### Learning Objectives

- Apply an algorithm to rewrite a trinomial as a four term polynomial
- Use factoring by grouping to factor a trinomial
- Use a shortcut to factor trinomials of the form [latex]x^2+bx+c[/latex]

In the last section we introduced the technique of factoring by grouping as a means to be able to factor a trinomial. Now we will actually get to the work of starting with a three term polynomial, and rewriting it as a four term polynomial so it can be factored.

We will start with factoring trinomials of the form [latex]x^{2}+bx+c[/latex] that don’t have a coefficient in front of the [latex]x^2[/latex] term.

Remember that when [latex]\left(x+2\right)[/latex] and [latex]\left(x+5\right)[/latex], are multiplied, the result is a four term polynomial and then it is simplified into a trinomial:

[latex]\left(x+2\right)\left(x+5\right)=x^2+5x+2x+10=x^2+7x+10[/latex]

Factoring is the reverse of multiplying, so let’s go in reverse and factor the trinomial [latex]x^{2}+7x+10[/latex]. The individual terms [latex]x^{2}[/latex], [latex]7x[/latex], and 10 share no common factors. If we rewrite the middle term as the sum of the two terms [latex]7x=5x+2x[/latex] then we can use the grouping technique:

[latex](x^{2}+5x)+(2x+10)[/latex]

Factor each pair: [latex]\begin{array}{l}x\underbrace{\left(x+5\right)}+2\underbrace{\left(x+5\right)}\\\text{common binomial factor}\end{array}[/latex]

Then pull out the common binomial factor: [latex]\left(x+5\right)\left(x+2\right)[/latex]

What would have happened if we had rewritten [latex]7x[/latex] as [latex]6x+x[/latex]?

[latex](x^{2}+6x)+(x+10)[/latex]

Factor each pair: [latex]x\left(x+6\right)+1\left(x+10\right)[/latex]

Then we don’t have a common factor of [latex]\left(x+5\right)[/latex] like we did before. There is a method to the madness of choosing how to rewrite the middle terms so that you will end up with a common binomial factor.

The following is a summary of the method, then we will show some examples of how to use it.

### Factoring Trinomials in the form [latex]x^{2}+bx+c[/latex]

To factor a trinomial in the form [latex]x^{2}+bx+c[/latex], find two integers, *r* and *s*, whose product is *c *and whose sum is *b*.

[latex]\begin{array}{l}r\cdot{s}=c\\\text{ and }\\r+s=b\end{array}[/latex]

Rewrite the trinomial as [latex]x^{2}+rx+sx+c[/latex] and then use grouping and the distributive property to factor the polynomial. The resulting factors will be [latex]\left(x+r\right)[/latex] and [latex]\left(x+s\right)[/latex].

For example, to factor [latex]x^{2}+7x+10[/latex], you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).

Look at factor pairs of 10:1 and 10, 2, and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite [latex]7x[/latex] as [latex]2x+5x[/latex], and continue factoring as in the example above. Note that you can also rewrite [latex]7x[/latex] as [latex]5x+2x[/latex]. Both will work.

Let’s factor the trinomial [latex]x^{2}+5x+6[/latex]. In this polynomial, the *b* part of the middle term is 5 and the *c* term is 6. A chart will help us organize possibilities. On the left, list all possible factors of the *c* term, 6; on the right you’ll find the sums.

Factors whose product is 6 | Sum of the factors |
---|---|

[latex]1\cdot6=6[/latex] | [latex]1+6=7[/latex] |

[latex]2\cdot3=6[/latex] | [latex]2+3=5[/latex] |

There are only two possible factor combinations, 1 and 6, and 2 and 3. You can see that [latex]2+3=5[/latex]. So [latex]2x+3x=5x[/latex], giving us the correct middle term.

### Example

Factor [latex]x^{2}+5x+6[/latex].

Note that if you wrote [latex]x^{2}+5x+6[/latex] as [latex]x^{2}+3x+2x+6[/latex] and grouped the pairs as [latex]\left(x^{2}+3x\right)+\left(2x+6\right)[/latex]; then factored, [latex]x\left(x+3\right)+2\left(x+3\right)[/latex], and factored out [latex]x+3[/latex], the answer would be [latex]\left(x+3\right)\left(x+2\right)[/latex]. Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well; they are equivalent answers.

In the following video, we present another example of how to use grouping to factor a quadratic polynomial.

Finally, let’s take a look at the trinomial [latex]x^{2}+x–12[/latex]. In this trinomial, the *c* term is [latex]−12[/latex]. So look at all of the combinations of factors whose product is [latex]−12[/latex]. Then see which of these combinations will give you the correct middle term, where *b* is 1.

Factors whose product is [latex]−12[/latex] | Sum of the factors |
---|---|

[latex]1\cdot−12=−12[/latex] | [latex]1+−12=−11[/latex] |

[latex]2\cdot−6=−12[/latex] | [latex]2+−6=−4[/latex] |

[latex]3\cdot−4=−12[/latex] | [latex]3+−4=−1[/latex] |

[latex]4\cdot−3=−12[/latex] | [latex]4+−3=1[/latex] |

[latex]6\cdot−2=−12[/latex] | [latex]6+−2=4[/latex] |

[latex]12\cdot−1=−12[/latex] | [latex]12+−1=11[/latex] |

There is only one combination where the product is [latex]−12[/latex] and the sum is 1, and that is when [latex]r=4[/latex], and [latex]s=−3[/latex]. Let’s use these to factor our original trinomial.

### Example

Factor [latex]x^{2}+x–12[/latex].

In the above example, you could also rewrite [latex]x^{2}+x-12[/latex] as [latex]x^{2}– 3x+4x–12[/latex] first. Then factor [latex]x\left(x – 3\right)+4\left(x–3\right)[/latex], and factor out [latex]\left(x–3\right)[/latex] getting [latex]\left(x–3\right)\left(x+4\right)[/latex]. Since multiplication is commutative, this is the same answer.

### Factoring Tips

Factoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored.

While there is no foolproof way to find the right combination on the first guess, there are some tips that can ease the way.

### Tips for Finding Values that Work

When factoring a trinomial in the form [latex]x^{2}+bx+c[/latex], consider the following tips.

Look at the *c* term first.

- If the
*c*term is a positive number, then the factors of*c*will both be positive or both be negative. In other words,*r*and*s*will have the same sign. - If the
*c*term is a negative number, then one factor of*c*will be positive, and one factor of*c*will be negative. Either*r*or*s*will be negative, but not both.

Look at the *b* term second.

- If the
*c*term is positive and the*b*term is positive, then both*r*and*s*are positive. - If the
*c*term is positive and the*b*term is negative, then both*r*and*s*are negative. - If the
*c*term is negative and the*b*term is positive, then the factor that is positive will have the greater absolute value. That is, if [latex]|r|>|s|[/latex], then*r*is positive and*s*is negative. - If the
*c*term is negative and the*b*term is negative, then the factor that is negative will have the greater absolute value. That is, if [latex]|r|>|s|[/latex],*r*is negative and*s*is positive.

After you have factored a number of trinomials in the form [latex]x^{2}+bx+c[/latex], you may notice that the numbers you identify for *r* and *s* end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews the three problems you have seen so far.

Trinomial | [latex]x^{2}+7x+10[/latex] | [latex]x^{2}+5x+6[/latex] | [latex]x^{2}+x-12[/latex] |
---|---|---|---|

r and s values |
[latex]r=+5,s=+2[/latex] | [latex]r=+2,s=+3[/latex] | [latex]r=+4,s=–3[/latex] |

Factored form | [latex]\left(x+5\right)\left(x+2\right)[/latex] | [latex]\left(x+2\right)\left(x+3\right)[/latex] | [latex](x+4)(x–3)[/latex] |

## The Shortcut

Notice that in each of the examples above, the *r* and *s* values are repeated in the factored form of the trinomial. So what does this mean? It means that in trinomials of the form [latex]x^{2}+bx+c[/latex] (where the coefficient in front of [latex]x^{2}[/latex] is 1), if you can identify the correct *r* and *s* values, you can effectively skip the grouping steps and go right to the factored form. For those of you that like shortcuts, let’s look at some examples where we use this idea.

In the next two examples, we will show how you can skip the step of factoring by grouping and move directly to the factored form of a product of two binomials with the r and s values that you find. The idea is that you can build factors for a trinomial in this form: [latex]x^2+bx+c[/latex] by finding r and s, then placing them in two binomial factors like this:

[latex]\left(x+r\right)\left(x+s\right)\text{ OR }\left(x+s\right)\left(x+r\right)[/latex]

### Example

Factor: [latex]y^2+6y-27[/latex]

We will show one more example so you can gain more experience.

### Example

Factor: [latex]-m^2+16m-48[/latex]

In the following video, we present two more examples of factoring a trinomial using the shortcut presented here.

## Summary

Trinomials in the form [latex]x^{2}+bx+c[/latex] can be factored by finding two integers, *r* and *s*, whose sum is *b* and whose product is *c.* Rewrite the trinomial as [latex]x^{2}+rx+sx+c[/latex] and then use grouping and the distributive property to factor the polynomial.