6.5.2 Factoring Trinomials with Leading Coefficients of 1

How to Factor a Trinomial of the Form $x^2+bx+c$

To figure out how we would factor a trinomial of the form $x^2+bx+c$, such as $x^2+5x+6$, and factor it to $(x+2)(x+3)$, let’s start with two general binomials of the form $(x+m)$ and $(x+n)$.

$(x+m)(x+n)$

Step 1 - FOIL to find the product.

$x^2+mx+nx+mn$

Step 2 - Factor the GCF from the middle terms.

$x^2+(m+n)x+mn$

Our trinomial is of the form $x^2+bx+c$.

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 the two numbers $m$ and $n$ multiply to $c$ and add to $b$.

Example 1

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

Step 1 - Write the factors as two binomials with the first terms $x$.

Write two sets of parentheses and put $x$ as the first term.

$x^2+11x+24$

$(x)(x)$

Step 2 - Find two numbers, $m$ and $n$, that

• multiply to $c$: $m·n=c$
• add to $b$: $m+n=b$

Find two numbers that multiply to 24 and add to 11.

Step 3 - Use $m$ and $n$ as the last terms of the factors.

Use 3 and 8 as the last terms of the binomials.

$(x+3)(x+8)$

Step 4 - Check by multiplying the factors.

$(x+3)(x+8)$

$x^2+8x+3x+24$

$x^2+11x+24✓$

Let’s summarize the steps we used to find the factors.

How to factor trinomials of the form $x^2+bx+c$:

Step 1 - Write the factors as two binomials with first terms x.

$$\begin{gathered} x^2+bx+c \\ (x)(x) \end{gathered}$$

Step 2 -Find two numbers, $m$ and $n$, that

• multiply to $c$: $m·n=c$
• add to $b$: $m+n=b$

Step 3 - Use $m$ and $n$ as the last terms of the factors.

$(x+m)(x+n)$

Step 4 - Check by multiplying the factors.

In the first example, 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.

How do you get a positive product and a negative sum? We use two negative numbers.

Example 2

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

Again, with the positive last term, $28$, and the negative middle term, $-11y$, we need two negative factors. Find two numbers that multiply to $28$ and add to $-11$.

Step 1 - Write the factors as two binomials with first terms $y$.

$(y)(y)$

Step 2 - Find two numbers that multiply to 28 and add to –11.

Step 3 - Use –4 and –7 as the last two terms of the binomials.

$(y-4)(y-7)$

Step 4 - Check.

$\begin{array}{c}(y-4)(y-7)\hfill \\ y^2-7y-4y+28\hfill \\ y^2-11y+28✓\hfill \end{array}$

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 when you choose factors to make sure you get the correct sign for the middle term, too. How do you get a negative product and a positive sum? We use one positive and one negative number. When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.

Example 3

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

Step 1 - Put the terms in decreasing degree order.

$x^2+2x-48$

Step 2 - Write the factors as two binomials with first terms $x$.

$(x)(x)$

Step 3 - Find two numbers that multiply to −48 and add to 2.

Since the middle term of the trinomial is positive, the sum of the factors must also be positive. This means that we are only looking at the factors in which the larger number is positive. This is why factors such as 1 and −48 are not included in the table.

Step 4 - Use –6 and 8 as the last terms of the binomials.

$(x-6)(x+8)$

Step 5 - Check.

$\begin{array}{c}(x-6)(x+8)\hfill \\ x^2–6x+8x–48\hfill \\ x^2+2x–48✓\hfill \end{array}$

Sometimes you’ll need to factor trinomials of the form $x^2+bxy+c{y}^2$ 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, $x·x$. 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 above.

Example 4

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.

Step 1 - Note that the first terms are $r$ and the last terms contain $s$.

$(rs)(rs)$

Step 2 - Find the numbers that multiply to −9 and add to −8.

Step 3 - Use 1 and –9 as coefficients of the last terms.

$(r+s)(r-9s)$

Step 4 - Check.

$\begin{array}{c}(r-9s)(r+s)\hfill \\ r^2+rs-9rs-9s^2\hfill \\ r^2-8rs-9s^2\hfill \end{array}$

It is important to keep in mind that 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. If there are no factor pairs that fit all the requirements, then the trinomial is prime.

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