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# Factoring Trinomials by Grouping

## Factoring a Trinomial of the Form ax2 + bx + c by Grouping

In this method, we rewrite the middle term of the trinomial using two terms. Then we factor by grouping.

 For example, weâ€™ll use this method to factor 15x2 - 16x + 4. Write the middle term, -16x, as -10x - 6x. Group the first two terms and group the last two terms. Factor 5x out of the first group; factor -2 out the second group. Finally, factor out (3x - 2). 15x2 - 16x + 4= 15x2 - 10x - 6x + 4 = (15x2 - 10x) + (-6x + 4) = 5x(3x - 2) + (-2)(3x - 2) = (3x - 2)(5x - 2)

The key to this method is knowing how to rewrite the middle term of the trinomial. The following procedure describes a way to do this.

Procedure â€” To Factor ax2 + bx + c by Grouping

Step 1 Factor out common factors (other than 1 or -1).

Step 2 List the values of a, b, and c. Then find two integers whose product is ac and whose sum is b. If no two such integers exist then the trinomial is not factorable over the integers.

Step 3 Replace the middle term, bx, with a sum or difference using the two integers found in Step 2.

Step 4 Factor by grouping. To check the factorization, multiply the binomial factors.

Note:

This method also works when a = 1. However, in those cases The Product-Sum method requires fewer steps.

Example 1

Factor: 6x2 + 7x + 2.

Solution

Step 1 Factor out common factors (other than 1 or -1).

There are no common factors other than 1 and -1.

Step 2 List the values of a, b, and c. Then find two integers whose product is ac and whose sum is b.

6x2 + 7x + 2 has the form ax2 + bx + c where a = 6, b = 7, and c = 2.

The product ac is 6 Â· 2 = 12.

Thus, find two integers whose product, ac, is 12 and whose sum, b, is 7.

â€¢ Since their product is positive, the integers must have the same sign.

â€¢ Since their sum is also positive, the integers must both be positive.

Here are the possibilities:

 Product 1 Â· 12 2 Â· 6 3 Â· 4 Sum 13 8 7
The integers 3 and 4 satisfy the requirements that their product is 12 and their sum is 7.
 Step 3 Replace the middle term, bx, with a sum or difference using the two integers found in Step 2. 6x2 + 7x + 2 Replace 7x with 3x + 4x. = 6x2 + 3x + 4x + 2 Step 4 Factor by grouping. Group the first pair of terms and group the second pair of terms. Factor 3x out of the first group; factor 2 out of the second group. Factor out the common factor (2x + 1). = (6x2 + 3x) + (4x + 2)= 3x(2x + 1) + 2(x + 1) = (2x + 1)(3x + 2)

The result is: 6x2 + 7x + 2 = (2x + 1)(3x + 2). You can multiply to check the factorization. We leave the check to you.

Note:

We replaced 7x with 3x + 4x.

If we switch 3x and 4x, we can still group and factor:

= 6x2 + 4x + 3x + 2

= (6x2 + 4x) + (3x + 2)

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

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

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