Lesson 72: Factoring Trinomials: x² + bx + c

New Concept

This procedure can be reversed to factor a trinomial into the product of two binomials.

What’s next

Next, you'll apply this pattern to factor trinomials by examining the signs of their terms and finding the correct factor pairs.

Property

To factor a trinomial $x^2 + bx + c$ where $c > 0$, find two numbers that have a product of $c$ and a sum of $b$. Both numbers will be positive if $b$ is positive, and both will be negative if $b$ is negative.

Explanation

Imagine you're a detective looking for two number suspects! Your clues are the $b$ and $c$ values. When $c$ is positive, you know the suspects are partners-in-crime with the same sign. If $b$ is positive, they are a pair of positive heroes. If $b$ is negative, they are a duo of negative villains. Find the pair that multiplies to $c$ and adds to $b$!

Examples

To factor $x^2 + 10x + 21$, find two positive numbers that multiply to 21 and add to 10. The numbers are 3 and 7, so the factors are $(x+3)(x+7)$.
To factor $x^2 - 11x + 24$, find two negative numbers that multiply to 24 and add to -11. The numbers are -3 and -8, so the factors are $(x-3)(x-8)$.

Let's see how finding two simple numbers can unlock the factors of a trinomial, which is an important key idea from this lesson.

Factor the trinomial $x^2 + 8x + 15$.

1. In this trinomial, $b$ is $8$ and $c$ is $15$. Because $b$ is positive, it must be the sum of two positive numbers that are factors of $c$.
2. Two pairs of positive numbers have a product of $15$.

3. Only one pair of these numbers has a sum of $8$.

4. The constant terms in the binomials are $3$ and $5$.

5. The factored form of $x^2 + 8x + 15$ is $(x+3)(x+5)$.

Property

To factor a trinomial $x^2 + bx + c$ where $c < 0$, find two numbers with opposite signs that have a product of $c$ and a sum of $b$. The number with the greater absolute value will have the same sign as $b$.

Explanation

When $c$ is negative, your two number suspects are rivals, one positive and one negative! They have to work together to multiply to $c$. To figure out who's who, look at $b$. The number with the bigger muscle (greater absolute value) gets the same sign as $b$. Find the rival pair that successfully multiplies to $c$ and adds up to $b$.

Examples

To factor $x^2 + 4x - 12$, find factors of -12 that sum to 4. The pair is 6 and -2, so the factors are $(x+6)(x-2)$.
To factor $x^2 - 3x - 28$, find factors of -28 that sum to -3. The pair is -7 and 4, so the factors are $(x-7)(x+4)$.

When the last term is negative, things get a little tricky; one factor is positive, and one is negative. This is the other key idea from this lesson.

Factor the trinomial $x^2 + 2x - 24$.

1. In this trinomial, $b$ is $2$ and $c$ is $-24$.
2. We need to find one positive and one negative number that have a product of $-24$. Let's list the pairs:

3. The sum of only one of these pairs is $2$.

4. The constant terms in the binomials are $-4$ and $6$. So,

Property

For a trinomial in the form $x^2 + bxy + cy^2$, find two terms that multiply to get $cy^2$ and add to get $bxy$. The factors will be in the form $(x + my)(x + ny)$.

Explanation

Seeing two variables might seem tricky, but it's just the same puzzle with a costume change! Ignore the $y$ for a moment and find two numbers that multiply to $c$ and add to $b$, just like always. Once you have your magic numbers, just attach a $y$ to each of them. Now they fit perfectly into the binomial factors. Easy peasy!

Examples

To factor $x^2 + 9xy + 14y^2$, find numbers that multiply to 14 and add to 9 (which are 2 and 7). The terms are $2y$ and $7y$, so the factors are $(x+2y)(x+7y)$.
To factor $x^2 - xy - 20y^2$, find numbers that multiply to -20 and add to -1 (which are -5 and 4). The terms are $-5y$ and $4y$, so the factors are $(x-5y)(x+4y)$.