# Factoring Trinomials into Binomials

Trinomials, in the form of $x^2 +bx+c$ and $ax^2+bx+c$, can often be factored into binomials. Factoring trinomials into binomials can make them much easier to solve (see lesson on solving quadratic equations by factoring).

There are many processes for factoring trinomials into binomials. We teach the one below because it eliminates guesswork and makes factoring trinomials in the form $ax^2+bx+c$ just as easy as factoring trinomials in the form $x^2 +bx+c$.

**We call this process the "Reverse FOIL" process.** By going through the reverse foil process, you will first break a trinomial into the four terms you get when you FOIL two binomials. You will then factor by grouping to find two binomials.

*Let's quickly review the FOIL process:*

Foil is an acronym that stands for first, outer, inner, last.

The basic process of using foil, is to take the two binomials and first multiply the "first" terms of each binomial together.

$\eqalign{(\color{red}{x}\color{black}+3)(\color{red}{x}\color{black}-5)\\\text{First:} \quad x \times x =\color{red}{ x^2}\\(x+3)(x-5)=\color{red}x^2\color{black} \text{...}}$

The basic process of using foil, is to take the two binomials and first multiply the "outer" terms of each binomial together.

$\eqalign{(\color{blue}{x}\color{black}+3)(x\color{blue}{-5}\color{black})\\\text{Outer:} \quad x \times -5 = \color{blue}{-5x}\\(x+3)(x-5)=\color{red}x^2\color{blue}-5x\color{black} \text{...}}$

The basic process of using foil, is to take the two binomials and first multiply the "Inner" terms of each binomial together.

$\eqalign{(x\color{green}{+3}\color{black})(\color{green}{x}\color{black}-5)\\\text{Inner:} \quad 3 \times x = \color{green}3x\\(x+3)(x-5)=\color{red}x^2 \color{blue}-5x\color{green}+3x \color{black}\text{...}}$

The basic process of using foil, is to take the two binomials and first multiply the "Last" terms of each binomial together.

$\eqalign{(x\color{purple}{+3}\color{black})(x\color{purple}{-5}\color{black})\\\text{Last:} \quad 3 \times -5 =\color{purple}{ -15}\\(x+3)(x-5)=\color{red}x^2\color{blue}-5x\color{green}+3x\color{purple}-15\\\text{Combine like terms:}\\x^2-2x-15}$

*Let's perform Reverse FOIL on that trinomial: $x^2-2x-15$*

First, let's get our labels correct, trinomials in standard form are written as: $ax^2+bx+c$.

In the equation above, $a=1$, $b=-2$, $c-15$.

First make a table with 2 columns. One column is the multiplication term in your trinomial and the other is the addition term.

Multiplication Term (First x Last) | Addition Term (Inner + Outer) |

= -15 | =-2 |

In the multiplication term column, write all of the pairs of factors that equal -15. Start with a $1 \times 15$, then go up one by one, so that you don't miss any factors. You are looking for a pair of factors that also adds up to the addition term. (In this case, because the Addition Term is negative, we know that the larger of the two factors is the negative number).

Multiplication Term (First x Last) | Addition Term (Inner + Outer) |

= -15 | =-2 |

$1 \times - 15$ | 14 NO |

$3 \times -5$ | -2 YES |

**3 and -5 work, they multiply to the Multiplication Term and add up to the Addition term. **

Now, we rewrite the trinomial, by breaking the $bx$ term into the two terms that combined in that term during the FOIL process.

$\eqalign{x^2-2x-15\\x^2+3x-5x-15}$

Now, we are going to do "factoring by grouping." We want to group the first two terms together and the second two terms together.

$\eqalign{x^2-2x-15\\x^2+3x-5x-15\\(x^2+3x)+(-5x-15)}$

Next, we want to factor a variable out of each of the binomials we just made. We're looking for variables to factor out that will leave us with two identical binomals. The first one is easy, only $x$ will factor out.

$\eqalign{x^2-2x-15\\x^2+3x-5x-15\\(x^2+3x)+(-5x-15)\\x(x+3)+(-5x-15)}$

What can we factor out of the second binomial to leave $(x+3)$? How about $-5$?

$\eqalign{x^2-2x-15\\x^2+3x-5x-15\\(x^2+3x)+(-5x-15)\\x(x+3)+(-5x-15)\\x(x+3)-5(x+3)}$

Finally, you group the two factors you factored out into one binomial, and multiply that by the other common binomial. And now, you have factored your trinomial!

$\eqalign{x^2-2x-15\\x^2+3x-5x-15\\(x^2+3x)+(-5x-15)\\x(x+3)+(-5x-15)\\x(x+3)-5(x+3)\\(x-5)(x+3)}$

This process may seem like a lot of work. But it's a great process when you have a coefficient on your $x^2$ term (in other words, when $a$ is not 1).

*Let's try the process on a trinomial in the form of $ax^2+bx+c$:* $6x^2+x-12$

First, let's get our labels correct, trinomials in standard form are written as: $ax^2+bx+c$.

In the equation above, $a=6$, $b=1$, $c-12$.

First make a table with 2 columns. One column is the multiplication term in your trinomial and the other is the addition term.

Multiplication Term (First x Last) | Addition Term (Inner + Outer) |

= -72 | =1 |

In the multiplication term column, write all of the pairs of factors that equal -72. Start with a $1 \times 72$, then go up one by one, so that you don't miss any factors. You are looking for a pair of factors that also adds up to the addition term. (In this case, because the Addition Term is positive, we know that the larger of the two factors is the positive number).

Multiplication Term (First x Last) | Addition Term (Inner + Outer) |

= -72 | =1 |

$-1 \times 72$ | 71 No |

$-2 \times 36$ | 34 No |

$-3 \times 24$ | 21 No |

$-4 \times 18$ | 14 No |

$-6 \times 12$ | 6 No |

$-8 \times 9$ | 1 YES |

**-8 and 9 work, they multiply to the Multiplication Term and add up to the Addition term. **

Now, we rewrite the trinomial, by breaking the $bx$ term into the two terms that combined in that term during the FOIL process.

$\eqalign{6x^2+x-12\\6x^2-8x+9x-12}$

Now, we are going to do "factoring by grouping." We want to group the first two terms together and the second two terms together.

$\eqalign{6x^2+x-12\\6x^2-3x+4x-12\\(6x^2-8x)+(9x-12)}$

Next, we want to factor a variable out of each of the binomials we just made. We're looking for variables to factor out that will leave us with two identical binomials. But, as the binomials are currently organized, I can't pull anything out to get the same binomial out of each (see attempt in last line below).

$\eqalign{6x^2+x-12\\6x^2-3x+4x-12\\(6x^2-8x)+(9x-12)\\2x(3x-4)+(9x-12)}$

What can we factor out of the second binomial to leave $(3x-4)$? How about $3$?

$\eqalign{6x^2+x-12\\6x^2-3x+4x-12\\(6x^2-8x)+(9x-12)\\2x(3x-4)+3(3x-4)}$

Finally, you group the two factors you factors out into one binomial, and multiply that by the other common binomial. And now, you have factored your trinomial!

$\eqalign{6x^2+x-12\\6x^2-3x+4x-12\\(6x^2-8x)+(9x-12)\\2x(3x-4)+3(3x-4)\\(2x+3)(3x-4)}$

#### Practice Problems:

## Factoring Trinomials into Binomials

Factor the following trinomials into binomials

- $x^2+3x+2$
- $x^2+2x-15$
- $x^2-6x+8$
- $x^2-36$
- $2x^2+11x+5$
- $3x^2-7x-6$
- $6x^2-6$
- $15x^2-5x-10$
- $18x^2-66x+20$
- $35x^2-29x+6$

#### Answer Key: