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Factorization of Expressions of the Form ax2 + bx + c, a ≠ 1

The below examples show that the method of factorizing ax2 + bx + c by breaking the middle term involves the following steps.

Steps:

1.Take the product of the constant term and the coefficient of x2, i.e., ac.

2. Break ac into two factors p, q whose sum is b, i.e., p + q = b.

3. Pair one of them, say px, with ax^2 and the other, qx, with c. Then factorize the expression.

Solved Examples on Factorization of Expressions of the Form ax^2 + bx + c, a ≠ 1:

1. Factorize: 6m2 + 7m + 2.

Solution:

Here, 6 × 2 = 12 = 3 × 4 and, 3 + 4 = 7 (= coefficient of m).

Therefore, 6m2 + 7m + 2 = 6m2 + 3m + 4m + 2

                                      = 3m(2m + 1) + 2(2m + 1)

                                      = (2m + 1)(3m + 2)

 

2. Factorize: 1 – 18x – 63x2

Solution:

The given expression is – 63x2 - 18x + 1

Here, (-63) × 1 = -63 = (-21) × (3), and -21 + 3 = -18(= coefficient of x).

Therefore, – 63x2 - 18x + 1 = – 63x2 – 21x + 3x + 1

                                         = -21x(3x + 1) + 1(3x + 1)

                                         = (3x + 1)(-21x + 1)

                                         = (1 + 3x)(1 – 21x).


3. Factorize: 6x2 – 7x – 5.

Solution:

6  × (-5) = -30 = (-10) × (3), and -10 + 3 = - 7 (= coefficient of x).

Therefore, 6x2 – 7x – 5 = 6x2 – 10x + 3x – 5

                                   = 2x(3x – 5) + 1(3x – 5)

                                   = (3x – 5)(2x + 1)


4. Factorize: 30m2 + 103mn – 7n2

Solution:

30  × (-7) = -210 = (105) × (-2), and 105 + (-2) = 103 (= coefficient of mn).

Therefore the given expression, 30m2 + 103mn – 7n2

                                          = 30m2 + 105mn – 2mn – 7n2

                                          = 15m(2m + 7n) – n(2m + 7n)

                                          = (2m + 7n)(15m – n)






9th Grade Math

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