SChand CLASS 9 Chapter 4 Factorisation Exercise 4(E)

 Exercise 4(E)

Question 1

7x² – 7

Sol :

$=7 x^{2}-7$

$=7\left(x^{2}-1\right)$

$=7\left[(x)^{2}-(1)^{2}\right]$

$=7(x+1)(x-1)$

Question 2

8 – 50y²z²

Sol :

$=8-50 y^{2} z^{2}$

$=2\left(4-25 y^{2} z^{2}\right)$

$=2\left[(2)^{2}-(5 y z)^{2}\right]$

$=2(2-5 y z)\left(2+5 y^{2}\right)$

Question 3

ab² – ac²

Sol :

$=a b^{2}-a c^{2}$

$=a\left[(b)^{2}-(c)^{2}\right]$

=a(b-c)(b+c)

Question 4

36x³ – x

Sol :

$=36 x^{3}-x$

$=x\left[(6 x)^{2}-(2)^{2}\right]$

=x(6 x-1)(6 x+1)

Question 5

x (x² – 1) + 7 (x² – 1)

Sol :

$x\left(x^{2}-1\right)+7\left(x^{2}-1\right)$

$=\left(x^{2}-1\right)\left(x^{2}+7\right)$

$=\left[(x)^{2}-(1)^{2}\right](x+7)$

=(x+1)(x-1)(x+7)

Question 6

t² (t – 3)² – (t – 3)²

Sol :

$\begin{aligned} & t^{2}(t-3)^{2}-(t-3)^{2} \\=&(t-3)^{2}\left(t^{2}-1\right) \\=&(t-3)^{2}\left[(t)^{2}-(1)^{2}\right] \\=&(t-3)^{2}(t+1)(t-1) \end{aligned}$

Question 7

5c² (c + 2)² – 45 (c + 2)²

Sol :

$=5 c^{2}(c+2)^{2}-45(c+2)^{2}$

$=5(c+2)^{2}\left[c^{2}-9\right]$

$=5(c+2)^{2}\left[(c)^{2}-(3)^{2}\right]$

$=5(c+2)^{2}(c+3)(c-3)$

Question 8

(a + b)² – 1

Sol :

$=(a+b)^{2}-1$

$=(a+b)^{2}-(1)^{2}$

=(a+b+h)(a+b-1)

Question 9

1 – (x – y)²

Sol :

$\begin{aligned} =&1-(x-y)^{2} \\=&(1)^{2}-(x-y)^{2} \\=&(1+x-y)(1-x-y) \end{aligned}$

Question 10

25a² – 16 (x – y)²

Sol :

$\begin{aligned} =& 25 a^{2}-16(x-y)^{2} \\=&(5 a)^{2}-[4(x-y)]^{2} \\=&[5 a+4(x-y)][5 a-4(x-y)] \end{aligned}$

Question 11

20 – 45 (m + n)²

Sol :

$\begin{aligned} =& 20-4 s(m+n)^{2} \\=& 5\left[4-9(m+n)^{2}\right] \\=& 5\left[(2)^{2}-(3(m+n))^{2}\right] \\=&5[2+3(m+n)][2-3(m+n)] \end{aligned}$

Question 12

$x^{4}-y^{4}$

Sol :

$=x^{4}-y^{4}$

$\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}$

$\left(x^{2}+y^{2}\right) \cdot\left(x^{2}-y^{2}\right)$

Question 13

$x^{4}-625$

Sol :

$=x^{4}-625$

$=\left(x^{2}\right)^{2}-(25)^{2}$

$=\left(x^{2}+25\right)^{2}\left(2 x^{2}-25\right)$

$=\left(x^{2}+25\right)\left[(x)^{2}-(5)^{2}\right]$

$=\left(x^{2}+25\right)(x+5)(x-5)$

Question 14

$xy^5 – yx^5$

Sol :

$=x y^{5}-y x^{5}$

$=x y\left[y^{4}-x^{4}\right]$

$=x y\left[\left(y^{2}\right)^{2}-\left(x^{2}\right)^{2}\right]$

$=x y\left(y^{2}-x^{2}\right)\left(y^{2}+x^{2}\right)$

$=x y\left[(y)^{2}-\left(x^{2}\right)\right]\left(y^{2}+x^{2}\right)$

$=x y(y+x)(y-x)\left(y^{2}+x^{2}\right)$

Question 15

 $81 x^{4}-2.56 y^{4}$

Sol :

$\begin{aligned} =& 81 x^{4}-2.56 y^{4} \\=&\left(9 x^{2}\right)^{2}-\left(16 y^{2}\right)^{2} \\=&\left(9 x^{2}-16 y^{2}\right)\left(9 x^{2}+16 y^{2}\right) \\=&\left[(3 x)^{2}-(4 y)^{2}\right]\left(9 x^{2}+16 y^{2}\right) \end{aligned}$

$=(3 x+4 y)(3 x-4 y)\left(9 x^{2}+16 y^{2}\right)$

Question 16

a² + ac + bc – b²

Sol :

$\begin{aligned}=&(a)^{2}-(b)^{2}+a c+b c \end{aligned}$

=(a+b)(a-b)+c(a+b)

=(a+b)(a-b+c)

Question 17

4a² – b² + 2a + b

Sol :

$\begin{aligned} =& 4 a^{2}-b^{2}+2 a+b \\=&(2 a)^{2}-(b)^{2}+2 a+b \end{aligned}$

=(2 a+b)(2 a-b)+1(2 a+b)

=(2 a+b)(2 a-b+1)

Question 18

x² + 3x – y² – 3y

Sol :

$\begin{aligned} & x^{2}+8 x-y^{2}-3 y \\=& x^{2}-y^{2}+3 x-3 y \end{aligned}$

=(x+y)(x-y)+3(x-y)

=(x+y)(x+y+3)

Question 19

a² + b² – 2ab – 4c²

Sol :

$=a^{2}+b^{2}-2 a b-4 c^{2}$

$=(a-b)^{2}-(2 c)^{2}$

=(a-b+2c)(a-b-2c)

Question 20

9x² – 6xy + y² – z²

Sol :

$\begin{aligned} =& 9 x^{2}-6 x y+y^{2}-z^{2} \\=&\left[(3 x)^{2}-2 \times 3 x \times y+\left(y\right)^{2}\right]-(2)^{2} \\=&(3 x-y)^{2}-(2)^{2} \end{aligned}$

Question 21

x² – 1 – 2a – a²

Sol :

$\begin{aligned} =& x^{2}-1-2 a-a^{2} \\=& x^{2}-\left(1+2 a+a^{2}\right) \\=&(x)^{2}-(1+a)^{2} \\=&(x+1+a)(x-1-a) \end{aligned}$

Question 22

4a² + b² – c² + 4ab

Sol :

$\begin{aligned} & 4 a^{2}+b^{2}-c^{2}+4 a b \\=& 4 a^{2}+b^{2}+4 a b-c^{2} \\=&(2 a)^{2}+\left(b^{2}\right)+2 \times 2 a \times b-(c)^{2} \\=&(2 a+b)^{2}-(c)^{2} \\=&(2 a+b+c)(2 a+b-c) \end{aligned}$

Question 23

x³ + 2x² – x – 2

Sol :

$\begin{aligned} =& x^{3}+2 x^{2}-x-2 \\=& x^{2}(x+2)-1(x+2) \\=&(x+2)\left(x^{2}-1\right) \\=&(x+2)\left[(x)^{2}-(1)^{2}\right] \\=&(x+2)(x+2)(x-1) \end{aligned}$

Question 24

1 + 2ab – (a² + b²)

Sol :

$=x+2 a b-\left(a^{2}+b^{2}\right)$

$=1+2 a b \cdot a^{2}-b^{2}$

$=1-\left(a^{2}+b^{2}-2 a b\right)$

$=(1)^{2}-(a-b)^{2}$

$=(1+a-b)(1-a+b)$

Question 25

$x^{2}+\frac{1}{x^{2}}-11 $

Sol :

$\begin{aligned} =& x^{2}+\frac{1}{x^{2}}-2-9 \\=&(x)^{2}-2+\frac{1}{(x)^{2}}-(3)^{2} \\=&\left(x-\frac{1}{x}\right)^{2}-(3)^{2} \\=&\left(x-\frac{1}{x}+3\right)\left(x-\frac{1}{x}-3\right) \end{aligned}$

Question 26

$x^{4}+3 x^{2}+4$

Sol :

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

$=x^{4}+4 x^{2}+4-x^{2}$

$=\left(x^{2}\right)^{2}+2 \times x^{2} \times 2+(2)^{2}-(x)^{2}$

$=\left(x^{2}+2\right)^{2}-(x)^{2}$

$=\left(x^{2}+2 \cdot x\right)\left(x^{2}+2+x\right)$

Question 27

Factorise the following :

(i) 4a² – b² + 2a + b

(ii) 9x² – 4 (y + 2x)²

(iii) 9 (x + y)² – x²

Sol :

(i) $\begin{aligned} & 4 a^{2}-b^{2}+2 a+b \\=&(2 a)^{2}-(b)^{2}+2 a+b \end{aligned}$

=(2a+b)(2a-b)+1(2a+b)

=(2a+b)(2a-b+1)

(ii) $\begin{aligned} & 9 x^{2}-4(y-2 x)^{2} \\=& 9 x^{2}-[2(y+2 x)]^{2} \\=&(3 x)^{2}-[2(y+2 x)]^{2} \end{aligned}$

=[3x+2(y+2x)][3x-2(y+2x)]

=(3x+2 y+4x)(3x-2y-4x)

=(7x+2y)(-x-2y)

=-(7x+2y)(x+2y)

(iii) $\begin{aligned} & 9(x+y)^{2}-x^{2} \\=&[3(x+y)]^{2}-x^{2}\end{aligned}$

=[3(x+y)+x][3(x+y)-x]

=(3x+3y+x(3x+3y-x)

=(4x+3y)(2x+3y)

Question 28

Factorise : x³ – 3x² – x + 3

Sol :

$=x^{3}-3 x^{2}-x+3$

$=x^{2}(x-3)-1(x-3)$

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

$=(x-3)\left[(x)^{2}-(1)^{2}\right]$

=(x-3)(x+1)(x-1)

Question 29

Factorise :

(a² – b²) (c² – d²) – 4abcd

Sol :

$\begin{aligned} =&\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)-4 a b c d \\=& a^{2} c^{2}-a^{2} d^{2}-b^{2} c^{2}+b^{2} d^{2}-4 a b c d \\=& a^{2} c^{2}+b^{2} d^{2}-2 a b c d-b^{2} c^{2}-2 a b c d-a^{2} d^{2} \end{aligned}$

$=(a c-b d)^{2}-(a d+b c)^{2}$

=(ac-bd+ad+bc)(ac-bd-ad-bc)

=[ac+ad-bc-bd][ac-ad-bc-bd]

=[a(c+d)-b(c-d)][a(c-d)-b(c+d)]

Question 30

Express (x² + 8x + 15) (x² – 8x + 15) as a difference of two squares.

Sol :

$=\left(x^{2}+8 x+15\right)\left(x^{2}-8 x+15\right)$

$=\left[\left(x^{2}+15\right)+8 x\right]\left[\left(x^{2}+15\right)-8 x\right]$

$=\left(x^{2}+15\right)^{2}-(8 x)^{2}$