Factor Trinomials

Factor trinomials of the form $x^2+bx+c$

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.

This tells us that to factor a trinomial of the form $x^2+bx+c$, we need two factors $(x+m)$ and $(x+n)$ where $m$ and $n$ multiply to $c$ and add to $b$.

Example. Factor $x^2+11x+24$.

Write two sets of parentheses and put $x$ as the first term. The factor pairs of $24$ have sums $1+24=25$, $2+12=14$, $3+8=11$, and $4+6=10$. Use $3$ and $8$ as the last terms.

In the first example, all terms were positive. With only the middle term negative, a positive product and negative sum require two negative numbers.

Example. Factor $y^2-11y+28$.

The negative factor pairs of $28$ have sums $-1+(-28)=-29$, $-2+(-14)=-16$, and $-4+(-7)=-11$. Therefore

If the last term is negative, its factors have opposite signs. Choose them carefully to get the correct sign for the middle term. Terms must be written in descending order.

Example. Factor $2x+x^2-48$.

First reorder it as $x^2+2x-48$. The factor-pair sums include $-1+48=47$, $-2+24=22$, $-3+16=13$, $-4+12=8$, and $-6+8=2$.

Trinomials may have two variables. The $y^2$ in the last term means the second terms of the binomial factors must each contain $y$.

Example. Factor $r^2-8rs-9s^2$.

The factor-pair sums for $-9$ are $8,-8,0$. Use $1,-9$:

Some trinomials are prime. The only way to be certain is to list all possibilities and show none work.

Example. Factor $u^2-9uv-12v^2$.

The factor-pair sums for $-12$ are $-11,11,-4,4,-1,1$. None is $-9$, so the trinomial is prime.

Factor trinomials of the form $ax^2+bx+c$ using trial and error

Remember to always check for a GCF first. Sometimes, after factoring the GCF, the leading coefficient becomes $1$.

Example. Factor completely: $4x^3+16x^2-20x$.

When there is no GCF, test factor pairs. For $3x^2+5x+2$, test $x,3x$ and both orders of $1,2$. Inner and outer products show

Example. Factor $3y^2+22y+7$ using trial and error.

There is no GCF. Test the sole factor pairs $y,3y$ and $1,7$:

Thus $3y^2+22y+7=(y+7)(3y+1)$.

When the middle term is negative and the last term positive, both binomial signs must be negative.

Example. Factor $6b^2-13b+5$ using trial and error.

Testing $b,6b$ and $2b,3b$ with negative factors of $5$ gives

So $6b^2-13b+5=(2b-1)(3b-5)$.

If an expression has no GCF, neither factor can have a common factor. This eliminates some combinations.

Example. Factor $18x^2-37xy+15y^2$ using trial and error.

Test $x,18x$, $2x,9x$, and $3x,6x$ with $-y,-15y$ and $-3y,-5y$, eliminating binomials with common factors. The correct product is

If the leading coefficient is negative, so is the GCF.

Example. Factor $-10y^4-55y^3-60y^2$.

Factor trinomials of the form $ax^2+bx+c$ using the “ac” method

The “ac” method, sometimes called the grouping method, extends the preceding method. It is very structured and always works.

Example. Factor $6x^2+7x+2$ using the “ac” method.

There is no GCF. Since $ac=12$, use $3,4$, which multiply to $12$ and add to $7$. Split the middle term and factor by grouping:

Example. Factor $10y^2-55y+70$ using the “ac” method.

Factor out $5$. Inside, $ac=28$ and $-4,-7$ multiply to $28$ and add to $-11$.

Factor using substitution

Sometimes a trinomial does not appear to be in $ax^2+bx+c$ form. A thoughtful substitution can make it fit. It is standard to use $u$. Look for a middle-term variable whose square is the variable part of the first term.

Example. Factor $x^4-4x^2-5$ by substitution.

Let $u=x^2$:

Sometimes the expression to substitute is not a monomial.

Example. Factor $(x-2)^2+7(x-2)+12$ by substitution.

Let $u=x-2$:

This could also be factored by multiplying out and combining like terms, but most students prefer substitution.

Key terms

trial and error — testing possible factor pairs until their product is the original trinomial. ac method — finding two numbers whose product is $ac$ and sum is $b$, splitting the middle term, and factoring by grouping. factoring by substitution — replacing a repeated expression with a variable to make a familiar factorable form. prime trinomial — a trinomial that cannot be factored over the integers.

This section is adapted from Intermediate Algebra 2e, Section 6.2: Factor Trinomials by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted worked-example tables and factoring diagrams as accessible math and tables, omitted the Be Prepared quiz, media link, and end-of-section exercises, and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.