Factorization of Trinomials of the Form x² + bx + c

A trinomial is an algebraic expression containing three terms.

For example, ax² + b  + c is a trinomial because it has three terms i.e. ax², bx, c  when a = 1 this expression becomes x² + b x  + c,  a simple trinomial.

Some quadratic expressions can be factorized by splitting the middle term.

Therefore, to factorize a trinomial of the form ax² + bx + c, look for pairs of factors of the constant term c that add up to b i.e. the coefficient of x.

Worked Example 6.3.1:

Factorize:

a. x² + 7x  + 10
b. x² + 3x  + 2
c. x² + 8x  + 15
d. y² + 9y + 18
e. 2x² + 13x + 6

Solution

a. x² + 7x  + 10

i. First multiply The first and the last term i.e. a × c

ii. Look for the factors of 10x², two factors when they are multiplied give 10x² and when added they give 7x, which is the middle term.

Factors of 10: 1, 2, 5 and 10

Factors of 10x² that add up to 7x are 2x and 5x

that is:

2x × 5x = 10x²
2x + 5x = 7x

⇒ \( \scriptsize 2x \: \times \: 5x = 10x^2 \)

⇒ \( \scriptsize 2x \: + \: 5x = 7x \)

So replace 7x with 2x + 5x

⇒  x² + 2x  + 5x + 10

Factorize by grouping

⇒ (x² + 2x ) + (5x + 10)

⇒ x (x + 2 ) + 5 (x + 2)

= (x + 5 )(x + 2)

b.   x² + 3x  + 2

Factors of 2: 1 and 2

∴ Factors of 2x² that add up to 3x are 2x and 1x

So replace 3x with 2x + 1x

⇒  x² + 2x  + 1x + 2

Factorize by grouping

⇒ (x² + 2x ) + (x  + 2)

⇒ x(x + 2) + 1(x  + 2)

= (x  + 1)(x  + 2)

c.   x² + 8x  + 15

Factors of 15: 1, 3, 5 and 15.

⇒  x² + 3x + 5x + 15

 ⇒ (x ² + 3x) + (5x + 15)

  ⇒ x(x + 3) + 5(x + 3)

= (x + 5)(x + 3)

d. y² + 9y + 18

Factors of 18: 1, 2, 3, 6, 9 and 18

⇒ y² + 3y + 6y + 18

⇒ (y² + 3y) + (6y + 18)

⇒ y (y + 3) + 6(y + 3)

= (y + 6)( y + 3)

e.  2x² + 13x + 6

Factors of 6: 1, 2, 3, 4, 6, and 12.

⇒  2x² + 12x + x + 6

⇒ (2x² + 12x ) + (x  + 6)

⇒ 2x (x + 6) + 1(x + 6)

= (2x + 1)(x + 6)

Worked Example 6.3.2:

Factorize:

i. 4x² – 3x – 22
ii. 2e² – 3e + 1

Solution

i. 4x² – 3x – 22

Factors of 88: 1, 2, 4, 8, 11, 22, 44, and 88

+ 8x – 11x = -3x or -11x + 8x = -3x

⇒ 4x² + 8x – 11x – 22

⇒ (4x² + 8x ) – (11x + 2)

⇒ 4x (x + 2) – 11 (x  + 2)

= (4x – 11) (x  + 2)

ii. 2e² – 3e + 1

⇒ 2e² – e – 2e + 1

⇒ (2e² – e) – (2e – 1)

⇒ e(2e – 1) – 1(2e – 1)

= (e – 1)(2e – 1)