Factoring Trinomials by Trial and Error – Method & Examples

Are you still struggling with the topic of factoring trinomials in Algebra? Well, no worries, because you are at the right place.

This article will introduce you to one of the simplest methods of factoring trinomials known as trial and error.

As the name suggests, trial and error factoring entails trying all possible factors until you find the right one.

Trial and error factoring is regarded as one of the best methods of factoring trinomials. It encourages students to develop their mathematical intuition and thus increase their conceptual understanding of the topic.

How to Unfoil trinomials?

Suppose we want to unfoil the general equation of a trinomial ax² + bx + c where a ≠ 1. Here are the steps to follow:

• Insert the factors of ax²in the 1^st positions of the two sets of brackets that represent the factors.
• Also, insert the possible factors of c into the 2^nd positions of brackets.
• Identify both the inner and outer products of the two sets of brackets.
• Keep on trying different factors until the sum of the two factors is equal to “bx.”

NOTE:

• If c is positive, both factors will have the same sign as “b”.
• If c is negative, one factor will have a negative sign.
• Never put in the same parentheses’ numbers with a common factor.

 

Trial and error factoring

Trial and error factoring, which is also referred to as reverse foil or unfoiling, is a method of factoring trinomials built upon different techniques such as foil, factoring by grouping, and some other concepts of factoring trinomials with a leading coefficient of 1.

Example 1

Use trial and error factoring to solve 6x² – 25x + 24

Solution

Paired factors of 6x² are x (6x) or 2x (3x), therefore our parentheses will be;

(x – ?) (6x – ?) or (2x – ?) (3x – ?)

Replace “bx” with possible paired factors of c. Try all paired factors of 24 that will produce -25 The possible choices are (1 & 24, 2 & 12, 3 & 8, 4 & 6). Therefore, the correct factoring is;

6x² – 25x + 24 ⟹ (2x – 3) (3x – 8)

Example 2

Factor x² – 5x + 6

Solution

The factors of the first term x², are x and x. Therefore, insert x in the first position of each parentheses.

x² – 5x + 6 = (x – ?) (x – ?)

Since last term is 6, therefore the possible choices of factors are:

(x + 1) (x + 6)
(x – 1) (x – 6)
(x + 3) (x + 2)
(x – 3) (x – 2)

The correct pair which gives -5x as the middle term is (x – 3) (x – 2). Hence,

(x – 3) (x – 2) is the answer.

Example 3

Factor x² – 7x + 10

Solution

Insert the factors of the first term in the first position of each parentheses.

⟹ (x -?) (x -?)

Try the possible pair of factors of the 10;

⟹ (-5) + (-2) = -7

Now replace the question marks in the parentheses with these two factors

⟹ (x -5) (x -2)

Hence, the correct factoring of x² – 7x + 10 is (x -5) (x -2)

Example 4

Factor 4x² – 5x – 6

Solution

(2x -?) (2x +?) and (4x -?) (x +?)

Try the possible pair of factors;

6 x² − 2x – 151 & 6, 2 & 3, 3 & 2, 6 & 1

Since the correct pair 3 and 2, therefore, (4x – 3) (x + 2) is our answer.

Example 5

Factor the trinomial x² − 2x – 15

Solution

Insert x in the first position of each parentheses.

(x -?) (x +?)

Find two numbers whose product and sum are -15 and -2, respectively. By trial and error, the possible combinations are:

15 and -1;

-1 and 15;

5 and -3;

-5 and 3;

Our correct combination is – 5 and 3. Therefore;

x² − 2x – 15 ⟹ (x -5) (x +3)

How to factor trinomials by grouping?

We can also factor trinomials by using a method of grouping. Let’s walk through the following steps to factor ax² + bx + c where a ≠1:

• Find the product of the leading coefficient “a” and the constant “c.”

⟹ a * c = ac

• Look for the factors of the “ac” that add to coefficient “b.”
• Rewrite bx as a sum or difference of the factors of ac that add to b.
• Now factor by grouping.

Example 6

Factor the trinomial 5x² + 16x + 3 by grouping.

Solution

Find the product of the leading coefficient and the last term.

⟹ 5 *3 = 15

Perform trial and error to find pair factors of 15 whose sum is the middle term (16). The correct pair is 1 and 15.

Rewrite the equation by replacing the middle term 16x by x and 15x.

5x² + 16x + 3⟹5x² + 15x + x + 3

Now, factor out by grouping

5x² + 15x + x + 3 ⟹ 5x (x + 3) + 1(x + 3)

⟹ (5x +1) (x + 3)

Example 7

Factor 2x² – 5x – 12 by grouping.

Solution

2x² – 5x – 12

= 2x² + 3x – 8x – 12

= x (2x + 3) – 4(2x + 3)

= (2x + 3) (x – 4)

Example 8

Factor 6x² + x – 2

Solution

Multiply the leading coefficient a and the constant c.

⟹ 6 * -2 = -12

Find two numbers whose product and sum are -12 and 1 respectively.

⟹ – 3 * 4

⟹ -3 + 4 = 1

Rewrite the equation by replacing the middle term -5x by -3x and 4x

⟹ 6x² -3x + 4x -2

Finally, factor out by grouping

⟹ 3x (2x –  1) + 2(2x – 1)

⟹ (3x + 2) (2x – 1)

Example 9

Factor 6y² + 11y + 4.

Solution

6y² + 11y + 4 ⟹ 6y² + 3y + y + 4

⟹ (6y² + 3y) + (8y + 4)

⟹ 3y (2y + 1) + 4(2y + 1)

= (2y + 1) (3y + 4)