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CBSE Class 9 Mathematics Question 8 of 16

Polynomials — Question 8

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8
Question

Question 8

Factorise each of the following:

(i) 8a3 + b3 + 12a2b + 6ab2

(ii) 8a3 - b3 - 12a2b + 6ab2

(iii) 27 - 125a3 - 135a + 225a2

(iv) 64a3 - 27b3 - 144a2b + 108ab2

(v) 27p3 - 1216\dfrac{1}{216} - 92\dfrac{9}{2}p2 + 14\dfrac{1}{4}p

Answer

(i) 8a3 + b3 + 12a2b + 6ab2

[∵ a3 + b3 + 3a2b + 3ab2 = (a + b)3]

= (2a)3 + (b)3 + 3(2a)2(b) + 3(2a)(b)2

= (2a + b)3

Hence, 8a3 + b3 + 12a2b + 6ab2 = (2a + b)(2a + b)(2a + b)

(ii) 8a3 - b3 - 12a2b + 6ab2

[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]

= (2a)3 - (b)3 - 3(2a)2(b) + 3(2a)(b)2

= (2a - b)3

Hence, 8a3 - b3 - 12a2b + 6ab2 = (2a - b)(2a - b)(2a - b)

(iii) 27 - 125a3 - 135a + 225a2

[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]

= (3)3 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2

= (3 - 5a)3

Hence, 27 - 125a3 - 135a + 225a2 = (3 - 5a)(3 - 5a)(3 - 5a)

(iv) 64a3 - 27b3 - 144a2b + 108ab2

[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]

= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2

= (4a - 3b)3

Hence, 64a3 - 27b3 - 144a2b + 108ab2 = (4a - 3b)(4a - 3b)(4a - 3b)

(v) 27p3 - 1216\dfrac{1}{216} - 92\dfrac{9}{2}p2 + 14\dfrac{1}{4}p

[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]

= (3p)3 - (16)\Big(\dfrac{1}{6}\Big)3 - 3(3p)2(16)\Big(\dfrac{1}{6}\Big) + 3(3p)(16)\Big(\dfrac{1}{6}\Big)2

= (3p−16)3\Big(3p - \dfrac{1}{6}\Big)^3

Hence, 27p3 - 1216\dfrac{1}{216} - 92\dfrac{9}{2}p2 + 14\dfrac{1}{4}p = (3p−16)\Big(3p - \dfrac{1}{6}\Big)\Big(3p - \dfrac{1}{6}\Big)(3p−16)\Big(3p - \dfrac{1}{6}\Big)

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Polynomials — Interactive Study Guide

Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification

Not a polynomial: √x (fractional power), 1/x = x−1 (negative power), x + 1/x.

Polynomial: 5 (constant), 3x + 2 (linear), x² − 1 (quadratic), 2x³ + x − 1 (cubic).

Remainder and Factor Theorems — Quick Guide

Remainder Theorem: When p(x) is divided by (x − a), remainder = p(a).
Factor Theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.

Watch the sign! Dividing by (x + 3) means a = −3. So remainder = p(−3).

Identity Mastery Checklist

See This PatternUse This Identity
a² + 2ab + b²= (a + b)²
a² − 2ab + b²= (a − b)²
a² − b²= (a + b)(a − b)
a³ + b³= (a + b)(a² − ab + b²)
a³ − b³= (a − b)(a² + ab + b²)
a + b + c = 0⇒ a³ + b³ + c³ = 3abc

Quick Self-Check

  1. Degree of 5x³ − 2x + 1? (3)
  2. Remainder when x² + 3x + 2 is divided by (x + 1)? (p(−1) = 1 − 3 + 2 = 0)
  3. Expand: (2a + 3b)² (= 4a² + 12ab + 9b²)
  4. Factorise: 8x³ − 27 (= (2x − 3)(4x² + 6x + 9))

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