Table of Contents
Factorization of Polynomials:
When a polynomial or an expression is expressed as a product of two or more polynomials, then each polynomial in the product is a factor of the given polynomial. The process of expressing a polynomial as a product of its factors is called factorization.
Factorization by taking out Common Factors:
If a factor is common to all the terms of a polynomial, then it is placed outside the brackets. The figures to be placed inside the brackets are obtained by dividing each term of the given polynomial by the common factor.
| Example 1: 2p + 2q = 2 (p + q) 4p2 + 16p = 4p (p + 4) 28x3 – 70x2 = 14x2 (2x – 5) x2y + xy2 = xy (x + y) pq + pqz = pq (1 + z) 20pq – 5q2 = 5q (4p – q) 9xy – 6y2 = 3y (3x – 2y) 5ab (ax2 + y2) – 6mn(ax2 + y2) = (ax2 + y2) (5ab – 6mn) 12x3y4z5 – 27x4y5z4 = 3x3y4z4 (4z – 9xy) 4a2b – 6ab2 + 8b2 = 2b (2a2 – 3ab + 4b) √8 x + √12 y + √20 z = √4 (√2 x + √3 y + √5 z) |
| Example 2: (i) 9a2 (b – 2c) + 6(b – 2c)2 = 3(b – 2c) [3a2 + 2(b – 2c)] = 3(b – 2c) (3a2 + 2b – 4c) (ii) 8 (3a + 2b)2 – 16 (3a + 2b) = 8 (3a + 2b) (3a + 2b – 2) (iii) 4x (a + b) – 3y (a + b) = (a + b) (4x – 3y) (iv) a2 (b – c) + b (c – b) = a2 (b – c) – b (b – c) = (b – c) (a2 – b) (v) 12 (a – b)3 + 18 (a – b)2 = 6 (a – b)2 [2 (a – b) + 3] = 6 (a – b)2 (2a – 2b + 3) |
Factorization by Grouping the Terms:
The terms in the given expression can be arranged in groups of two or three to get a common factor. The given expression can then be factorized by taking out the common factor.
| Example: (i) a2 + bc + ab + ac = (a2 + ab) + (bc + ac) = a (a + b) + c (b + a) = a (a + b) + c (a + b) = (a + b) (a + c) (ii) 2xy – 3ab + 2bx – 3ay = 2xy + 2bx – 3ay – 3ab = 2x (y + b) – 3a (y + b) = (y + b) (2x – 3a) (iii) a3x – a2b (x – y) – ab2 (y – z) – b3z = a3x – a2bx + a2by – ab2y + ab2z – b3z = a2x (a – b) + aby (a – b) + b2z (a – b) = (a – b) (a2x + aby + b2z) (iv) a(2a + 3b – 4c) – 6bc = 2a2 + 3ab – 4ac – 6bc = a (2a + 3b) – 2c (2a + 3b) = (2a + 3b) (a – 2c) (v) ab (x2 + 1) + x (a2 + b2) = abx2 + ab + a2x + b2x = abx2 + a2x + b2x + ab = ax (bx + a) + b (bx + a) = (bx + a) (ax + b) (vi) (ax + by)2 + (bx – ay)2 = a2x2 + 2abxy + b2y2 + b2x2 – 2abxy + a2y2 = a2x2 + b2y2 + b2x2 + a2y2 = a2x2 + b2x2 + b2y2 + a2y2 = x2 (a2 + b2) + y2 (a2 + b2) = (a2 + b2) (x2 + y2) (vii) abc2 + (ac – b)c – c = c [abc + (ac – b) – 1] = c [abc + ac – b – 1] = c [ac (b + 1) – 1 (b + 1)] = c [(b + 1) (ac – 1)] = c (b + 1) (ac – 1) |
Factorization of the difference between two Squares:
When the given expression is written as the difference between two squares, it is factorized using the formula a2 – b2 = (a + b) (a – b).
| Example: (i) x2 – 1/36 = (x)2 – (1/6)2 = (x + 1/6) (x – 1/6) (ii) 16a2 – 25b2 = (4a)2 – (5b)2 = (4a + 5b) (4a – 5b) (iii) a4 – 81 = (a2)2 – (9)2 = (a2 + 9) (a2 – 9) = (a2 + 9) [(a)2 – (3)2] = (a2 + 9) (a + 3) (a – 3) (iv) x4 + x2 + 1 = x4 + 2x2 + 1 – x2 = (x2 + 1)2 – (x)2 = (x2 + 1 + x) (x2 + 1 – x) (v) 36x2 – 84x + 49 – 25y2 = (36x2 – 84x + 49) – (25y2) = (6x – 7)2 – (5y)2 = (6x – 7 + 5y) (6x – 7 – 5y) |
Factorization of Trinomial ax2 + bx + c by splitting the middle term:
Let the given trinomial be ax2 + bx + c. The terms of the trinomial should be arranged in a descending or ascending order of the powers of x. In order to factorize ax2 + bx + c, we should find two numbers p and q whose sum equals the coefficient of x i.e. b and the product equals the product of the coefficients of x2, and the constant term i.e. (a) x (c).
We split the middle-term bx as px + qx and then factorize it by grouping.
| Example: (i) x2 – 21x – 72 Here, a = 1, b = -21 and c = -72 We have (-24) + (3) = -21 and (-24) x (3) = -72 ⇒ x2 – 21x – 72 = x2 – 24x + 3x – 72 = x (x – 24) + 3 (x – 24) = (x – 24) (x + 3) (ii) 9x2 + 24xy + 16y2 = (3x)2 + 2 (3x) (4y) + (4y)2 = (3x + 4y)2 = (3x + 4y) (3x + 4y) (iii) x6 – 8x3 + 16 = (x3)2 – 2(x3) (4) + (4)2 = (x3 – 4)2 = (x3 – 4) (x3 – 4) (iv) a2 – 2 + 1/a2 = (a)2 – 2 (a) (1/a) + (1/a)2 = (a – 1/a)2 = (a – 1/a) ) (a – 1/a) (v) 30x2 + 7x – 15 = 30x2 + 25x – 18x – 15 = 5x (6x + 5) – 3 (6x + 5) = (6x + 5) (5x – 3) (vi) 4 – x (5 – x) = 4 – 5x + x2 = 4 – 4x – x + x2 = 4 (1 – x) – x (1 – x) = (1 – x) (4 – x) (vii) x2/5 + 2x – 15 = 1/5 (x2 + 10x – 75) = 1/5 (x2 + 15x – 5x – 75) = 1/5 [x (x + 15) – 5 (x + 15)] = 1/5 (x + 15) (x – 5) (viii) px2 + (4p2 – 3q)x – 12pq = px2 + 4p2x – 3qx – 12pq = px (x + 4p) – 3q (x + 4p) = (x + 4p) (px – 3q) (ix) 3 (6x2 + 5x)2 – 10 (6x2 + 5x) – 8 Put (6x2 + 5x) = a ⇒ 3a2 – 10a – 8 = 3a2 – 12a + 2a – 8 = 3a (a – 4) + 2 (a – 4) = (a – 4) (3a + 2) = (6x2 + 5x – 4) (18x2 + 15x + 2) = [6x2 + 8x – 3x – 4] [18x2 + 12x + 3x + 2] = [2x (3x + 4) – 1 (3x + 4)] [6x (3x + 2) + 1 (3x + 2)] = (3x + 4) (2x – 1) (3x + 2) (6x + 1) (x) x2 + 10√3x + 63 = x2 + 7√3x + 3√3x + 63 = x (x + 7√3) + 3√3 (x + 7√3) = (x + 7√3) (x + 3√3) (xi) x4 + 9x2– 52 = x4 + 13x2 -4x2 – 52 = x2 (x2 + 13) – 4 (x2 + 13) = (x2 + 13) (x2 – 4) = (x2 + 13) (x + 2) (x – 2) |
Factorization of the sum and the difference of the Two Cubes:
The sum and the difference of the two cubes can be factorized using:
| a3 + b3 = (a + b) (a2 – ab + b2) a3 – b3 = (a – b) (a2 + ab + b2) |
| Example: (i) 64a3 + 27b3 = (4a)3 + (3b)3 = (4a + 3b) [(4a)2 – (4a) (3b) + (3b)2] = (4a + 3b) (16a2 – 12ab + 9b2) (ii) (2a + 3b)3 + (2a – 3b)3 = [(2a + 3b) + (2a – 3b)] [(2a + 3b)2 – (2a + 3b) (2a – 3b) + (2a – 3b)2] = (4a) [4a2 + 9b2 + 12ab – (4a2 – 9b2) + 4a2 + 9b2 – 12ab] = 4a (4a2 + 27b2) (iii) a6 – b6 = (a3)2 – (b3)2 = (a3 – b3) (a3 + b3) = (a – b) (a2 + ab + b2) (a + b) (a2 – ab + b2) = (a – b) (a + b) (a2 + ab + b2) (a2 – ab + b2) (iv) x6 + 7x3 – 8 = x6 + 8x3 – x3 – 8 = x3 (x3 + 8) – 1 (x3 + 8) = (x3 + 8) (x3 – 1) = (x + 2) (x2 – 2x + 4) (x – 1) (x2 + x + 1) = (x – 1) (x + 2) (x2 – 2x + 4) (x2 + x + 1) |
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