Rightarrow 3x=-7$ only if 7+3x divides 3x3+7x$3x^{3}+7x$ leaving no reaminder.

Let p(x)=3x3+7x$p(x)=3x^{3}+7x$

7+3x=03x=7x=73$7+3x=0\Rightarrow 3x=-7\Rightarrow x=-\frac{7}{3}$ Remainder=3(73)3+7(73)=3439493=49090$∴ Remainder=3(-\frac{7}{3})^{3}+7(-\frac{7}{3})=-\frac{343}{9}-\frac{49}{3}=-\frac{490}{9}\neq 0$

$∴$7+3x is not a factor of 3x3+7x$3x^{3}+7x$

Exercise – 4

Q.1. Determine which of the following polynomials has (x + 1) a factor:

(i)
x3+x2+x+1$x^{3} + x^{2}+ x + 1$

Let p(x)= x3+x2+x+1$x^{3} + x^{2}+ x + 1$

The zero of x+1 is -1.

p(1)=(1)3+(1)2+(1)+1=1+11+1=0$p(-1)=(-1)^{3} + (-1)^{2}+ (-1) + 1\\ =-1+1-1+1=0$

$∴$By factor theorem, x+1 is a factor of x3+x2+x+1$x^{3} + x^{2}+ x + 1$

(ii) x4 + x3 + x2 + x + 1

Let p(x)= x4+x3+x2+x+1$x^{4} + x^{3}+x^{2} +x + 1$

The zero of x+1 is -1.

p(1)=(1)4+(1)3+(1)2+(1)+1=11+11+1=10$p(-1)=(-1)^{4} + (-1)^{3}+ (-1)^{2} + (-1)+1\\ =1-1+1-1+1=1\neq 0$ $∴$By factor theorem, x+1 is a factor of x4+x3+x2+x+1$x^{4} + x^{3}+x^{2} +x + 1$

(iii) x4 + 3x3 + 3x2 + x + 1

Let p(x)= x4+3x3+3x2+x+1$x^{4} +3 x^{3}+3x^{2} +x + 1$

The zero of x+1 is -1.

p(1)=(1)4+3(1)3+3(1)2+(1)+1=13+31+1=10$p(-1)=(-1)^{4} + 3(-1)^{3}+ 3(-1)^{2} + (-1)+1\\ =1-3+3-1+1=1\neq 0$ $∴$By factor theorem, x+1 is a factor of x4+3x3+3x2+x+1$x^{4} + 3x^{3}+3x^{2} +x + 1$

(iv) x3x2(2+2)x+2$x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}$

Let p(x)= x3x2(2+2)x+2$x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}$

The zero of x+1 is -1.

p(1)=(1)3(1)2(2+2)(1)+2$p(-1)=(-1)^{3} – (-1)^{2} – (2 + \sqrt{2})(-1) + \sqrt{2}$

$∴$By factor theorem, x+1 is not a factor of x3x2(2+2)x+2$x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}$

Q.2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i)
p(x)=2x3+x22x1$p(x) = 2x^{3} + x^{2} – 2x – 1$, g(x) = x + 1

Answer: p(x)=2x3+x22x1$p(x) = 2x^{3} + x^{2} – 2x – 1$, g(x) = x + 1

g(x)=0

x+1=0x=1$\Rightarrow x+1=0\Rightarrow x=-1$

$∴$Zero of g(x) is -1.

Now, p(1)=2(1)3+(1)22(1)1=2+1+21=0$p(-1) = 2(-1)^{3} + (-1)^{2} – 2(-1) – 1\\ =-2+1+2-1=0$

$∴$By factor theorem, g(x) is a factor of p(x).
(ii) p(x)=x3+3x2+3x+1$p(x) = x^{3}+ 3x^{2} + 3x + 1$, g(x) = x + 2

Answer: p(x)=x3+3x2+3x+1$p(x) = x^{3} + 3x^{2} +3x + 1$, g(x) = x + 2

g(x)=0

x+2=0x=2$\Rightarrow x+2=0\Rightarrow x=-2$

$∴$Zero of g(x) is -2.

Now, p(2)=(2)3+3(2)2+3(2)+1=8+126+1=1=0$p(-2) = (-2)^{3} + 3(-2)^{2} +3(-2) + 1\\ =-8+12-6+1=-1\neq=0$

$∴$By factor theorem, g(x) is not a factor of p(x).
(iii) p(x)=x34x2+x+6$p(x) = x^{3} – 4 x^{2} + x + 6$, g(x) = x – 3

Answer: p(x)=x34x2+x+6$p(x) = x^{3} – 4x^{2} +x + 6$, g(x) = x -3

g(x)=0

x3=0x=3$\Rightarrow x-3=0\Rightarrow x=3$

$∴$Zero of g(x) is 3.

Now, p(3)=(3)34(3)2+(3)+6=2736+3+6=0$p(3) = (3)^{3} -4(3)^{2} +(3) + 6\\ =27-36+3+6=0$

$∴$By factor theorem, g(x) is a factor of p(x).

Q.3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i)p(x)=x2+x+k$p(x) = x^{2}+ x + k$

Answer: If x-1 is a factor of p(x), then p(1)=0   |By Factor Theorem (1)2+(1)+k=01+1+k=0$\Rightarrow (1)^{2}+ (1) + k=0\Rightarrow 1+1+k=0$

2+k=0k=2$\Rightarrow2+k=0\Rightarrow k=-2$

(ii) p(x)=2x2+kx+2$p(x) = 2x^{2} + kx + \sqrt{2}$

Answer: If x-1 is a factor of p(x), then p(1)=0

2(1)2+k(1)+2=02+k+2=0k=(2+2)$\Rightarrow 2(1)^{2}+ k(1) +\sqrt{2} =0\Rightarrow 2+k+\sqrt{2}=0\Rightarrow k=-(2+\sqrt{2})$

(iii) p(x)=kx22x+1$p(x) = kx^{2} – \sqrt{2}x + 1$

Answer: If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

k(1)22(1)+1=0k=21$\Rightarrow k(1)^{2}-\sqrt{2}(1)+1=0\\ \Rightarrow k=\sqrt{2}-1$
(iv) p(x)=kx23x+k$p(x) = kx^{2} – 3x + k$

Answer: If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

k(1)23(1)+k=0k3+k=02k3=0k=32$\Rightarrow k(1)^{2} – 3(1) + k=0\Rightarrow k-3+k=0\\ \Rightarrow 2k-3=0\\ \Rightarrow k=\frac{3}{2}$

Q.4. Factorize:

(i) 12x27x+1$12x^{2} – 7x + 1$

12x27x+1=12x24x3x+1=4x(3x1)1(3x1)=(4x1)(3x1)$12x^{2} – 7x + 1=12x^{2} – 4x-3x + 1\\ =4x(3x-1)-1(3x-1) =(4x-1)(3x-1)$

Let p(x)= 12x27x+1$12x^{2} – 7x + 1$

Then p(x)= 12(x2712x+112)=12q(x)$12(x^{2} – \frac{7}{12}x + \frac{1}{12})=12q(x)$

Where q(x)= x2712x+112$x^{2} – \frac{7}{12}x + \frac{1}{12}$

By trial, we find that

q(13)=(13)2712(13)+112=47+336=036=0$q(\frac{1}{3})=(\frac{1}{3})^{2} – \frac{7}{12}(\frac{1}{3}) + \frac{1}{12}\\ =\frac{4-7+3}{36}=\frac{0}{36} =0$

$∴$By Factor Theorem,

(x13)$(x-\frac{1}{3})$ is a factor of q(x).

Similarly, by trial, we find that

q(14)=(14)2712(14)+112=37+418=048=0$q(\frac{1}{4})=(\frac{1}{4})^{2} – \frac{7}{12}(\frac{1}{4}) + \frac{1}{12}\\ =\frac{3-7+4}{18}=\frac{0}{48} =0$

$∴$By Factor Theorem,

(x14)$(x-\frac{1}{4})$ is a factor of q(x).

Therefore, 12x27x+1=12(x13)(x14)=12(3x13)(4x14)=(3x1)(4x1)$12x^{2}-7x+1=12(x-\frac{1}{3})(x-\frac{1}{4})\\ =12(\frac{3x-1}{3})(\frac{4x-1}{4})=(3x-1)(4x-1)$

(ii)2x2+7x+3$2x^{2} + 7x + 3$

2x2+7x+3=2x2+6x+x+3=2x(x+3)+1(x+3)=(x+3)(2x+1)$2x^{2} + 7x + 3=2x^{2} + 6x+x + 3\\ =2x(x+3) +1 (x + 3)\\ =(x+3)(2x+1)$

Let p(x)= 2x2+7x+3$2x^{2} + 7x + 3$

Then p(x)= 2(x2+72x+32)=2q(x)$2(x^{2} + \frac{7}2{}x + \frac{3}{2})=2q(x)$
Where q(x)= x2+72x+32$x^{2} + \frac{7}2{}x + \frac{3}{2}$

By trial, we find that

q(3)=(3)272(3)+32=9212+32=0$q(-3)=(-3)^{2} – \frac{7}{2}(-3) + \frac{3}{2}\\ =9-\frac{21}{2}+\frac{3}{2}=0$

$∴$By Factor Theorem,

(x3),i.e.(x+3)$(x-3),i.e.(x+3)$ is a factor of q(x).

Similarly, by trial, we find that

q(12)=(12)2+712(12)+32=1374+34=0$q(-\frac{1}{2})=(-\frac{1}{2})^{2} +\frac{7}{12}(-\frac{1}{2}) + \frac{3}{2}\\ =\frac{1}{3}-\frac{7}{4}+\frac{3}{4} =0$

$∴$By Factor Theorem,

(x(12)),i.e.(x+12)isafactorofq(x).Therefore,2x2+7x+3=2(x+3)(x+12)=2(x+3)(2x+12)=(x+3)(2x+1)$(x-(-\frac{1}{2})),i.e.(x+\frac{1}{2}) \;is\; a\; factor\; of\; q(x).\\ Therefore, 2x^{2}+7x+3=2(x+3)(x+\frac{1}{2})\\ =2(x+3)(\frac{2x+1}{2})\\ =(x+3)(2x+1)$

By Simpler method,

(iii) $$6x^{2}+5x-6 =6x^{2}+ 9x – 4x – 6 =3x (2x + 3) – 2 (2x + 3) = (2x + 3) (3x – 2)$$

(iv)3x2 – x – 4 =3x2x4=3×24x+3x4=x(3x4)+1(3x4)=(3x4)(x+1)$3x^{2} – x – 4 \\ =3×2 – 4x + 3x – 4 \\ = x (3x – 4) + 1 (3x – 4)\\ = (3x – 4) (x + 1)$

Q.5. Factorize:

(i)
x32x2x+2$x^{3} – 2x^{2} – x + 2$

Answer: Let p(x)=x32x2x+2$p(x) = x^{3} – 2x^{2} – x + 2$
Factors of 2 are ±1 and ± 2
By trial method, we find that
p(1) = 0
So, (x+1) is factor of p(x)
Now,
p(x)=x32x2x+2$p(x) = x^{3} – 2x^{2} – x + 2$

p(1)=(1)32(1)2(1)+2=11+1+2=0$p(-1) = (-1)^{3} – 2(-1)^{2} – (-1) + 2=-1-1+1+2=0$

Therefore, (x+1) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x23x+2)=(x+1)(x2x2x+2)=(x+1)(x(x1)2(x1))=(x+1)(x1)(x+2)$(x+1) (x^{2} – 3x + 2)\\ = (x+1) (x^{2} – x – 2x + 2)\\ = (x+1) (x(x-1) -2(x-1))\\ = (x+1) (x-1) (x+2)$

(ii) x33x29x5$x^{3} – 3x^{2} – 9x – 5$
Answer: Let p(x) = x33x29x5$x^{3} – 3x^{2} – 9x – 5$
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = x33x29x5$x^{3} – 3x^{2} – 9x – 5$

p(5) = (5)33(5)29(5)5=12575455=0$(5)^{3} – 3(5)^{2} – 9(5) – 5=125-75-45-5=0$

Therefore, (x-5) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x5)(x2+2x+1)=(x5)(x2+x+x+1)=(x5)(x(x+1)+1(x+1))=(x5)(x+1)(x+1)$(x-5) (x^{2} + 2x + 1)\\ = (x-5) (x^{2} + x + x + 1)\\ = (x-5) (x(x+1) +1(x+1))\\ = (x-5) (x+1) (x+1)$

(iii) x3+13x2+32x+20$x^{3} + 13x^{2} + 32x + 20$

Answer: Let p(x) = x3+13x2+32x+20$x^{3} + 13x^{2} + 32x + 20$

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) = x3+13x2+32x+20$x^{3} + 13x^{2} + 32x + 20$

p(-1) = (1)3+13(1)2+32(1)+20=1+1332+20=0$(-1)^{3} + 13(-1)^{2} + 32(-1) + 20=-1+13-32+20=0$

Therefore, (x+1) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x2+12x+20)=(x+1)(x2+2x+10x+20)=(x5)x(x+2)+10(x+2)=(x5)(x+2)(x+10)$(x+1) (x^{2} + 12x + 20)\\ = (x+1) (x^{2} + 2x + 10x + 20)\\ = (x-5) {x(x+2) +10(x+2)}\\ = (x-5) (x+2) (x+10)$

(iv) 2y3+y22y1$2y^{3} + y^{2} – 2y – 1$
Answer: Let p(y) = 2y3+y22y1$2y^{3} + y^{2} – 2y – 1$

Factors of ab = 2×(1)$2\times (-1)$= -2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) = 2y3+y22y1$2y^{3} + y^{2} – 2y – 1$

p(1) = 2(1)3+(1)22(1)1=2+12=0$2(1)^{3} + (1)^{2} – 2(1) – 1=2+1-2=0$

Therefore, (y-1) is the factor of  p(y)

Now, Dividend = Divisor × Quotient + Remainder

(y1)(2y2+3y+1)=(y1)(2y2+2y+y+1)=(y1)(2y(y+1)+1(y+1))=(y1)(2y+1)(y+1)$(y-1) (2y^{2} + 3y + 1)\\ = (y-1) (2y^{2} + 2y + y + 1)\\ = (y-1) (2y(y+1) +1(y+1))\\ = (y-1) (2y+1) (y+1)$

Exercise – 5

Q.1.Use suitable identities to find the following products:

(i)(x + 4) (x + 10)

Answer: (x+4)(x+10)=x2+(4+10)x+(4×10)=x2+14x+40$(x + 4) (x + 10) = x^{2} + (4 + 10)x + (4 \times 10)\\ = x^{2} + 14x + 40$
(ii)(x + 8) (x – 10)

Answer: (x+8)(x10)=x2+(8+(10))x+(8×(10))=x2+(810)x80=x22x80$(x + 8) (x -10 ) = x^{2} + (8+(- 10))x + (8\times(-10))\\ = x^{2} +(8-10)x – 80\\ =x^{2}-2x-80$

(iii)(3x + 4) (3x – 5)

Answer: (3x+4)(3x5)=(3x)2+4+(5)3x+4×(5)=9x2+3x(45)20=9x23x20$(3x + 4)(3x -5) = (3x) ^{2} + {4 + (-5)}3x + {4×(-5)}\\ = 9x^{2} + 3x(4 – 5) – 20\\ = 9x^{2} – 3x – 20$
(iv)(y2+32)(y232)$(y^{2} + \frac{3}{2}) (y^{2} – \frac{3}{2})$

Answer: (y2+32)(y232)=(y2)2(32)2=y494$(y^{2} + \frac{3}{2}) (y^{2} – \frac{3}{2}) = (y^{2})^{2} – (\frac{3}{2})^{2}\\ = y^{4} – \frac{9}{4}$

Q.2. Evaluate the following products without multiplying directly:

(i)103 × 107

Answer:   103×107=(100+3)×(100+7)=(100)2+(3+7)(100+(3×7))=10000+1000+21=11021$103 \times 107 = (100 + 3)\times (100 + 7)\\ = (100)^{2} + (3 + 7)(100 + (3 \times 7))\\ = 10000 + 1000 + 21 \\ = 11021$

(ii) 95 × 96

Answer: 95×96=(90+5)×(90+6)=(90)2+90(5+6)+(5×6)=8100+990+30=9120$95 \times 96 = (90 + 5) \times (90 + 6)\\ = (90)^{2} + 90(5 + 6) + (5 \times 6) \\ = 8100 + 990 + 30 = 9120$

(iii) 104 × 96
Answer: 104×96=(100+4)×(1004)=(100)2(4)2=1000016=9984$104 \times 96 = (100 + 4) \times (100 – 4)\\ = (100)^{2} – (4)^{2}\\ = 10000 – 16 \\ = 9984$

Q.3. Factorize the following using appropriate identities:

(i)9x2+6xy+y2$9x^{2} + 6xy + y^{2}$

Answer: 9x2+6xy+y2=(3x)2+(2×3x×y)+y2=(3x+y)2=(3x+y)(3x+y)$9x^{2} + 6xy + y^{2} \\ = (3x) ^{2} + (2\times 3x\times y) + y^{2}\\ = (3x + y)^{2} \\ = (3x + y) (3x + y)$

(ii) 4y24y+1$4y^{2}- 4y + 1$

Answer: 4y24y+1=(2y)2(2×2y×1)+12=(2y1)2=(2y1)(2y1)$4y^{2}- 4y + 1 \\ =(2y)^{2} – (2\times 2y\times 1) + 1^{2}\\ = (2y – 1)^{2}\\ = (2y – 1) (2y – 1)$

(iii) x2y2100$x2 – \frac{y^{2}}{100}$

Answer: x2(y10)2=(xy10)(x+y10)$x^{2} – (\frac{y}{10})^{2} = (x – \frac{y}{10}) (x + \frac{y}{10})$

Q.4. Expand each of the following, using suitable identities:

(i) (x+2y+4z)2$(x + 2y + 4z)^{2}$

(ii) (2xy+z)2$(2x- y + z)^{2}$

(iii) (2x+3y+2z)2$(-2x+3y+2z)^{2}$

(iv) [14a12b+1]2$[\frac{1 }{4}a – \frac{1}{2} b + 1]^{2}$

(i) (x+2y+4z)2$(x + 2y + 4z)^{2}$
(x+2y+4z)2=x2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)=x2+4y2+16z2+4xy+16yz+8xz$(x + 2y + 4z)^{2} \\ = x^{2} + (2y)^{2} + (4z)^{2} + (2\times x\times 2y) + (2\times 2y\times 4z) + (2\times 4z\times x)\\ = x^{2} + 4y^{2} + 16z2 + 4xy + 16yz + 8xz$
(ii)  (2xy+z)2$(2x- y + z)^{2}$