Lesson 2: Factor Trinomials of the Form x2+bx+c

New Concept

Factoring trinomials like $x^2 + bx + c$ is about finding two numbers that multiply to $c$ and add to $b$. This skill helps you 'undo' multiplication to find the original binomial factors, including those with two variables.

What’s next

You've got the core idea! Next up, we’ll work through several examples and then you'll apply the technique on interactive practice cards and challenge problems.

Property

To factor a trinomial of the form $x^2 + bx + c$ means to start with the product and end with the factors, $(x+m)(x+n)$. To get the correct factors, we find two numbers $m$ and $n$ whose product is $c$ and sum is $b$.

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

1. Write the factors as two binomials with first terms $x$: $(x \quad)(x \quad)$.
2. Find two numbers $m$ and $n$ that multiply to $c$ ($m \cdot n = c$) and add to $b$ ($m+n = b$).
3. Use $m$ and $n$ as the last terms of the factors: $(x+m)(x+n)$.
4. Check by multiplying the factors.

Examples

• To factor $x^2 + 9x + 20$, we need two numbers that multiply to 20 and add to 9. The numbers are 4 and 5. So, the factors are $(x+4)(x+5)$.
• To factor $a^2 - 8a + 15$, we need two numbers that multiply to 15 and add to -8. The numbers are -3 and -5. Thus, the factors are $(a-3)(a-5)$.
• To factor $p^2 + 3p - 10$, we need two numbers that multiply to -10 and add to 3. The numbers are 5 and -2. Therefore, the factors are $(p+5)(p-2)$.

Property

When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors. For $x^2 + bx + c = (x+m)(x+n)$:

When $c$ is positive, $m$ and $n$ have the same sign.

• If $b$ is positive: $m, n$ are positive. Example: $x^2 + 5x + 6 = (x+2)(x+3)$.
• If $b$ is negative: $m, n$ are negative. Example: $x^2 - 5x + 6 = (x-2)(x-3)$.

When $c$ is negative, $m$ and $n$ have opposite signs. The sign of the factor with the larger absolute value matches the sign of $b$.

• Example: $x^2 + x - 12 = (x+4)(x-3)$.
• Example: $x^2 - x - 12 = (x-4)(x+3)$.

Property

To factor trinomials of the form $x^2 + bxy + cy^2$, the process is similar to factoring $x^2 + bx + c$. 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 $y$. The factors will have the form $(x+my)(x+ny)$. You still need to find two numbers, $m$ and $n$, that multiply to $c$ and add to $b$.

Examples

• To factor $a^2 + 8ab + 15b^2$, find two numbers that multiply to 15 and add to 8 (which are 3 and 5). The factors are $(a+3b)(a+5b)$.
• To factor $r^2 - 10rs + 21s^2$, find two numbers that multiply to 21 and add to -10 (which are -3 and -7). The factors are $(r-3s)(r-7s)$.
• To factor $m^2 - mn - 30n^2$, find two numbers that multiply to -30 and add to -1 (which are -6 and 5). The factors are $(m-6n)(m+5n)$.

Explanation

This is a variation where the second variable, $y$, is part of the middle and last terms. Think of it the same way: find two numbers that multiply to $c$ and add to $b$, but remember to include $y$ in the last term of each binomial factor.

Property

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. A trinomial of the form $x^2 + bx + c$ is considered prime if there are no integers $m$ and $n$ such that $m \cdot n = c$ and $m+n = b$.

Examples

• Consider $x^2 + 5x + 10$. The factors of 10 are (1,10) and (2,5). Their sums are 11 and 7, neither of which is 5. So, the trinomial is prime.
• Consider $y^2 - 7y + 8$. The negative factors of 8 are (-1,-8) and (-2,-4). Their sums are -9 and -6, neither of which is -7. So, it is prime.
• Consider $z^2 + 2z - 6$. The factors of -6 are (1,-6), (-1,6), (2,-3), and (-2,3). Their sums are -5, 5, -1, and 1, none of which is 2. Therefore, it is prime.

Explanation

Just like a prime number cannot be factored into smaller whole numbers, a prime trinomial cannot be factored into simpler binomials with integer coefficients. This happens when no pair of factors for $c$ adds up to $b$.