Factor Trinomials of the Form $ax^2+bx+c$

Recognize a preliminary strategy for factoring

Let’s summarize where we are so far with factoring polynomials. In the first two sections of this chapter, we used three methods of factoring: factoring the GCF, factoring by grouping, and factoring a trinomial by “undoing” FOIL. More methods will follow as you continue in this chapter, as well as later in your studies of algebra.

How will you know when to use each factoring method? It will help to organize the factoring methods into a strategy that can guide you to the correct method.

As you start to factor a polynomial, always ask first, “Is there a greatest common factor?” If there is, factor it first.

The next thing to consider is the type of polynomial. How many terms does it have? Is it a binomial? A trinomial? Or does it have more than three terms?

• If it is a trinomial where the leading coefficient is one, $x^2+bx+c$, use the “undo FOIL” method.
• If it has more than three terms, try the grouping method. This is the only method to use for polynomials of more than three terms.

Some polynomials cannot be factored. They are called “prime.”

Use the preliminary strategy to completely factor a polynomial. A polynomial is factored completely if, other than monomials, all of its factors are prime.

Example. Identify the best method to use to factor each polynomial: (a) $6y^2-72$, (b) $r^2-10r-24$, (c) $p^2+5p+pq+5q$.

For (a), we first ask if there is a greatest common factor. Yes, it is $6$:

Inside the parentheses is a binomial, and we have no method to factor binomials yet.

For (b), there is no common factor. It is a trinomial with leading coefficient $1$, so we “undo” FOIL.

For (c), there is no common factor. It has more than three terms, so we factor using grouping.

Factor trinomials of the form $ax^2+bx+c$ with a GCF

Now that we have organized what we’ve covered so far, we are ready to factor trinomials whose leading coefficient is not $1$, trinomials of the form $ax^2+bx+c$.

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes $1$ and you can factor it by the methods in the last section. Let’s do a few examples to see how this works. Watch out for the signs in the next two examples.

Example. Factor completely: $2n^2-8n-42$.

Use the preliminary strategy. The GCF is $2$, so factor it out:

Inside the parentheses is a trinomial whose leading coefficient is $1$, so undo FOIL. We look for two numbers that multiply to $-21$ and add to $-4$: those are $3$ and $-7$.

Check by multiplying:

Example. Factor completely: $4y^2-36y+56$.

The GCF is $4$, so factor it out:

Inside the parentheses is a trinomial with leading coefficient $1$, so undo FOIL. We need two numbers that multiply to $14$ and add to $-9$. Since the last term is positive and the middle term is negative, both factors must be negative: $-2$ and $-7$.

In the next example the GCF will include a variable.

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

The GCF is $4u$, so factor it out:

Inside the parentheses is a trinomial with leading coefficient $1$, so undo FOIL. We need two numbers that multiply to $-5$ and add to $4$: those are $-1$ and $5$.

Factor trinomials using trial and error

What happens when the leading coefficient is not $1$ and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the trial and error method.

Let’s factor the trinomial $3x^2+5x+2$. From our earlier work we expect this will factor into two binomials:

We know the first terms of the binomial factors will multiply to give us $3x^2$. The only factors of $3x^2$ are $1x,\ 3x$. We can place them in the binomials as $(x\ \ )(3x\ \ )$.

We know the last terms of the binomials will multiply to $2$. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of $2$ are $1$ and $2$, but it will make a difference whether we write $1,\ 2$ or $2,\ 1$. So we consider both cases:

To decide which is correct, we multiply the inner and outer terms and add them. Since the middle term of the trinomial is $5x$, the factors in the first case work: the outer term $2x$ plus the inner term $3x$ gives $5x$. Let’s FOIL to check:

Our result is $3x^2+5x+2 = (x+1)(3x+2)$.

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

The trinomial is already in descending order. The only factor pair of the first term $3y^2$ is $1y,\ 3y$, so we put them in the binomials: $(y\ \ )(3y\ \ )$. The only factor pair of the last term $7$ is $1,\ 7$. We test the possible combinations until the correct product is found:

The correct factors are those whose product is the original trinomial:

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

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

The trinomial is already in descending order. The first term $6b^2$ has factor pairs $1b,\ 6b$ and $2b,\ 3b$. Since the last term $5$ is positive, its factors must both be positive or both be negative; the coefficient of the middle term is negative, so we use the negative factors of $5$, namely $-1,\ -5$. Consider all the combinations:

The correct factors are those whose product is the original trinomial:

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

Example. Factor completely using trial and error: $14x^2-47x-7$.

The trinomial is already in descending order. The first term $14x^2$ has factor pairs $1x,\ 14x$ and $2x,\ 7x$. Since the last term $-7$ is negative, one factor must be positive and one negative: $1,\ -7$ or $-1,\ 7$ (and their reverse orders). Because the original trinomial has no common factor, neither binomial factor may contain a common factor — that rules out several combinations. The one whose product is the original trinomial is:

Example. Factor completely using trial and error: $18n^2-37n+15$.

The first term $18n^2$ has factor pairs $1n,\ 18n$; $2n,\ 9n$; and $3n,\ 6n$. The last term $15$ is positive and the middle term is negative, so we use the negative factors of $15$. Testing the combinations, the one whose product is the original trinomial is:

Don’t forget to look for a GCF first.

Example. Factor completely using trial and error: $10y^4+55y^3+60y^2$.

Notice the greatest common factor, and factor it first:

Now factor the trinomial $2y^2+11y+12$ by trial and error. The correct factors are those whose product is the trinomial; remember to include the factor $5y^2$:

Factor trinomials using the “ac” method

Another way to factor trinomials of the form $ax^2+bx+c$ is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is, step-by-step), and it always works!

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

There is no greatest common factor. Find the product $ac$:

Find two numbers $m$ and $n$ that multiply to $ac=12$ and add to $b=7$. Both factors must be positive: $3$ and $4$, since $3 \cdot 4 = 12$ and $3+4=7$.

Split the middle term $7x$ as $3x+4x$:

Factor by grouping:

Check by multiplying:

When the third term of the trinomial is negative, the factors of the third term will have opposite signs.

Example. Factor using the “ac” method: $8u^2-17u-21$.

There is no greatest common factor. Find the product $ac$:

Find two numbers that multiply to $-168$ and add to $-17$. The larger factor must be negative: $7$ and $-24$, since $7(-24)=-168$ and $7+(-24)=-17$.

Split the middle term $-17u$ as $7u-24u$:

Factor by grouping:

Example. Factor using the “ac” method: $2x^2+6x+5$.

There is no greatest common factor. Find the product $ac$:

We need two numbers that multiply to $10$ and add to $6$:

There are no factors that multiply to $10$ and add to $6$. The polynomial is prime — it cannot be factored.

Don’t forget to look for a common factor!

Example. Factor completely: $10y^2-55y+70$.

The GCF is $5$. Factor it out, being careful to keep the factor of $5$ all the way through the solution:

The trinomial inside the parentheses has a leading coefficient that is not $1$. Factor it (by trial and error or the “ac” method) to get:

Key terms

trial and error — a method of factoring $ax^2+bx+c$ by listing the factor pairs of the first and last terms, then testing the possible binomial combinations until the product matches the trinomial. “ac” method — a structured method of factoring $ax^2+bx+c$ by finding two numbers that multiply to $ac$ and add to $b$, splitting the middle term with them, and factoring by grouping. prime polynomial — a polynomial that cannot be factored (other than monomial factors).

This section is adapted from Elementary Algebra 2e, Section 7.3: Factor Trinomials of the Form ax²+bx+c by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the worked-example step tables as prose with typeset display arrays and the trial-and-error “possible factors” work as markdown tables; described the factoring-strategy flowchart in prose; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.