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)4p ^{2} + 16p = 4p (p + 4)28x ^{3} – 70x^{2} = 14x^{2} (2x – 5)x ^{2}y + xy^{2} = xy (x + y)pq + pqz = pq (1 + z)20pq – 5q ^{2} = 5q (4p – q)9xy – 6y ^{2} = 3y (3x – 2y)5ab (ax ^{2} + y^{2}) – 6mn(ax^{2} + y^{2}) = (ax^{2} + y^{2}) (5ab – 6mn)12x ^{3}y^{4}z^{5} – 27x^{4}y^{5}z^{4} = 3x^{3}y^{4}z^{4} (4z – 9xy)4a ^{2}b – 6ab^{2} + 8b^{2} = 2b (2a^{2} – 3ab + 4b)√8 x + √12 y + √20 z = √4 (√2 x + √3 y + √5 z) |

Example 2:(i) 9a ^{2} (b – 2c) + 6(b – 2c)^{2}= 3(b – 2c) [3a ^{2} + 2(b – 2c)]= 3(b – 2c) (3a^{2} + 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) a ^{2} (b – c) + b (c – b)= a ^{2} (b – c) – b (b – c) = (b – c) (a^{2} – 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) a ^{2} + bc + ab + ac= (a ^{2} + 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) a ^{3}x – a^{2}b (x – y) – ab^{2} (y – z) – b^{3}z= a ^{3}x – a^{2}bx + a^{2}by – ab^{2}y + ab^{2}z – b^{3}z= a ^{2}x (a – b) + aby (a – b) + b^{2}z (a – b)= (a – b) (a^{2}x + aby + b^{2}z)(iv) a(2a + 3b – 4c) – 6bc = 2a ^{2} + 3ab – 4ac – 6bc= a (2a + 3b) – 2c (2a + 3b) = (2a + 3b) (a – 2c)(v) ab (x ^{2} + 1) + x (a^{2} + b^{2})= abx ^{2} + ab + a^{2}x + b^{2}x= abx ^{2} + a^{2}x + b^{2}x + ab= ax (bx + a) + b (bx + a) = (bx + a) (ax + b)(vi) (ax + by) ^{2} + (bx – ay)^{2}= a ^{2}x^{2} + 2abxy + b^{2}y^{2} + b^{2}x^{2} – 2abxy + a^{2}y^{2}= a ^{2}x^{2} + b^{2}y^{2} + b^{2}x^{2} + a^{2}y^{2} = a ^{2}x^{2} + b^{2}x^{2} + b^{2}y^{2} + a^{2}y^{2} = x ^{2} (a^{2} + b^{2}) + y^{2} (a^{2} + b^{2})= (a^{2} + b^{2}) (x^{2} + y^{2})(vii) abc ^{2} + (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 a^{2} – b^{2} = (a + b) (a – b).

Example: (i) x ^{2} – 1/36= (x) ^{2} – (1/6)^{2}= (x + 1/6) (x – 1/6)(ii) 16a ^{2} – 25b^{2}= (4a) ^{2} – (5b)^{2}= (4a + 5b) (4a – 5b)(iii) a ^{4} – 81= (a ^{2})^{2} – (9)^{2} = (a ^{2} + 9) (a^{2} – 9)= (a ^{2} + 9) [(a)^{2} – (3)^{2}]= (a^{2} + 9) (a + 3) (a – 3)(iv) x ^{4} + x^{2} + 1= x ^{4} + 2x^{2} + 1 – x^{2}= (x ^{2} + 1)^{2} – (x)^{2}= (x^{2} + 1 + x) (x^{2} + 1 – x)(v) 36x ^{2} – 84x + 49 – 25y^{2}= (36x ^{2} – 84x + 49) – (25y^{2})= (6x – 7) ^{2} – (5y)^{2}= (6x – 7 + 5y) (6x – 7 – 5y) |

### Factorization of Trinomial ax^{2} + bx + c by splitting the middle term:

Let the given trinomial be ax^{2} + 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 ax^{2} + 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 x^{2,} 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) x ^{2} – 21x – 72Here, a = 1, b = -21 and c = -72 We have (-24) + (3) = -21 and (-24) x (3) = -72 ⇒ x ^{2} – 21x – 72= x ^{2} – 24x + 3x – 72= x (x – 24) + 3 (x – 24) = (x – 24) (x + 3)(ii) 9x ^{2} + 24xy + 16y^{2}= (3x) ^{2 }+ 2 (3x) (4y) + (4y)^{2}= (3x + 4y) ^{2}= (3x + 4y) (3x + 4y)(iii) x ^{6} – 8x^{3} + 16= (x ^{3})^{2} – 2(x^{3}) (4) + (4)^{2}= (x ^{3} – 4)^{2}= (x^{3} – 4) (x^{3} – 4)(iv) a ^{2} – 2 + 1/a^{2}= (a) ^{2} – 2 (a) (1/a) + (1/a)^{2}= (a – 1/a) ^{2}= (a – 1/a) ) (a – 1/a)(v) 30x ^{2} + 7x – 15= 30x ^{2} + 25x – 18x – 15= 5x (6x + 5) – 3 (6x + 5) = (6x + 5) (5x – 3)(vi) 4 – x (5 – x) = 4 – 5x + x ^{2}= 4 – 4x – x + x ^{2}= 4 (1 – x) – x (1 – x) = (1 – x) (4 – x)(vii) x ^{2}/5 + 2x – 15= 1/5 (x ^{2} + 10x – 75)= 1/5 (x ^{2} + 15x – 5x – 75)= 1/5 [x (x + 15) – 5 (x + 15)] = 1/5 (x + 15) (x – 5)(viii) px ^{2} + (4p^{2} – 3q)x – 12pq= px ^{2} + 4p^{2}x – 3qx – 12pq= px (x + 4p) – 3q (x + 4p) = (x + 4p) (px – 3q)(ix) 3 (6x ^{2} + 5x)^{2} – 10 (6x^{2} + 5x) – 8Put (6x ^{2} + 5x) = a⇒ 3a ^{2} – 10a – 8= 3a ^{2} – 12a + 2a – 8= 3a (a – 4) + 2 (a – 4) = (a – 4) (3a + 2) = (6x ^{2} + 5x – 4) (18x^{2} + 15x + 2)= [6x ^{2} + 8x – 3x – 4] [18x^{2} + 12x + 3x + 2]= [2x (3x + 4) – 1 (3x + 4)] [6x (3x + 2) + 1 (3x + 2)] = (3x + 4) (2x – 1) (3x + 2) (6x + 1)(x) x ^{2} + 10√3x + 63= x ^{2} + 7√3x + 3√3x + 63= x (x + 7√3) + 3√3 (x + 7√3) = (x + 7√3) (x + 3√3)(xi) x ^{4} + 9x^{2}– 52= x ^{4} + 13x^{2} -4x^{2} – 52= x ^{2} (x^{2} + 13) – 4 (x^{2} + 13)= (x ^{2} + 13) (x^{2} – 4)= (x^{2} + 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:

a^{3} + b^{3} = (a + b) (a^{2} – ab + b^{2})a ^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2}) |

Example:(i) 64a ^{3} + 27b^{3}= (4a) ^{3} + (3b)^{3}= (4a + 3b) [(4a) ^{2} – (4a) (3b) + (3b)^{2}]= (4a + 3b) (16a^{2} – 12ab + 9b^{2})(ii) (2a + 3b) ^{3} + (2a – 3b)^{3}= [(2a + 3b) + (2a – 3b)] [(2a + 3b) ^{2} – (2a + 3b) (2a – 3b) + (2a – 3b)^{2}]= (4a) [4a ^{2} + 9b^{2} + 12ab – (4a^{2} – 9b^{2}) + 4a^{2} + 9b^{2} – 12ab]= 4a (4a^{2} + 27b^{2})(iii) a ^{6} – b^{6}= (a ^{3})^{2} – (b^{3})^{2}= (a ^{3} – b^{3}) (a^{3} + b^{3})= (a – b) (a ^{2} + ab + b^{2}) (a + b) (a^{2} – ab + b^{2})= (a – b) (a + b) (a^{2} + ab + b^{2}) (a^{2} – ab + b^{2})(iv) x ^{6} + 7x^{3} – 8= x ^{6} + 8x^{3} – x^{3} – 8= x ^{3} (x^{3} + 8) – 1 (x^{3} + 8)= (x ^{3} + 8) (x^{3} – 1)= (x + 2) (x ^{2} – 2x + 4) (x – 1) (x^{2} + x + 1)= (x – 1) (x + 2) (x^{2} – 2x + 4) (x^{2} + x + 1) |

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