Frank Solutions for Class 9 Maths Chapter 5 Factorisation provide students with detailed answers prepared by subject experts at BYJU’S, who have vast experience in the field. The solutions in simple language help students prepare comprehensively for the final examinations. Our topmost faculty have designed the answers in such a way that students can clear their doubts instantly.
Chapter 5 deals with the study of Factorisation. When a number is split into factors or divisors, it is known as factorisation. This chapter plays a vital role in the subject as it is continued in higher classes also. By practising these solutions, students can enhance their problem-solving and time-management skills, which are essential to ace the annual exam. In order to score good marks, students can download Frank Solutions for Class 9 Maths Chapter 5 Factorisation PDF from the link below and practise regularly.
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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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