# Ncert Solutions For Class 9 Maths Ex 2.4

## Ncert Solutions For Class 9 Maths Chapter 2 Ex 2.4

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