Factorise the following using the identity a² - b² = (a + b) (a - b).
Solution:
(i) x² - 9
Given, x² - 9 = x² - (3)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = x and b = 3 the expression can be written as,
x² - 9 = x² - (3)²
= (x + 3)(x - 3)
(ii) 4x² - 25y²
Given, 4x² - 25y² = (2x)² - (5y)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = 2x and b = 5y the expression can be written as,
4x² - 25y² = (2x)² - (5y)²
= (2x + 5y)(2x - 5y)
(iii) 4x² - 49y²
Given, 4x² - 25y² = (2x)² - (7y)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = 2x and b = 7y the expression can be written as,
4x² - 49y² = (2x)² - (7y)²
= (2x + 7y)(2x - 7y)
(iv) 3a²b³ - 27a⁴b
Given, 3a²b³ - 27a⁴b = 3a²b[b² - 9a²] = 3a²b[(b)² - (3a)²]
Now, using the identity: a² - b² = (a + b) (a - b), where a = ab and b = 3a² the expression can be written as,
3a²b³ - 27a⁴b = 3a²b[b² - 9a²]
= 3a²b[(b)² - (3a)²]
= 3a²b[(b + 3a) (b - 3a)]
(v) 28ay² - 175ax²
Given, 28ay² - 175ax² = 7a[4y² - 25x²] = 7a[(2y)² - (5x)²]
Now, using the identity: a² - b² = (a + b) (a - b), where a = 2y and b = 5x the expression can be written as,
28ay² - 175ax² = 7a[4y² - 25x²]
= 7a[(2y)² - (5x)²]
= 7a[(2y + 5x) (2y - 5x)]
(vi) 9x² - 1
Given, 9x² - 1 = (3x)² - (1)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = 3x and b = 1 the expression can be written as,
9x² - 1 = (3x)² - (1)²
= (3x + 1) (3x - 1)
(vii) 25ax² - 25a
Given, 25ax² - 25a = 25a(x² - 1)
Now, using the identity: a² - b² = (a + b) (a - b), where a = x and b = 1 the expression can be written as,
25a(x² - 1) = 25a[(x)² - (1)²]
= 25a[(x + 1)(x - 1)]
(viii) (x²/9) - (y²/25)
Given, (x²/9) - (y²/25) = (x/3)² - (y/5)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = x/3 and b = y/5 the expression can be written as,
(x²/9) - (y²/25) = (x/3)² - (y/5)²
= [(x/3) + (y/5)][(x/3) - (y/5)]
(ix) 2p²/25 - 32q²
Given, 2p²/25 - 32q² = 2[(p²/25) - 16q²] = 2[(p/5)² - (4q)²]
Now, using the identity: a² - b² = (a + b) (a - b), where a = p/5 and b = 4q the expression can be written as,
2p²/25 - 32q² = 2[(p/5)² - (4q)²]
= 2[(p/5) + 4q][(p/5) - 4q]
(x) 49x² - 36y²
Given, 49x² - 36y² = (7x)² - (6y)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = 7x and b = 6y the expression can be written as,
49x² - 36y² = (7x)² - (6y)²
= (7x + 6y)(7x - 6y)
(xi) y³ - y/9
Given, y³ - y/9 = y[y² - (1/3)²]
Now, using the identity: a² - b² = (a + b) (a - b), where a = y and b = 1/3 the expression can be written as,
y³ - y/9 = y[y² - (1/3)²]
= y[(y + (1/3)][y - (1/3)]
(xii) (x²/25) - 625
Given, (x²/25) - 625 = (x/5)² - 25²
Now, using the identity: a² - b² = (a + b) (a - b), where a = x/5 and b = 25 the expression can be written as,
(x²/25) - 625 = (x/5)² - 25²
= [(x/5) + 25][(x/5) - 25]
(xiii) (x²/8) - (y²/18)
Given, (x²/8) - (y²/18) = (1/2)[(x²/4) - (y²/9)]
Now, using the identity: a² - b² = (a + b) (a - b), where a = x/2 and b = y/3 the expression can be written as,
(x²/8) - (y²/18) = (1/2)[(x²/4) - (y²/9)]
= (1/2)[(x/2)² - (y/3)²]
= (1/2)[(x/2) + (y/3)][(x/2) - (y/3)]
(xiv) (4x²/9) - (9y²/16)
Given, (4x²/9) - (9y²/16) = (2x/3)² - (3y/4)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = 2x/3 and b = 3y/4 the expression can be written as,
(4x²/9) - (9y²/16) = (2x/3)² - (3y/4)²
= [(2x/3) + (3y/4)][(2x/3) - (3y/4)]
(xv) (x³y/9) - (xy³/16)
Given, (x³y/9) - (xy³/16) = xy[(x²/9) - (y²/16)] = xy[(x/3)² - (y/4)²]
Now, using the identity: a² - b² = (a + b) (a - b), where a = x/3 and b = y/4 the expression can be written as,
(x³y/9) - (xy³/16) = xy[(x/3)² - (y/4)²]
= xy[(x/3) + (y/4)][(x/3) - (y/4)]
(xvi) 1331x³y - 11y³x
Given, 1331x³y - 11y³x = 11xy[121x² - y²] = 11xy[(11x)² - (y)²]
Now, using the identity: a² - b² = (a + b) (a - b), where a = 11x and b = y the expression can be written as,
1331x³y - 11y³x = 11xy[(11x)² - (y)²]
= 11xy[(11x + y)(11x - y)]
(xvii) (1/36)a²b² - (16/49)b²c²
Given, (1/36)a²b² - (16/49)b²c² = (ab/6)² - (4bc/7)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = ab/6 and b = 4bc/7 the expression can be written as,
(1/36)a²b² - (16/49)b²c² = (ab/6)² - (4bc/7)²
= [(ab/6) + (4bc/7)][(ab/6) - (4bc/7)]
= b[(a/6) + (4c/7)] b[(a/6) - (4c/7)]
= b²[(a/6) + (4c/7)][(a/6) - (4c/7)]
(xviii) a⁴ - (a - b)⁴
Given, a⁴ - (a - b)⁴ = (a²)² - ((a - b)²)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = a² and b = (a - b)² the expression can be written as,
a⁴ - (a - b)⁴ = (a²)² - ((a - b)²)²
= (a² + (a - b)²) (a² - (a - b)²)
= (a² + a² - 2ab + b²) {[a + (a - b)][a - (a - b)]}
= (a² + a² - 2ab + b²) {[a + a - b][a - a + b]}
= (2a² - 2ab + b²)(2a - b) (b)
(xix) x⁴ - 1
Given, x⁴ - 1 = (x²)² - (1)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = x² and b = 1 the expression can be written as,
x⁴ - 1 = (x²)² - (1)²
= (x² + 1) (x² - 1)
= (x² + 1) (x + 1) (x - 1)
(xx) y⁴ - 625
Given, y⁴ - 625 = (y²)² - (25)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = y² and b = 25 the expression can be written as,
y⁴ - 625 = (y²)² - (25)²
= (y² + 25)(y² - 25)
= (y² + 25)(y² - 5²)
= (y² + 25)(y + 5)(y - 5)
(xxi) p⁵ - 16p
Given, p⁵ - 16p = p(p⁴ - 16) = p[(p²)² - 4²]
Now, using the identity: a² - b² = (a + b) (a - b), where a = p² and b = 4 the expression can be written as,
p⁵ - 16p = p[(p²)² - 4²]
= (p² + 4)(p² - 4)
= (p² + 4)(p² - 2²)
= (p² + 4)(p + 2)(p - 2)
(xxii) 16x⁴ - 81
Given, 16x⁴ - 81 = (4x²)² - 9²
Now, using the identity: a² - b² = (a + b) (a - b), where a = 4x² and b = 9 the expression can be written as,
16x⁴ - 81 = (4x²)² - 9²
= (4x² + 9)(4x² - 9)
= (4x² + 9)[(2x)² - (3)²]
= (4x² + 9)(2x + 3)(2x - 3)
(xxiii) x⁴ - y⁴
Given, x⁴ - y⁴ = (x²)² - (y²)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = x² and b = y² the expression can be written as,
x⁴ - y⁴ = (x²)² - (y²)²
= (x² + y²)(x² - y²)
= (x² + y²)(x + y)(x - y)
(xxiv) y⁴ - 81
Given, y⁴ - 81 = (y²)² - 9²
Now, using the identity: a² - b² = (a + b) (a - b), where a = y² and b = 9 the expression can be written as,
y⁴ - 81 = (y²)² - 9²
= (y² + 9)(y² - 9)
= (y² + 9)(y² - 3²)
= (y² + 9)(y + 3)(y - 3)
(xxv) 16x⁴ - 625y⁴
Solution: Given, 16x⁴ - 625y⁴ = (4x²)² - (25y²)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = 4x² and b = 25y² the expression can be written as,
16x⁴ - 625y⁴ = (4x²)² - (25y²)²
= (4x² + 25y²)(4x² - 25y²)
= (4x² + 25y²)[(2x)² - (5y)²]
= (4x² + 25y²)(2x + 5y)(2x - 5y)
(xxvi) (a - b)² - (b - c)²
Given, (a - b)² - (b - c)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = (a - b) and b = (b - c) the expression can be written as,
(a - b)² - (b - c)²
= [(a - b) + (b - c)][(a - b) - (b - c)]
= [a - b + b - c][a - b - b + c]
= (a - c)(a - 2b + c)
(xxvii) (x + y)⁴ - (x - y)⁴
Given, (x + y)⁴ - (x - y)⁴ = [(x + y)²]² - [(x - y)²]²
Now, using the identity: a² - b² = (a + b) (a - b), where a = (x + y)² and b = (x - y)² the expression can be written as,
(x + y)⁴ - (x - y)⁴ = [(x + y)²]² - [(x - y)²]²
= [(x + y)² + (x - y)²] [(x + y)² - (x - y)²]
= [(x² + 2xy + y²) + (x² - 2xy + y²)] [(x + y) + (x - y)][(x + y) - (x - y)]
= [x² + 2xy + y² + x² - 2xy + y²][x + y + x - y][x + y - x + y]
= [2x² + 2y²][2x][2y]
= 2[x² + y²][4xy]
= 8xy [x² + y²]
(xxviii) x⁴ - y⁴ + x² - y²
Given, x⁴ - y⁴ + x² - y² = [(x²)² - (y²)²] + [x² - y²]
Now, using the identity: a² - b² = (a + b) (a - b), the expression can be written as,
= [(x² + y²)(x² - y²)] + (x² - y²)
= (x² - y²) [(x² + y²) + 1]
= (x + y) (x - y) [(x² + y²) + 1]
(xxix) 8a³ - 2a
Given, 8a³ - 2a = 2a[4a² - 1] = 2a[(2a)² - 1²]
Now, using the identity: a² - b² = (a + b) (a - b), where a = 2a and b = 1 the expression can be written as,
8a³ - 2a = 2a[4a² - 1] = 2a[(2a)² - 1²]
= 2a[(2a + 1)(2a - 1)]
(xxx) x² - (y²/100)
Given, x² - (y²/100) = (x)² - (y/10)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = x and b = y/10 the expression can be written as,
x² - (y²/100) = (x)² - (y/10)²
= [x + (y/10)][x - (y/10)]
(xxxi) 9x² - (3y + z)²
Given, 9x² - (3y + z)² = (3x)² - (3y + z)²
Now, using the identity: a² - b² = (a + b) (a - b), where a = 3x and b = (3y + z) the expression can be written as,
9x² - (3y + z)² = (3x)² - (3y + z)²
= [(3x) + (3y + z)][(3x) - (3y + z)]
✦ Try This: Factorise the following using the identity a² - b² = (a + b) (a - b).
(i) p² - 16, (ii) (x²/9) - y², (iii)2p³/25 - 32pq²/121
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9
NCERT Exemplar Class 8 Maths Chapter 7 Problem 92
Factorise the following using the identity a² - b² = (a + b) (a - b). (i) x² - 9, (ii) 4x² - 25y², (iii) 4x² - 49y², (iv) 3a²b³ - 27a⁴b, (v) 28ay² - 175ax², (vi) 9x² - 1, (vii) 25ax² - 25a, (viii) (x²/9) - (y²/25), (ix) 2p²/25 - 32q², (x) 49x² - 36y², (xi) y³ - y/9, (xii) (x²/25) - 625, (xiii) (x²/8) - (y²/18), (xiv) (4x²/9) - (9y²/16), (xv) (x³y/9) - (xy³/16), (xvi) 1331x³y - 11y³x, (xvii) (1/36)a²b² - (16/49)b²c², (xviii) a⁴ - (a - b)⁴, (xix) x⁴ - 1, (xx) y⁴ - 625, (xxi) p⁵ - 16p, (xxii) 16x⁴ - 81, (xxiii) x⁴ - y⁴, (xxiv) y⁴ - 81, (xxv) 16x⁴ - 625y⁴, (xxvi) (a - b)² - (b - c)², (xxvii) (x + y)⁴ - (x - y)⁴, (xxviii) x⁴ - y⁴ + x² - y², (xxix) 8a³ - 2a, (xxx) x² - (y²/100), (xxxi) 9x² - (3y + z)²
Summary:
Factorising the following using the identity a² - b² = (a + b) (a - b) we get, (i) x² - 9 = (x + 3)(x - 3), (ii) 4x² - 25y² = (2x + 5y)(2x - 5y), (iii) 4x² - 49y² = (2x + 7y)(2x - 7y), (iv) 3a²b³ - 27a⁴b = 3a²b[(b + 3a) (b - 3a)], (v) 28ay² - 175ax² = 7a[(2y + 5x) (2y - 5x)], (vi) 9x² - 1 = (3x + 1) (3x - 1), (vii) 25ax² - 25a = 25a[(x + 1)(x - 1)], (viii) (x²/9) - (y²/25) = [(x/3) + (y/5)][(x/3) - (y-5)], (ix) 2p²/25 - 32q² = 2[(p/5) + 4q][(p/5) - 4q], (x) 49x² - 36y² = (7x + 6y)(7x - 6y), (xi) y³ - y/9 = y[(y + (1/3)][y - (1/3)], (xii) (x²/25) - 625 = [(x/5) + 25][(x/5) - 25], (xiii) (x²/8) - (y²/18) = (1/2)[(x/2) + (y/3)][(x/2) - (y/3)], (xiv) (4x²/9) - (9y²/16) = [(2x/3) + (3y/4)][(2x/3) - (3y/4)], (xv) (x³y/9) - (xy³/16) = xy[(x/3) + (y/4)][(x/3) - (y/4)], (xvi) 1331x³y - 11y³x = 11xy[(11x + y)(11x - y)], (xvii) (1/36)a²b² - (16/49)b²c² = b²[(a/6) + (4c/7)][(a/6) - (4c/7)], (xviii) a⁴ - (a - b)⁴ = (2a² - 2ab + b²)(2a - b) (b), (xix) x⁴ - 1 = (x² + 1) (x + 1) (x - 1), (xx) y⁴ - 625 = (y² + 25)(y + 5)(y - 5), (xxi) p⁵ - 16p = p(p² + 4)(p + 2)(p - 2), (xxii) 16x⁴ - 81= (4x² + 9)(2x + 3)(2x - 3), (xxiii) x⁴ - y⁴ = (x² + y²)(x + y)(x - y), (xxiv) y⁴ - 81 = (y² + 9)(y + 3)(y - 3), (xxv) 16x⁴ - 625y⁴ = (4x² + 25y²)(2x + 5y)(2x - 5y), (xxvi) (a - b)² - (b - c)² = (a - c)(a - 2b + c), (xxvii) (x + y)⁴ - (x - y)⁴ = 8xy [x² + y²], (xxviii) x⁴ - y⁴ + x² - y² = (x + y) (x - y) [(x² + y²) + 1], (xxix) 8a³ - 2a = 2a[(2a + 1)(2a - 1)], (xxx) x² - (y²/100) = [x + (y/10)][x - (y/10)], (xxxi) 9x² - (3y + z)² = [(3x) + (3y + z)][(3x) - (3y + z)]
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