When you multiply two binomials like $(x + 2)(x + 5)$, you get $x^2 + 7x + 10$. Notice where the numbers went: $2 + 5 = 7$ and $2 \cdot 5 = 10$. Factoring a trinomial $x^2 + bx + c$ means finding those two numbers again. It also builds on factoring out the GCF, which you should always try first.
Why it works
$$(x + p)(x + q) = x^2 + qx + px + pq = x^2 + (p + q)x + pq$$ So for $x^2 + bx + c$ to factor this way, we need $p + q = b$ and $pq = c$. In the area model, the two middle rectangles are $px$ and $qx$, and the corner rectangle is $pq$. Signs follow from arithmetic: if $c > 0$, $p$ and $q$ have the same sign (the sign of $b$); if $c < 0$, they have opposite signs.
Key idea
To factor $x^2 + bx + c$, find two numbers with product $c$ and sum $b$; then $x^2 + bx + c = (x + p)(x + q)$. Check by multiplying.
Worked examples
Worked example: Both negative
Factor $x^2 - 9x + 20$.
- Product $20 > 0$ and sum $-9 < 0$: both numbers are negative.
- Pairs for $20$: $(-1, -20)$, $(-2, -10)$, $(-4, -5)$. Only $-4 + (-5) = -9$.
Answer: $(x - 4)(x - 5)$
Worked example: Opposite signs
Factor $x^2 + 3x - 28$.
- Product $-28 < 0$: opposite signs. The sum $3$ is positive, so the larger number is positive.
- $7 \cdot (-4) = -28$ and $7 + (-4) = 3$.
Answer: $(x + 7)(x - 4)$
Worked example: Two variables
Factor $x^2 + 5xy + 6y^2$.
- Same idea with $y$ along for the ride: numbers with product $6$ and sum $5$ are $2$ and $3$.
Answer: $(x + 2y)(x + 3y)$
Common mistake
Finding a pair with the right product but ignoring signs, so the middle term comes out wrong: $x^2 + 3x - 28 = (x - 7)(x + 4)$ gives a middle term of $-3x$. Always multiply your answer back out and compare the middle term.