Frank Solutions for Chapter 5 Factorisation Class 9 Mathematics ICSE
Exercise 5.1
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)
Answer
(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)
Answer
(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)
(o) 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,
= {(1/25 x2) + 16x2 + (8/5)} – {12x + (3/5 x)}
= {(1/5 x)2 + (4x)2 + (2) (1/5 x) (4x)} – {12x + (3/5 x)}
= {(1/5x) + 4x}2 – 3 {4x + (1/5 x)}
= {(1/5 x) + 4x}2 – 3{(1/5 x) + 4x}
We get,
= {(1/5 x) + 4x) {(1/5 x) + 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
Answer
(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
Answer
(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
Answer
(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
Answer
(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
Answer
(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
Answer
(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)
Answer
(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
(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
Answer
(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)
Exercise 5.2
1. 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) Factorise the following by splitting the middle term: 7x2 + 40x – 12
Answer
(a)
(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
Answer
(a)
(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
Answer
(a)
(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
Answer
(a)
Exercise 5.3
1. 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 – 1/9n2
(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
Answer
(a)
(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
Answer
(a)
(b)
(c)
3. 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)
Answer
(a)
4. 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
Answer
(a)
