Factorization of Quadratic Trinomials

In factorization of quadratic trinomials there are two forms: 

(i) First form: x2 + px + q

(ii) Second form: ax2 + bx + c

(i) Factorization of trinomial of the form x^2 + px + q:

Suppose we are given a quadratic trinomial x2 + px + q.

Then, we use the identity:

x2 + (a + b) × + ab = (x + a)(x + b).


Solved examples on factorization of quadratic trinomials of the form x^2 + px + q:



1. Factorize the algebraic expression of the form x2 + px + q:

(i) x2 - 7x + 12

Solution:

The given expression is x2 - 7x + 12

Find two numbers whose sum = -7 and product = 12

Clearly, such numbers are (-4) and (-3).

Therefore, x2 - 7x + 12 = x2 - 4x - 3x + 12

                                = x(x - 4) -3 (x - 4) 

                                = (x - 4)(x - 3).


(ii) x2 + 2x - 15

Solution:

The given expression is x2 + 2x - 15

To factorize the given quadratic trinomial, we have to find two numbers a and b, such that a + b = 2 and ab = -15

Clearly, 5 + (-3) = 2 and 5 × (-3) = -15

Therefore such numbers are 5 and -3

Now, splitting the middle term 2x of the given quadratic trinomial x2 + 2x -15, we get,

x2 + 5x - 3x -15

= x(x +5) - 3(x + 5)

= (x + 5) (x - 3)

 

(ii) Factorization of trinomial of the form ax^2 + bx + c:

In order to factorize the expression ax2 + bx + c we have to find the two numbers p and q, such that

p + q = b and p × q = ac


Solved examples on factorization of quadratic trinomials of the form ax^2 + bx + c:

2. Factorize the algebraic expression of the form ax2 + bx + c:

(i) 15x2 - 26x + 8

Solution:

The given expression is 15x2 - 26x + 8.

Find two numbers whose sum = -26 and product = (15 × 8) = 120.

Clearly, such numbers are -20 and -6.

Therefore, 15x2 - 26x + 8 = 15x2 - 20x - 6x + 8

                                   = 5x(3x - 4) - 2(3x - 4) 

                                   = (3x - 4)(5x - 2).


(ii) 3q2 – q – 4

Solution:

Here, two numbers m and n are such that their sum m + n = -1 and their product m × n = 3 × (-4) i.e. m × n = - 12

Clearly, such numbers are -4 and 3

Now, splitting the middle term –q of the given quadratic trinomial 3q2 – q – 4 we get,

3q2 - 4q + 3q – 4

= q(3q – 4) + 1(3q – 4)

= (3q – 4)(q + 1)





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