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

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 start with the product and end up with the factors. Let’s look at an example of multiplying binomials to refresh your memory.

To factor the trinomial means to start with the product, $x^2+5x+6$, and end with the factors, $(x+2)(x+3)$. You need to think about where each of the terms in the trinomial came from.

The first term came from multiplying the first term in each binomial. So to get $x^2$ in the product, each binomial must start with an $x$.

The last term in the trinomial came from multiplying the last terms in each binomial. So the last terms must multiply to $6$. What two numbers multiply to $6$? The factors of $6$ could be $1$ and $6$, or $2$ and $3$. How do you know which pair to use?

Consider the middle term. It came from adding the outer and inner terms. So the numbers that must have a product of $6$ will need a sum of $5$. We’ll test both possibilities in the table below.

We see that $2$ and $3$ are the numbers that multiply to $6$ and add to $5$. So we have the factors of $x^2+5x+6$. They are $(x+2)(x+3)$.

You should check this by multiplying.

Looking back, we started with $x^2+5x+6$, which is of the form $x^2+bx+c$, where $b=5$ and $c=6$. We factored it into two binomials of the form $(x+m)$ and $(x+n)$. To get the correct factors, we found two numbers $m$ and $n$ whose product is $c$ and sum is $b$.

Example. Factor: $x^2+7x+12$.

Find two numbers that multiply to $12$ and add to $7$:

Check by multiplying the factors:

Example. Factor: $u^2+11u+24$.

Notice that the variable is $u$, so the factors will have first terms $u$. Find two numbers that multiply to $24$ and add to $11$:

Use $3$ and $8$ as the last terms of the binomials: $(u+3)(u+8)$. Check by multiplying:

Example. Factor: $y^2+17y+60$.

Write the factors as two binomials with first terms $y$, then find two numbers that multiply to $60$ and add to $17$:

Use $5$ and $12$ as the last terms: $(y+5)(y+12)$. Check by multiplying:

Factor trinomials with $b$ negative, $c$ positive

In the examples so far, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

Remember: to get a negative sum and a positive product, the numbers must both be negative.

Example. Factor: $t^2-11t+28$.

With the positive last term, $28$, and the negative middle term, $-11t$, we need two negative factors. Find two numbers that multiply to $28$ and add to $-11$:

Use $-4$ and $-7$ as the last terms of the binomials: $(t-4)(t-7)$. Check by multiplying:

Factor trinomials with $c$ negative

Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.

Remember: to get a negative product, the numbers must have different signs.

Example. Factor: $z^2+4z-5$.

To get a negative last term, multiply one positive and one negative. We need factors of $-5$ that add to positive $4$:

Notice we listed both $1, -5$ and $-1, 5$ to make sure we got the sign of the middle term correct. Use $-1$ and $5$ as the last terms: $(z-1)(z+5)$. Check by multiplying:

Let’s make a minor change to the last trinomial and see what effect it has on the factors.

Example. Factor: $z^2-4z-5$.

This time, we need factors of $-5$ that add to $-4$:

Use $1$ and $-5$ as the last terms of the binomials: $(z+1)(z-5)$. Check by multiplying:

Notice that the factors of $z^2-4z-5$ are very similar to the factors of $z^2+4z-5$. It is very important to make sure you choose the factor pair that results in the correct sign of the middle term.

Example. Factor: $q^2-2q-15$.

The factors will be two binomials with first terms $q$. You can use $3, -5$ as the last terms of the binomials, giving $(q+3)(q-5)$.

Check by multiplying:

When a trinomial is prime

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

Example. Factor: $y^2-6y+15$.

The factors will be two binomials with first terms $y$. Since the last term, $15$, is positive and the middle term is negative, we look for two negative factors of $15$:

As shown in the table, none of the factors add to $-6$; therefore, the expression is prime.

Putting it together

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

First we put the terms in decreasing degree order: $x^2+2x-48$. The factors will be two binomials with first terms $x$. Since the last term is negative, we need factors with different signs that add to $2$:

Use $-6, 8$ as the last terms of the binomials: $(x-6)(x+8)$. Check by multiplying:

Let’s summarize the method we just developed to factor trinomials of the form $x^2+bx+c$.

Factor Trinomials of the Form $x^2+bxy+cy^2$

Sometimes you’ll need to factor trinomials of the form $x^2+bxy+cy^2$ with two variables, such as $x^2+12xy+36y^2$. The first term, $x^2$, is the product of the first terms of the binomial factors, $x \cdot x$. The $y^2$ in the last term means that the second terms of the binomial factors must each contain $y$. To get the coefficients $b$ and $c$, you use the same process summarized in the previous objective.

Example. Factor: $x^2+12xy+36y^2$.

Note that the first terms are $x$ and the last terms contain $y$: $(x\ \_\ y)(x\ \_\ y)$. Find the numbers that multiply to $36$ and add to $12$:

Use $6$ and $6$ as the coefficients of the last terms: $(x+6y)(x+6y)$. Check by multiplying:

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

We need $r$ in the first term of each binomial and $s$ in the second term. The last term of the trinomial is negative, so the factors must have opposite signs. Find the numbers that multiply to $-9$ and add to $-8$:

Use $1, -9$ as coefficients of the last terms: $(r+s)(r-9s)$. Check by multiplying:

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

We need $u$ in the first term of each binomial and $v$ in the second term. The last term is negative, so the factors must have opposite signs. Find the numbers that multiply to $-12$ and add to $-9$:

Note there are no factor pairs that give us $-9$ as a sum. The trinomial is prime.

Key terms

factor a trinomial — to write a trinomial $x^2+bx+c$ as a product of two binomials $(x+m)(x+n)$, where $m$ and $n$ multiply to $c$ and add to $b$. prime trinomial — a trinomial that cannot be written as a product of two binomials with integer coefficients (no factor pair of $c$ adds to $b$).

This section is adapted from Elementary Algebra 2e, Section 7.2: Factor Trinomials of the Form $x^2+bx+c$ 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 typeset display arrays and the factor-pair searches as markdown tables; 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, replacing the two prime-trinomial “Try Its” (which cannot be typed into the answer box) with factorable drills plus a note to reason the prime cases through by hand.