7.2 Factor quadratic trinomials with leading coefficient 1

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, $\left(x+2\right)\left(x+3\right)$ . 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 term 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 and summarize the results in [link] —the table will be very helpful when you work with numbers that can be factored in many different ways.

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 $\left(x+2\right)\left(x+3\right)$ .

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 $\left(x+m\right)\text{and}\left(x+n\right)$ .

To get the correct factors, we found two numbers m and n whose product is c and sum is b .

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

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

Solution

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