"

Chapter 7: Factoring

7.3 Factoring Trinomials where a = 1

Factoring expressions with three terms, or trinomials, is a very important type of factoring to master, since this kind of expression is often a quadratic and occurs often in real life applications. The strategy to master these is to turn the trinomial into the four-term polynomial problem type solved in the previous section. The tool used to do this is central to the Master Product Method. To better understand this, consider the following example.

Example 7.3.1

Factor the trinomial x2+2x24.

Start by multiplying the coefficients from the first and the last terms. This is 124, which yields −24.

The next task is to find all possible integers that multiply to −24 and their sums.

multiply to 24sum of these integers1242321210380546026428351221024123

Look for the pair of integers that multiplies to −24 and adds to 2, so that it matches the equation that you started with.

For this example, the pair is 46, which adds to 2.

Now take the original trinomial x2+2x24 and break the 2x into 4x and 6x.

Rewrite the original trinomial as x24x+6x24.

Now, split this into two binomials as done in the previous section and factor.

x24x yields x(x4) and 6x24 yields 6(x4).

x24x+6x24 becomes x(x4)+6(x4).

x(x4)+6(x4) factors to (x4)(x+6).

x2+2x24=(x4)(x+6)

Example 7.3.2

Factor the trinomial x2+9x+18.

Start by multiplying the coefficients from the first and the last terms. This is 118, which yields 18.

The next task is to find all possible integers that multiply to 18 and their sums.

multiply to 18sum of these integers118192911369639921118119

Look for the pair of integers that multiplies to 18 and adds to 9, so that it matches the equation that you started with.

For this example, the pair is 36, which adds to 9.

Now take the original trinomial x2+9x+18 and break the 9x into 3x and 6x.

Rewrite the original trinomial as x2+3x+6x+18.

Now, split this into two binomials as done in the previous section and factor.

x2+3x yields x(x+3) and 6x+18 yields 6(x+3).

x2+3x+6x+18 becomes x(x+3)+6(x+3).

x(x+3)+6(x+3) factors to (x+3)(x+6).

x2+9x+18=(x+3)(x+6)

Please note the following is also true:

multiply to 18sum of these integers118192911369639921118119

This means that solutions can be found where the middle term is 19x, 11x, 9x, 19x, 11x or 9x.

Questions

Factor each of the following trinomials.

  1. p2+17p+72
  2. x2+x72
  3. n29n+8
  4. x2+x30
  5. x29x10
  6. x2+13x+40
  7. b2+12b+32
  8. b217b+70
  9. u28uv+15v2
  10. m23mn40n2
  11. m2+2mn8n2
  12. x2+10xy+16y2
  13. x211xy+18y2
  14. u29uv+14v2
  15. x2+xy12y2
  16. x2+14xy+45y2

Answer Key 7.3

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book