Frank Solutions for Class 9 Maths Chapter 5 Factorisation

Frank Solutions for Class 9 Maths Chapter 5 Factorisation help students with detailed answers prepared by experts, having vast experience in the education industry. These solutions in a simple language help students prepare comprehensively for their final examinations. Our topmost experts have designed the answers in such a way that students clear their doubts instantly. In order to score more marks, students can make use of Frank Solutions for Class 9 Maths Chapter 5 Factorisation PDF, from the links which are given below.

Chapter 5 deals with the study of Factorisation. When a number is split into factors or divisors, then it is known as factorisation. This chapter plays a vital role in Mathematics as it is continued in further classes. The solutions help students to speed up their problem solving and time management skills, which are essential to ace the exam.

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Access Frank Solutions for Class 9 Maths Chapter 5 Factorisation

1. Factorise the following by taking out the common factors:

(a) 4x2y3 – 6x3y2 – 12xy2

(b) 5a (x2 – y2) + 35b (x2 – y2)

(c) 2x5y + 8x3y2 – 12x2y3

(d) 12a3 + 15a2b – 21ab2

(e) 24m4n6 + 56m6n4 – 72m2n2

(f) (a – b)2 – 2(a – b)

(g) 2a (p2 + q2) + 4b (p2 + q2)

(h) 81 (p + q)2 – 9p – 9q

(i) (mx + ny)2 + (nx – my)2

(j) 36 (x + y)3 – 54 (x + y)2

(k) p (p2 + q2 – r2) + q (r2 – q2 – p2) – r (p2 + q2 – r2)

Solution:

(a) 4x2y3 – 6x3y2 – 12xy2

Here,

The common factor is 2xy2

Dividing throughout by 2xy2

We get,

{(4x2y3) / (2xy2)} – {(6x3y2) / (2xy2)} – {(12xy2) / (2xy2)}

= 2xy – 3x2 – 6

Hence,

4x2y3 – 6x3y2 – 12xy2 = 2xy2 (2xy – 3x2 – 6)

(b) 5a (x2 – y2) + 35b (x2 – y2)

Here,

The common factor is 5 (x2 – y2)

Dividing throughout by 5(x2 – y2),

We get,

{5a (x2 – y2) / 5 (x2 – y2)} + {35b (x2 – y2) / 5 (x2 – y2)}

= a + 7b

Hence,

5a (x2 – y2) + 35b (x2 – y2) = 5 (x2 – y2) (a + 7b)

(c) 2x5y + 8x3y2 – 12x2y3

Here,

The common factor is 2x2y

Dividing throughout by 2x2y, We get,

{(2x5y) / (2x2y)} + {8x3y2) / (2x2y)} – {(12x2y3) / (2x2y)}

= x3+ 4xy – 6y2

Hence,

2x5y + 8x3y2 – 12x2y3 = 2x2y (x3 + 4xy – 6y2)

(d) 12a3 + 15a2b – 21ab2

Here,

The common factor is 3a

Dividing throughout by 3a,

We get,

{(12a3) / (3a)} + {(15a2b) / (3a)} – {(21ab2) / (3a)}

= 4a2 + 5ab – 7b2

Hence,

12a3 + 15a2b – 21ab2 = 3a (4a2 + 5ab – 7b2)

(e) 24m4n6 + 56m6n4 – 72m2n2

Here,

The common factor is 8m2n2

Dividing throughout by 3a,

We get,

{(24m4n6) / (8m2n2)} + {(56m6n4) / (8m2n2)} – {(72m2n2) / (8m2n2)}

= 3m2n4 + 7m4n2 – 9

Hence,

24m4n6 + 56m6n4 – 72m2n2 = 8m2n2 (3m2n4 + 7m4n2 – 9)

(f) (a – b)2 – 2(a – b)

Here,

The common factor is (a – b)

Dividing throughout by (a – b),

We get,

{(a – b)2 / (a – b)} – {2 (a – b) / (a – b)}

= a – b – 2

Hence,

(a – b)2 – 2(a – b) = (a – b) (a – b – 2)

(g) 2a (p2 + q2) + 4b (p2 + q2)

Here,

The common factor is 2(p2 + q2),

Dividing throughout by 2(p2 + q2),

We get,

{2a (p2 + q2) / 2 (p2 + q2)} + {4b (p2 + q2) / 2 (p2 + q2)}

= a + 2b

Hence,

2a (p2 + q2) + 4b (p2 + q2) = 2 (p2 + q2) (a + 2b)

(h) 81 (p + q)2 – 9p – 9q

= 81 (p + q)2 – 9 (p + q)

Here,

The common factor is 9(p + q)

Dividing throughout by 9(p + q),

We get,

{81 (p + q)2 / 9 (p + q)} – {9 (p + q) / 9 (p + q)}

= 9 (p + q) – 1

Hence,

81 (p + q)2 – 9p – 9q = 9 (p + q) {9 (p + q) – 1}

(i) (mx + ny)2 + (nx – my)2

= m2x2 + n2y2 + 2mnxy + n2x2 + m2y2 – 2mnxy

= m2x2 + n2y2 + n2x2 + m2y2

= m2x2 + n2x2 + m2y2 + n2y2

= x2 (m2 + n2) + y2 (m2 + n2)

Here,

The common factor is (m2 + n2)

Dividing throughout by (m2 + n2),

We get,

{x2 (m2 + n2) / (m2 + n2)} + {y2 (m2 + n2) / (m2 + n2)}

= x2 + y2

Hence,

(mx + ny)2 + (nx – my)2 = (m2 + n2) (x2 + y2)

(j) 36 (x + y)3 – 54 (x + y)2

Here,

The common factor is 18 (x + y)2

Dividing throughout by 18 (x + y)2,

We get,

{36 (x + y)3 / 18 (x + y)2} – {54 (x + y)2 / 18 (x + y)2}

= 2 (x + y) – 3

Hence,

36 (x + y)3 – 54 (x + y)2 = 18 (x + y)2 {2 (x + y) – 3}

(k) p (p2 + q2 – r2) + q (r2 – q2 – p2) – r (p2 + q2 – r2)

= p (p2 + q2 – r2) – q (-r2 + q2 + p2) – r (p2 + q2 – r2)

= p (p2 + q2 – r2) – q (p2 + q2 – r2) – r (p2 + q2 – r2)

Here,

The common factor is (p2 + q2 – r2)

Dividing throughout by (p2 + q2 – r2),

We get,

{p (p2 + q2 – r2) / (p2 + q2 – r2)} – {q (p2 + q2 – r2) / (p2 + q2 – r2)} – {r (p2 + q2 – r2) / (p2 + q2 – r2)}

= p – q – r

Hence,

p (p2 + q2 – r2) + q (r2 – q2 – p2) – r (p2 + q2 – r2) = (p2 + q2 – r2) (p – q – r)

2. Factorise the following by grouping the terms:

(a) 15xy – 9x – 25y + 15

(b) 15x2 + 7y – 3x – 35xy

(c) 9 + 3xy + x2y + 3x

(d) 8 (2a + b)2 – 8a – 4b

(e) x (x – 4) – x + 4

(f) 2m3 – 5n2 – 5m2n + 2mn

(g) 4abx2 + 49aby2 + 14xy (a2 + b2)

(h) 9x3 + 6x2y2 – 4y3 – 6xy

(i) 3ax2 – 5bx2 + 9az2 + 6ay2 – 10by2 – 15bz2

(j) 8x3 – 24x2y + 54xy2 – 162y3

(k) 2a + b + 3c – d + (2a + b)3 + (2a + b)2 (3c – d)

(l) xy (a2 + 1) + a (x2 + y2)

(m) p2x2 + (px2 + 1)x + p

(n) x2 – (p + q)x + pq

(0) p2 + (1 / p2) – 2 – 5p + (5 / p)

(p) x + y + m (x + y)

(q) (1 / 25x2) + 16x2 + (8 / 5) – 12x – (3 / 5x)

(r) 2p (a2 – 2b2) – 14p + (a2 – 2b2)2 – 7 (a2 – 2b2)

Solution:

(a) 15xy – 9x – 25y + 15

On grouping the terms, we get,

= (15xy – 9x) – (25y + 15)

= 3x (5y – 3) – 5 (5y – 3)

We get,

= (5y – 3) (3x – 5)

(b) 15x2 + 7y – 3x – 35xy

= 15x2 – 3x – 35xy + 7y

On grouping the terms, we get,

= (15x2 – 3x) – (35xy – 7y)

= 3x (5x – 1) – 7y (5x – 1)

We get,

= (5x – 1) (3x – 7y)

(c) 9 + 3xy + x2y + 3x

= 9 + 3xy + 3x + x2y

On grouping the terms, we get,

= (9 + 3xy) + (3x + x2y)

= 3 (3 + xy) + x (3 + xy)

We get,

= (3 + xy) (3 + x)

(d) 8 (2a + b)2 – 8a – 4b

On grouping the terms, we get,

= 8 (2a + b)2 – (8a + 4b)

= 8 (2a + b)2 – 4 (2a + b)

= 4 (2a + b) {2 (2a + b) – 1}

We get,

= 4 (2a + b) (4a + 2b – 1)

(e) x (x – 4) – x + 4

On grouping the terms, we get,

= x (x – 4) – 1 (x – 4)

= (x – 4) (x – 1)

(f) 2m3 – 5n2 – 5m2n + 2mn

= 2m3 + 2mn – 5m2n – 5n2

On grouping the terms, we get,

= (2m3 + 2mn) – (5m2n + 5n2)

= 2m (m2 + n) – 5n (m2 + n)

We get,

= (m2 + n) (2m – 5n)

(g) 4abx2 + 49aby2 + 14xy (a2 + b2)

= 4abx2 + 49aby2 + 14a2xy + 14b2xy

On grouping the terms, we get,

= (4abx2 + 14a2xy) + (14b2xy + 49aby2)

= 2ax (2bx + 7ay) + 7by (2bx + 7ay)

We get,

= (2bx + 7ay) (2ax + 7by)

(h) 9x3 + 6x2y2 – 4y3 – 6xy

= 9x3 + 6x2y2 – 6xy – 4y3

On grouping the terms, we get,

= (9x3 + 6x2y2) – (6xy + 4y3)

= 3x2 (3x + 2y2) – 2y (3x + 2y2)

We get,

= (3x + 2y2) (3x2 – 2y)

(i) 3ax2 – 5bx2 + 9az2 + 6ay2 – 10by2 – 15bz2

= 3ax2 + 6ay2 + 9az2 – 5bx2 – 10by2 – 15bz2

On grouping the terms, we get,

= (3ax2 + 6ay2 + 9az2) – (5bx2 + 10by2 + 15bz2)

= 3a (x2 + 2y2 + 3z2) – 5b (x2 + 2y2 + 3z2)

We get,

= (x2 + 2y2 + 3z2) (3a – 5b)

(j) 8x3 – 24x2y + 54xy2 – 162y3

On grouping the terms, we get,

= (8x3 – 24x2y) + (54xy2 – 162y3)

= 8x2 (x – 3y) + 54y2 (x – 3y)

We get,

= (x – 3y) (8x2 + 54y2)

(k) 2a + b + 3c – d + (2a + b)3 + (2a + b)2 (3c – d)

On grouping the terms, we get,

= (2a + b + 3c – d) + {(2a + b)3 + (2a + b)2 (3c – d)}

= 1 (2a + b + 3c – d) + (2a + b)2 {(2a + b + 3c – d)}

We get,

= (2a + b + 3c – d) {1 + (2a + b)2}

(l) xy (a2 + 1) + a (x2 + y2)

= a2xy + xy + ax2 + ay2

On grouping the terms, we get,

= (a2xy + ax2) + (ay2 + xy)

= ax (ay + x) + y (ay + x)

We get,

= (ay + x) (ax + y)

(m) p2x2 + (px2 + 1)x + p

= p2x2 + px3 + x + p

On grouping the terms, we get,

= (p2x2 + px3) + (p + x)

= px2 (p + x) + 1 (p + x)

We get,

= (p + x) (px2 + 1)

(n) x2 – (p + q)x + pq

= x2 – px – qx + pq

On grouping the terms, we get,

= (x2 – px) – (qx + pq)

= x (x – p) – q(x – p)

We get,

= (x – p) (x – q)

(0) p2 + (1 / p2) – 2 – 5p + (5 / p)

= {p2 + (1 / p2) – 2} – {5p – (5 / p)}

= {(p)2 + (1 / p)2 – 2 (p) (1 / p)} – {5p – (5 / p)}

= {p – (1 / p)}2 – 5 {p – (1 / p)}

We get,

= {p – (1 / p)} {p – (1 / p) – 5}

(p) x + y + m (x + y)

On grouping the terms, we get,

= (x + y) + m (x + y)

= (x + y) (1 + m)

(q) (1 / 25x2) + 16x2 + (8 / 5) – 12x – (3 / 5x)

On grouping the terms, we get,

= {(1 / 25x2) + 16x2 + (8 / 5)} – {12x + (3 / 5x)}

= {(1 / 5x)2 + (4x)2 + (2) (1 / 5x) (4x)} – {12x + (3 / 5x)}

= {(1 / 5x) + 4x}2 – 3 {4x + (1 / 5x)}

= {(1 / 5x) + 4x}2 – 3{(1 / 5x) + 4x}

We get,

= {(1 / 5x) + 4x) {(1 / 5x) + 4x – 3}

(r) 2p (a2 – 2b2) – 14p + (a2 – 2b2)2 – 7 (a2 – 2b2)

= 2p (a2 – 2b2) + (a2 – 2b2)2 – 14p – 7 (a2 – 2b2)

On grouping the term, we get,

= {2p (a2 – 2b2) + (a2 – 2b2)2 } – {14p + 7 (a2 – 2b2)}

= (a2 – 2b2) (2p + a2 – 2b2) – 7 (2p + a2 – 2b2)

We get,

= (2p + a2 – 2b2) (a2 – 2b2 – 7)

3. Factorise the following by splitting the middle term:

(a) x2 + 6x + 8

(b) x2 – 11x + 24

(c) x2 + 5x – 6

(d) p2 – 12p – 64

(e) y2 – 2y – 24

(f) 3x2 + 19x – 14

(g) 15a2 – 14a – 16

(h) 12 + x – 6x2

(i) 7x2 + 40x – 12

Solution:

(a) x2 + 6x + 8

By splitting the middle term, we get,

= x2 + 4x + 2x + 8

= x (x + 4) + 2 (x + 4)

We get,

= (x + 4) (x + 2)

(b) x2 – 11x + 24

By splitting the middle term, we get,

= x2 – 8x – 3x + 24

= x (x – 8) – 3 (x – 8)

We get,

= (x – 8) (x – 3)

(c) x2 + 5x – 6

By splitting the middle term, we get,

= x2 + 6x – x – 6

= x (x + 6) – 1 (x + 6)

We get,

= (x + 6) (x – 1)

(d) p2 – 12p – 64

By splitting the middle term, we get,

= p2 – 16p + 4p – 64

= p (p – 16) + 4 (p – 16)

We get,

= (p – 16) (p + 4)

(e) y2 – 2y – 24

By splitting the middle term, we get,

= y2 – 6y + 4y – 24

= y (y – 6) + 4 (y – 6)

We get,

= (y – 6) (y + 4)

(f) 3x2 + 19x – 14

By splitting the middle term, we get,

= 3x2 + 21x – 2x – 14

= 3x (x + 7) – 2 (x + 7)

We get,

= (x + 7) (3x – 2)

(g) 15a2 – 14a – 16

By splitting the middle term, we get,

= 15a2 – 24a + 10a – 16

= 3a (5a – 8) + 2 (5a – 8)

We get,

= (5a – 8) (3a + 2)

(h) 12 + x – 6x2

By splitting the middle term, we get,

= 12 + 9x – 8x – 6x2

= 3 (4 + 3x) – 2x (4 + 3x)

We get,

= (4 + 3x) (3 – 2x)

(i) 7x2 + 40x – 12

By splitting the middle term, we get,

= 7x2 + 42x – 2x – 12

= 7x (x + 6) – 2 (x + 6)

We get,

= (x + 6) (7x – 2)

4. Factorise the following:

(a) 5x2 – 17xy + 6y2

(b) 9x2 – 22xy + 8y2

(c) 2x3 + 5x2y – 12xy2

(d) x2y2 + 15xy – 16

(e) (2p + q)2 – 10p – 5q – 6

(f) y2 + 3y + 2 + by + 2b

(g) x3y3 – 8x2y2 + 15xy

(h) 6√3x2 – 19x + 5√3

(i) 2√5x2 – 7x – 3√5

Solution:

(a) 5x2 – 17xy + 6y2

= 5x2 – 15xy – 2xy + 6y2

= 5x (x – 3y) – 2y (x – 3y)

We get,

= (x – 3y) (5x – 2y)

(b) 9x2 – 22xy + 8y2

= 9x2 – 18xy – 4xy + 8y2

= 9x (x – 2y) – 4y (x – 2y)

We get,

= (x – 2y) (9x – 4y)

(c) 2x3 + 5x2y – 12xy2

= 2x3 + 8x2y – 3x2y – 12xy2

= 2x2 (x + 4y) – 3xy (x + 4y)

We get,

= (x + 4y) (2x2 – 3xy)

= (x + 4y) x (2x – 3y)

= x (x + 4y) (2x – 3y)

(d) x2y2 + 15xy – 16

= x2y2 + 16xy – 1xy – 16

= xy (xy + 16) -1 (xy + 16)

We get,

= (xy + 16) (xy – 1)

(e) (2p + q)2 – 10p – 5q – 6

= (2p + q)2 – 10p – 5q – 6

= (2p + q)2 – 5 (2p + q) – 6

= (2p + q)2 – 6 (2p + q) + (2p + q) – 6

= (2p + q) (2p + q – 6) + 1 (2p + q– 6)

We get,

= (2p + q – 6) (2p + q + 1)

(f) y2 + 3y + 2 + by + 2b

= y2 + y + 2y + 2 + by + 2b

= y2 + y + by + 2y + 2 + 2b

= y (y + 1 + b) + 2 (y + 1 + b)

We get,

= (y + 1 + b) (y + 2)

(g) x3y3 – 8x2y2 + 15xy

= x3y3 – 3x2y2 – 5x2y2 + 15xy

= x2y2 (xy – 3) – 5xy (xy – 3)

= (xy – 3) (x2y2 – 5xy)

= (xy – 3) xy (xy – 5)

We get,

= xy (xy – 3) (xy – 5)

(h) 6√3x2 – 19x + 5√3

= 6√3x2 – 10x – 9x + 5√3

= 2x (3√3x – 5) – √3 (3√3x – 5)

We get,

= (3√3x – 5) (2x – √3)

(i) 2√5x2 – 7x – 3√5

= 2√5x2 – 10x + 3x – 3√5

= 2√5x (x – √5) + 3 (x – √5)

We get,

= (x – √5) (2√5x + 3)

5.Factorise the following:

(a) 5 (3x + y)2 + 6 (3x + y) – 8

(b) 5 – 4 (a – b) – 12 (a – b)2

(c) (3a – 2b)2 + 3 (3a – 2b) – 10

(d) (a2 – 2a)2 – 23 (a2 – 2a) + 120

(e) (x + 4)2 – 5xy – 20y – 6y2

(f) 7 (x – 2)2 – 13 (x – 2) – 2

(g) 12 – (y + y2) (8 – y – y2)

(h) (p2 + p)2 – 8 (p2 + p) + 12

Solution:

(a) 5 (3x + y)2 + 6 (3x + y) – 8

= 5 (3x + y)2 + 10 (3x + y) – 4 (3x + y) – 8

= 5 (3x + y) (3x + y + 2) – 4 (3x + y + 2)

We get,

= (3x + y + 2) {5 (3x + y) – 4}

(b) 5 – 4 (a – b) – 12 (a – b)2

= 5 – 10 (a – b) + 6 (a – b) – 12 (a – b)2

= 5 {1 – 2(a – b)} + 6 (a – b) {1 – 2 (a – b)}

= [5 + 6 (a – b)] [1 – 2 (a – b)]

We get,

= (5 + 6a – 6b) (1 – 2a + 2b)

(c) (3a – 2b)2 + 3 (3a – 2b) – 10

= (3a – 2b)2 + 5 (3a – 2b) – 2 (3a – 2b) – 10

= (3a – 2b) (3a – 2b + 5) – 2 (3a – 2b + 5}

We get,

= (3a – 2b + 5) (3a – 2b – 2)

(d) (a2 – 2a)2 – 23 (a2 – 2a) + 120

= (a2 – 2a)2 – 15 (a2 – 2a) – 8 (a2 – 2a) + 120

= (a2 – 2a) (a2 – 2a – 15) – 8(a2 – 2a – 15)

We get,

= (a2 – 2a – 15) (a2 – 2a – 8)

= (a2 – 5a + 3a – 15) (a2 – 4a + 2a – 8)

= [a (a – 5) + 3 (a – 5)] [a (a – 4) + 2 (a – 4)]

= (a – 5) (a + 3){(a – 4) (a + 2)}

= (a + 2) (a + 3) (a – 4) (a – 5)

(e) (x + 4)2 – 5xy – 20y – 6y2

= (x + 4)2 – 5y (x + 4) – 6y2

= (x + 4)2 – 6y (x + 4) + y (x + 4) – 6y2

= (x + 4) (x + 4 – 6y) + y (x + 4 – 6y)

= (x + 4 – 6y) (x + 4 + y)

We get,

= (x – 6y + 4) (x + y + 4)

(f) 7 (x – 2)2 – 13 (x – 2) – 2

= 7 (x – 2)2 – 14 (x – 2) + 1 (x – 2) – 2

= 7 (x – 2) (x – 2 – 2) + 1 (x – 2 – 2)

= 7 (x – 2) (x – 4) + 1 (x – 4)

We get,

= (x – 4) {7 (x – 2) + 1}

= (x – 4) (7x – 14 + 1)

= (x – 4) (7x – 13)

(g) 12 – (y + y2) (8 – y – y2)

Let us consider (y + y2) = a, we get,

= 12 – a (8 – a)

= 12 – 8a + a2

= 12 – 6a – 2a + a2

= 6 (2 – a) – a (2 – a)

= (2 – a) (6 – a)

= {2 – (y + y2)} {6 – (y + y2)}

= (2 – y – y2) (6 – y – y2)

= (2 – 2y + y – y2) (6 – 3y + 2y – y2)

= {2 (1 – y) + y (1 – y)} {3 (2 – y) + y (2 – y)}

= (1 – y) (2 + y) (2 – y) (3 + y)

We get,

= (y – 1) (y + 2) (y – 2) (y + 3)

(h) (p2 + p)2 – 8 (p2 + p) + 12

= (p2 + p)2 – 6 (p2 + p) – 2 (p2 + p) + 12

= (p2 + p) (p2 + p – 6) – 2 (p2 + p – 6)

= (p2 + p – 6) (p2 + p – 2)

= (p2 + 3p – 2p – 6) (p2 + 2p – p – 2)

= {p (p + 3) – 2 (p + 3)}{p (p + 2) – 1 (p + 2)}

= {(p + 3) (p – 2)} {(p + 2) (p – 1)}

We get,

= (p + 3) (p – 2) (p + 2) (p – 1)

6. Factorise the following:

(a) (y2 – 3y) (y2 – 3y + 7) + 10

(b) (t2 – t) (4t2 – 4t – 5) – 6

(c) 12 (2x – 3y)2 – 2x + 3y – 1

(d) 6 – 5x + 5y + (x – y)2

(e) 2x2 + (x / 6) – 1

(f) p4 + 23p2q2 + 90q4

(g) 2a3 + 5a2b – 12ab2

Solution:

(a) (y2 – 3y) (y2 – 3y + 7) + 10

Taking (y2 – 3y) = a, we get,

= a (a + 7) + 10

= a2 + 7a + 10

= a2 + 5a + 2a + 10

= a (a + 5) + 2 (a + 5)

We get,

= (a + 5) (a + 2)

= (y2 – 3y + 5) (y2 – 3y + 2)

= (y2 – 3y + 5) (y2 – 2y – y + 2)

On further calculation, we get,

= (y2 – 3y + 5) {y (y – 2) – 1 (y – 2)}

= (y2 – 3y + 5) {(y – 2) (y – 1)}

We get,

= (y – 1) (y – 2) (y2 – 3y + 5)

(b) (t2 – t) (4t2 – 4t – 5) – 6

= (t2 – t) {4 (t2 – t) – 5} – 6

Taking (t2 – t) = a, we get,

= a (4a – 5) – 6

= 4a2 – 5a – 6

By splitting the middle term, we get,

= 4a2 – 8a + 3a – 6

= 4a (a – 2) + 3 (a – 2)

We get,

= (a – 2) (4a + 3)

= (t2 – t – 2) {4 (t2 – t) + 3}

= (t2 – 2t + t – 2) (4t2 – 4t + 3)

= {t (t – 2) + 1 (t – 2)} (4t2 – 4t + 3)

= {(t – 2) (t + 1)} (4t2 – 4t + 3)

We get,

= (t + 1) (t – 2) (4t2 – 4t + 3)

(c) 12 (2x – 3y)2 – 2x + 3y – 1

12 (2x – 3y)2 – 1 (2x – 3y) – 1

Taking (2x – 3y) = a, we get,

= 12a2 – a – 1

By splitting the middle term, we get,

= 12a2 – 4a + 3a – 1

= 4a (3a – 1) + 1 (3a – 1)

= (3a – 1) (4a + 1)

Now,

Put a = (2x – 3y)

= {3 (2x – 3y) – 1} {4 (2x – 3y) + 1}

We get,

= (6x – 9y – 1) (8x – 12y + 1)

(d) 6 – 5x + 5y + (x – y)2

= 6 – 5 (x – y) + (x – y)2

= 6 – 3 (x – y) – 2 (x – y) + (x – y)2

= 3 {2 – (x – y)} – (x – y) {2 – (x – y)}

= 3 (2 – x + y) – (x – y) (2 – x + y)

We get,

= (2 – x + y) (3 – x + y)

(e) 2x2 + (x / 6) – 1

= (1 / 6) {12x2 + x – 6}

= (1 / 6) {12x2 + 9x – 8x – 6}

= (1 / 6) {3x (4x + 3) – 2 (4x + 3)}

We get,

= (1 / 6) {(4x + 3) (3x – 2)}

= (1 / 6) (4x + 3) (3x – 2)

(f) p4 + 23p2q2 + 90q4

By splitting the middle term, we get,

= p4 + 18p2q2 + 5p2q2 + 90q4

= p2 (p2 + 18q2) + 5q2 (p2 + 18q2)

We get,

= (p2 + 18q2) (p2 + 5q2)

(g) 2a3 + 5a2b – 12ab2

By splitting the middle term, we get,

= 2a3 + 8a2b – 3a2b – 12ab2

= 2a2 (a + 4b) – 3ab (a + 4b)

We get,

= (a + 4b) (2a2 – 3ab)

= (a + 4b) a (2a – 3b)

We get,

= a (a + 4b) (2a – 3b)

7. Factorise the following by the difference of two squares:

(a) x2 – 16

(b) 64x2 – 121y2

(c) 441 – 81y2

(d) x6 – 196

(e) 625 – b2

(f) m2 – (1 / 9) n2

(g) 8xy2 – 18x3

(h) 16a4 – 81b4

(i) a (a – 1) – b (b – 1)

(j) (x + y)2 – 1

(k) x2 + y2 – z2 – 2xy

(l) (x – 2y)2 – z2

Solution:

(a) x2 – 16

= x2 – 42

We get,

= (x – 4) (x + 4)

(b) 64x2 – 121y2

= (8x)2 – (11y)2

We get,

= (8x – 11y) (8x + 11y)

(c) 441 – 81y2

= (21)2 – (9y)2

= (21 – 9y) (21 + 9y)

= 3 (7 – 3y) 3 (7 + 3y)

We get,

= 9 (7 – 3y) (7 + 3y)

(d) x6 – 196

= (x3)2 – (14)2

We get,

= (x3 – 14) (x3 + 14)

(e) 625 – b2

= (25)2 – (b)2

We get,

= (25 – b) (25 + b)

(f) m2 – (1 / 9) n2

= m2 – {(1 / 3) n}2

We get,

= {m – (1 / 3) n} {m + (1 / 3) n}

(g) 8xy2 – 18x3

= 2x (4y2 – 9x2)

= 2x {(2y)2 – (3x)2}

= 2x {(2y – 3x) (2y + 3x)}

We get,

= 2x (2y – 3x) (2y + 3x)

(h) 16a4 – 81b4

= (4a2)2 – (9b2)2

= (4a2 – 9b2) (4a2 + 9b2)

= {(2a)2 – (3b)2} (4a2 + 9b2)

= {(2a – 3b) (2a + 3b)} (4a2 + 9b2)

We get,

= (2a – 3b) (2a + 3b) (4a2 + 9b2)

(i) a (a – 1) – b (b – 1)

= a2 – a – b2 + b

= a2 – b2 – a + b

= (a2 – b2) – (a – b)

= (a – b) (a + b) – (a – b)

We get,

= (a – b) (a+ b – 1)

(j) (x + y)2 – 1

= (x + y)2 – (1)2

We get,

= (x + y + 1) (x + y – 1)

(k) x2 + y2 – z2 – 2xy

= x2 + y2 – 2xy – z2

= (x2 + y2 – 2xy) – z2

= (x – y)2 – (z)2

We get,

= (x – y – z) (x – y + z)

(l) (x – 2y)2 – z2

= (x – 2y)2 – (z)2

We get,

= (x – 2y – z) (x – 2y + z)

8. Factorise the following:

(a) 9 (a – b)2 – (a + b)2

(b) 25 ( x – y)2 – 49 (c – d)2

(c) (2a – b)2 – 9 (3c – d)2

(d) b2 – 2bc + c2 – a2

(e) x2 + (1 / x2) – 2

(f) (x2 + y2 – z2)2 – 4x2y2

(g) a2 + b2 – c2 – d2 + 2ab – 2cd

(h) 4xy – x2 – 4y2 + z2

(i) 4x2 – 12ax – y2 – z2 – 2yz + 9a2

(j) (x + y)3 – x – y

(k) y4 + y2 + 1

(l) (a2 – b2) (c2 – d2) – 4abcd

Solution:

(a) 9 (a – b)2 – (a + b)2

= {3 (a – b)}2 – (a + b)2

= {3 (a – b) – (a + b)}{3 (a – b) + (a + b)}

∵ (a2 – b2) = (a – b) (a + b)

= (3a – 3b – a – b) (3a – 3b + a + b)

We get,

= (2a – 4b) (4a – 2b)

= 2 (a – 2b) 2 (2a – b)

= 4 (a – 2b) (2a – b)

(b) 25 (x – y)2 – 49 (c – d)2

= {5 (x – y)}2 – {7 (c – d)}2

= {5 (x – y) – 7 (c – d)} {5 (x – y) + 7 (c – d)}

∵ (a2 – b2) = (a – b) (a + b)

We get,

= (5x – 5y – 7c + 7d) (5x – 5y + 7c – 7d)

(c) (2a – b)2 – 9 (3c – d)2

= (2a – b)2 – {3 (3c – d)}2

= {(2a – b) – 3(3c – d)} {(2a – b) + 3 (3c – d)}

∵ (a2 – b2) = (a – b) (a + b)

We get,

= (2a – b – 9c + 3d) (2a – b + 9c – 3d)

(d) b2 – 2bc + c2 – a2

= (b2 – 2bc + c2) – a2

= (b – c)2 – (a)2 [∵ (a – b)2 = a2 – 2ab + b2]

We get,

= (b – c – a)(b – c + a) [∵ (a2 – b2) = (a – b) (a + b)]

(e) x2 + (1 / x2) – 2

= x2 + (1 / x2) – 2 × x × (1 / x)

We get,

= {x – (1 / x)}2

= {x – (1 / x)} {x – (1 / x)}

(f) (x2 + y2 – z2)2 – 4x2y2

= (x2 + y2 – z2)2 – (2xy)2

= (x2 + y2 – z2 – 2xy) (x2 + y2 – z2 + 2xy)

[∵ (a2 – b2) = (a – b) (a + b)]

= {(x2 + y2 – 2xy) – z2} {(x2 + y2 + 2xy) – z2}

We get,

= {(x – y)2 – z2} {(x + y)2 – z2}

[∵ (a2 – b2) = (a – b) (a + b)]

= {(x – y – z) (x – y + z)}{(x + y – z) (x + y + z)}

= (x – y – z) (x – y + z) (x + y – z) (x + y + z)

(g) a2 + b2 – c2 – d2 + 2ab – 2cd

= (a2 + b2 + 2ab) – (c2 + d2 + 2cd)

= (a + b)2 – (c + d)2

We get,

= (a + b + c + d) (a + b – c – d)

(h) 4xy – x2 – 4y2 + z2

= z2 – x2 – 4y2 + 4xy

= z2 – (x2 + 4y2 – 4xy)

On further calculation, we get,

= z2 – (x – 2y)2

= {z – (x – 2y)} {z + (x – 2y)}

We get,

= (z – x + 2y) (z + x – 2y)

(i) 4x2 – 12ax – y2 – z2 – 2yz + 9a2

= (4x2 – 12ax + 9a2) – (y2 + z2 + 2yz)

We get,

= (2x – 3a)2 – (y + z)2

= {(2x – 3a) + (y + z)} {(2x – 3a) – (y + z)}

We get,

= (2x – 3a + y + z) (2x – 3a – y – z)

(j) (x + y)3 – x – y

This can be written as,

= (x + y) (x + y)2 – (x + y)

Taking (x + y) as common, we get,

= (x + y) {(x + y)2 – 1}

= (x + y) {(x + y + 1) (x + y – 1)}

We get,

= (x + y) (x + y + 1) (x + y – 1)

(k) y4 + y2 + 1

= y4 + 2y2 + 1 – y2

= (y2 + 1)2 – y2

We get,

= (y2 + 1 + y) (y2 + 1 – y)

(l) (a2 – b2) (c2 – d2) – 4abcd

On simplification, we get,

= a2c2 – a2d2 – b2c2 + b2d2 – 4abcd

= a2c2 + b2d2 – 2abcd – a2d2 – b2c2 – 2abcd

= (a2c2 + b2d2 – 2abcd) – (a2d2 + b2c2 + 2abcd)

= (ac – bd)2 – (ad + bc)2

= {(ac – bd) + (ad + bc)} {(ac – bd) – (ad + bc)}

We get,

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

9. Express each of the following as the difference of two squares:

(a) (x2 – 2x + 3) (x2 + 2x + 3)

(b) (x2 – 2x + 3) (x2 – 2x – 3)

(c) (x2 + 2x – 3) (x2 – 2x + 3)

Solution:

(a) (x2 – 2x + 3) (x2 + 2x + 3)

= (x2 + 3 – 2x) (x2 + 3 + 2x)

= {(x2 + 3) – 2x}{(x2 + 3) + 2x}

[∵ (a2 – b2) = (a – b) (a + b)]

Hence,

= (x2 + 3)2 – (2x)2

We get,

= (x2 + 3)2 – 4x2

(b) (x2 – 2x + 3) (x2 – 2x – 3)

= {(x2 – 2x) + 3}{(x2 – 2x) – 3}

[∵ (a2 – b2) = (a – b) (a + b)]

Hence,

= (x2 – 2x)2 – (3)2

= (x2 – 2x)2 – 9

(c) (x2 + 2x – 3) (x2 – 2x + 3)

= {x2 + (2x – 3)}{x2 – (2x – 3)}

[∵ (a2 – b2) = (a – b) (a + b)]

Hence,

= (x2)2 – (2x – 3)2

We get,

= x4 – (2x – 3)2

10. Factorise:

(a) y2 + (1 / 4y2) + 1 – 6y – (3 / y)

(b) 4a2 + (1 / 4a2) – 2 – 6a + (3 / 2a)

(c) x4 + y4 – 6x2y2

(d) 4x4 + 25y4 + 19x2y2

(e) p2 + (1 / p2) – 3

(f) 5x2 – y2 – 4xy + 3x – 3y

Solution:

(a) y2 + (1 / 4y2) + 1 – 6y – (3 / y)

= {y2 + (1 / 4y2) + 1} – {6y + (3 / y)}

= {y + (1 / 2y)}2 – 6 {y + (1 / 2y)}

We get,

= {y + (1 / 2y)}{y + (1 / 2y) – 6}

(b) 4a2 + (1 / 4a2) – 2 – 6a + (3 / 2a)

= {4a2 + (1 / 4a2) – 2} – {6a – (3 / 2a)}

= {2a – (1 / 2a)}2 – 3 {2a – (1 / 2a)}

We get,

= {2a – (1 / 2a)} {2a – (1 / 2a) – 3}

(c) x4 + y4 – 6x2y2

This can be written as,

= (x2)2 + (y2)2 – 2x2y2 – 4x2y2

= {(x2)2 + (y2)2 – 2x2y2} – (4x2y2)

= (x2 – y2)2 – (2xy)2

We get,

= (x2 – y2 – 2xy) (x2 – y2 + 2xy)

(d) 4x4 + 25y4 + 19x2y2

This can be written as,

= 4x4 + 25y4 + 20x2y2 – x2y2

= (2x2)2 + (5y2)2 + 2 × (2x2) × (5y2) – x2y2

= {(2x2)2 + (5y2)2 + 2 × (2x2) × (5y2)} – x2y2

We get,

= (2x2 + 5y2)2 – (xy)2

= (2x2 + 5y2 – xy) (2x2 – 5y2 + xy)

(e) p2 + (1 / p2) – 3

This can be written as,

= p2 + (1 / p2) – 2 – 1

= {(p2 + (1 / p2) – 2 × p × (1 / p)} – 1

We get,

= {p – (1 / p)}2 – (1)2

= {p – (1 / p) + 1}{p – (1 / p) – 1}

(f) 5x2 – y2 – 4xy + 3x – 3y

This can be written as,

= x2 + 4x2 – y2 – 4xy + 3x – 3y

= (x2 – y2) + (4x2 – 4xy) + (3x – 3y)

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

Taking (x – y) as common, we get,

= (x – y) {(x + y) + 4x + 3}

= (x – y) (x + y + 4x + 3)

We get,

= (x – y) (5x + y + 3)

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