Now that you can factor quadratics, you can solve quadratic equations. Factoring turns a sum into a product, and products have a special property when they equal zero.
Why it works
If $a \cdot b = 0$, then $a = 0$ or $b = 0$. Why? If $a$ is not zero, divide both sides by $a$: $b = \frac{0}{a} = 0$. So at least one factor must be zero. This works only for zero: $a \cdot b = 6$ allows $2 \cdot 3$, $1 \cdot 6$, $12 \cdot \frac{1}{2}$, and endless other pairs, so it tells you nothing about each factor.
Key idea
To solve a quadratic equation by factoring: get $0$ on one side, factor, set each factor equal to $0$, and solve each small equation.
Worked examples
Worked example: Factor and solve
Solve $x^2 - 2x - 15 = 0$.
- Factor: $(x - 5)(x + 3) = 0$.
- $x - 5 = 0$ or $x + 3 = 0$.
- Check: $25 - 10 - 15 = 0$ and $9 + 6 - 15 = 0$.
Answer: $x = 5$ or $x = -3$
Worked example: Get zero first
Solve $x^2 = 4x$.
- Do not divide by $x$: that would lose a solution. Subtract $4x$: $x^2 - 4x = 0$.
- Factor: $x(x - 4) = 0$, so $x = 0$ or $x = 4$.
Answer: $x = 0$ or $x = 4$
Worked example: Not zero on the right
Solve $(x - 2)(x - 3) = 6$.
- The right side is not $0$, so the property does not apply yet. Multiply out: $x^2 - 5x + 6 = 6$.
- Subtract $6$: $x^2 - 5x = 0$, so $x(x - 5) = 0$.
Answer: $x = 0$ or $x = 5$
Common mistake
Setting each factor equal to $6$ when $(x - 2)(x - 3) = 6$, which gives $x = 8$ and $x = 9$. Check: $(8 - 2)(8 - 3) = 30$, not $6$. The zero product property needs a zero; move everything to one side first.