The FOIL process changes a product of 2 binomials into a polynomial. The reverse
process starts

with a polynomial and finds the 2 binomials whose product will be that
polynomial. The process of

changing a polynomial into a product of 2 binomials is called **factoring a
polynomial.** This unit will

look at factoring two types of trinomials.

A second degree **Trinomial** that has1x^2 as the first term and
**ends with a positive number.**

A second degree **Trinomial **that has1x^2 as the first term
**ends with a negative number.**

**
**The technique for factoring each type of trinomial is similar. We will start
with the first type.

Factoring **Trinomials** that start with a1x^2 and

**end** with a **positive last term into**(x ± a ) (x ± b)

The results of the FOIL of
(x + 5) (x + 3) i s a trinomial with a first, middle and last term.

The **First Term of the trinomial** is
x ^2 . This first term comes from the products of the two first terms

of the binomials(first_{1} ± last_{1}) (first_{1} ± last _{2} ). The only way get a product of
x ^2 in the trinomial is to

have both first terms of the binomials be x.
(x __ ) (x __ )

The** Last Term of the trinomial** is **positive **15. It comes from the product of the
two last terms of

of the binomials
(x ± last_{1} ) (x ± last _{2}) . If the last term of the trinomial is **positive** 15
then the last two

terms in the binomials
(x ± last_{1}) (x ± last_{2} ) could be +3 and +5 or -3 and -5 but we don't know

which one to use.

In either case when you FOIL the two middle terms of the trinomial will have the
**same signs** and the

middle term of the trinomial will be found by **adding** the two last terms of the
binomials.

The** Middle Term of the trinomial** is + 8. It comes from** adding** the last two terms
of the binomials

(x ± last_{1}) (x ± last_{2} ). Only the +3 and +5 **add** to give +8 s so the last two
terms in the binomials

must be +3 and +5.

The binomials whose product is
x ^2 + 8x +15 must be
(x + 3) (x + 5) so

(x + 3) (x + 5)** are the factors of**
x ^2 + 8x +15

Factoring trinomials that start with a
1x^2 and **end** with a with a **positive last term** into
(x ± a ) (x ± b)

Factor x^2 + 10x + 24 into (x ± a ) (x ± b)

The two last terms of
(x ± last_{1}) (x ± last_{2} ) must **multiply** to give 24 and **add** to give +10

**Step 1:** The only two first terms that multiply to give x^2 are x and x. Put them
in the **first **places in

both of the parenthesis.

**Step 2:** List all the ways the last term can be **multiplied to give 24**.

**Step 3:** Find the pair of products that will **Add to give + 10.** (**+ 4 and + 6
**work)

**Step 4:** Put these two terms into the last places in the parenthesis.
(x + 4) (x +6)

**Step 5: **Check your answer by FOIL.

Factor x ^2 −11x + 30 into (x ± a ) (x ± b)

The two last terms of
(x ± last_{1}) (x ± last_{2} ) must multiply to give 30 and add to give –11

**Step 1**: The only two first terms that multiply to give x^2 are x and x. Put them
in the** first **places in

both of the parenthesis.

**Step 2:** List all the ways the last term can be **multiplied to give 30.**

**Step 3:** Find the pair of products that will **Add to give – 11** (**– 5 and – 6** work)

**Step 4:** Put these two terms into the **last **places in the parenthesis.
(x −5) (x − 6)

**Step 5:** Check your answer by FOIL.

Factoring trinomials that start with a
1x^2 and **end** with a with a **positive last term** into
(x ± a ) (x ± b)

Example 1 |
Example 2 |
Example 3 |

last terms must multiply to 20 | last terms must multiply to 18 | last terms must multiply to 12 |

last terms must add to – 9 | last terms must add to + 9 | last terms must add to –7 |

–4 and –5 add to – 9 | + 3 and + 6 add to + 9 | –3 and –4 add to –7 |

(x − 4) (x −5) | (x + 3) (x + 6) | (x − 3) (x − 4) |

Example 4 |
Example 5 |
Example 6 |

last terms must multiply to 20 | last terms must multiply to 18 | last terms must multiply to 12 |

last terms must add to +12 | last terms must add to – 19 | last terms must add to + 8 |

+ 2 and + 10 add to + 12 | – 1 and – 18 add to – 19 | + 2 and + 6 add to + 8 |

(x + 2) (x +10) | (x −1) (x −18) | (x + 2) (x + 6) |

Example 7 |
Example 8 |
Example 9 |

last terms must multiply to 20 | last terms must multiply to 18 | last terms must multiply to 12 |

last terms must add to – 21 | last terms must add to +11 | last terms must add to –13 |

– 1 and – 20 add to – 21 | + 2 and + 9 add to +11 | –1 and –12 add to –13 |

(x −1) (x −20) | (x + 2) (x + 9) | (x −1) (x −12) |

Factoring **Trinomials** that start with a
1x^2 and

**end **with a **negative last term into**
(x ± a ) (x ± b)

The last section discussed how to factor Trinomials with a positive last term
like
x ^2 + 8x +15

The rules for factoring Trinomials with a **positive** last term were

1. The **product of the last terms in**
(x ± last_{1}) (x ± last_{2} ) must be +15

2. The **sum of the last terms in**
(x ± last_{1}) (x ± last_{2} ) must be –8

The rules for Factoring Trinomials with a **negative** last term are similar

**Factoring Trinomials:** Factor
x ^2 − 2x −15 into
(x ± a ) (x ± b) .

To determine a rule for finding a and b we FOIL (x − b)(x + a) to get

using the distributive property on the middle terms we get

The **First Term of the trinomial** is
x ^2 . This first term comes from the products of the two first terms

of the binomials
(first_{1} ± last_{1}) (first_{1} ± last _{2} ). The only way get a product of
x ^2 in the trinomial is to

have both first terms of the binomials be x.
(x __ ) (x __ )

The **Last Term of the trinomial** is **negative** 15. It comes from the product of the
two last terms of

of the binomials
(x ± last_{1} ) (x ± last _{2}) . If the last term of the trinomial is negative 15
then the two last

terms in the binomials
(x ± last_{1}) (x ± last_{2} ) could be +3 and –5 or –3 and ≠5 but we don't know

which one to use.

In either case when you FOIL the two middle terms of the trinomial will have
different **signs** and the

middle term of the trinomial will be found by subtracting the two last terms of
the binomials.

The **Middle Term of the trinomial **is –2. It comes from
**subtracting** the last two
terms of the

binomials
(x ± last_{1}) (x ± last_{2} ). Only the +3 and –5 **subtract **to give –8 s so the two
last terms in

the binomials must be +3 and –5.

The binomials whose product is
x ^2 − 2x −15 must be
(x + 3) (x − 5)

(x + 3) (x − 5) **are the factors of**
x ^2 − 2x −15

Factoring trinomials that start with a
1x^2 and **end **with a with a negative **last term**