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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.

In this article, you will be introduced 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 because, it encourages students to develop their mathematical intuition and thus increasing 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 1st positions of the two sets of brackets that represent the factors.
  • Also insert the possible factors of c into the 2ng positions of brackets.
  • Identify the both the inner and outer products of the two sets of brackets.
  • Keep on trying different factors until the sum of the two factors are 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 that is 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 by 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?

Trinomials can also be factored 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)

Practice Questions

Solve the following trinomials by any suitable method:

  1. 3x2– 8x – 60
  2. x2– 21x + 90
  3. x2 – 22x + 117
  4. x2 – 9x + 20
  5. x2 + x – 132
  6. 30a2+ 57ab – 168b2
  7. x2 + 5x – 104
  8. y2 + 7y – 144
  9. z2+ 19z – 150
  10. 24x2 + 92xy + 60y2
  11. y2 + y – 72
  12. x2+ 6x – 91
  13. x2– 4x -7
  14. x2 – 6x – 135
  15. x2– 11x – 42
  16. x2 – 12x – 45
  17. x2 – 7x – 30
  18. x2 – 5x – 24
  19. 3x2 + 10x + 8
  20. 3x2 + 14x + 8
  21. 2x2 + x – 45
  22. 6x2 + 11x – 10
  23. 3x2 – 10x + 8
  24. 7x2+ 79x + 90

Answers

  1. (3x + 10) (x – 6)
  2. (x – 15) (x – 6)
  3. (x – 13) (x – 9)
  4. (x – 5) (x – 4)
  5. (x + 12) (x – 11)
  6. 3(5a – 8b) (2a + 7b)
  7. (x + 13) (x – 8)
  8. (y + 16) (y – 9)
  9. (z + 25) (z – 6)
  10. 4(x + 3y) (6x + 5y)
  11. (y + 9) (y – 8)
  12. (x + 13) (x – 7)
  13. (x – 11) (x + 7)
  14. (x – 15) (x + 9)
  15. (x – 14) (x + 3)
  16. (x – 15) (x + 3)
  17. (x – 10) (x + 3)
  18. (x – 8) (x + 3)
  19. (x + 2) (3x + 4)
  20. (x + 4) (3x + 2)
  21. (x + 5) (2x – 9)
  22. (2x + 5) (3x – 2)
  23. (x – 2) (3x – 4)
  24. (7x + 9) (x + 10)

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