Factoring Quadratic Trinomials

Introduction

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials. In mathematics, factorization (also factorisation in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. In this article we shall discuss about factoring quadratic trinomial.

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Example Problem on Factoring Trinomials

Problem 1:

Factoring trinomials

15 – 2x – x2.

Solution: Writing in the standard form,

15 – 2x – x2 = –x2 – 2x + 15

= (–1) (x2 + 2x – 15).

Here, we find –15 = 5 × –3, 5 + (–3) = 2

Hence, we get 15 – 2x – x2 = (–1) [(x+5) {x + (–3)}]

= (–1) (x +5)(x – 3)

= (x + 5) ( 3 – x).

Problem 2:

Factoring  trinomials

x2 – x – 132.

Solution: We find –132 = (–12) × (11), (–12) + 11 = –1.

Hence we get x2 – x – 132 = [x + (–12)] (x + 11) = (x – 12) (x + 11).

Next, we consider the quadratic polynomial ax2 + bx + c where a, b, c are integers and a ≠0,1. If we are able to find two integers p and q such that pq = ac and p + q = b. Then

ax2 + bx + c =`1/a ` (a2x2 + abx + ac)

=`1/a` [a2x2 + a(p+q)x + pq]= `1/a` [a2x2 + apx + aqx + pq]=`1/a` [ax (ax + p) + q(ax + p)]

=`1/a` (ax + p) (ax + q)

Thus, we are able to factorize the expression

Problem 3:

Factoring trinomial

2x2+ 7x + 3.

Solution: Here a = coefficient of x2 = 2

b = coefficient of x = 7

c = constant term = 3

We find a × c = 2 × 3 = 6 = 6 × 1, 6 + 1 = 7 = b. Hence

2x2 + 7x + 3 = 21 (2x + 6) (2x + 1) =(x+3)(2x+1).

Instead of applying the final result of the rule, we can also do the factorization by splitting the middle term and grouping as follows:

2x2 + 7x + 3 = 2x2 + (6 + 1)x + 3

= 2x2 + 6x + x + 3

= 2x(x + 3) + (1)(x+3) = (2x+1) (x+3).

Practice Problem on Factoring Quadratic Trinomials

Problem 1:

Factorize 8a2 + 2a – 3.

Answer:

(4a + 3) (2a – 1)

Problem 2:

Factorize 6 + 11/2 x + x2.

Answer:

1/2 (x + 4) (2x + 3).