Solving Quadratics by Factoring
Quadratic Equations
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a is not zero. Solving by factoring is generally the fastest method — factor the quadratic expression and set each factor equal to zero, applying the zero-product property.
The zero-product property states that if the product of two numbers is zero, at least one of them must be zero. This simple fact is the bridge between factoring and solving.
This lesson focuses on solving quadratic equations by factoring, including special patterns and equations that require rearranging into standard form first.
A quadratic equation has the standard form $ax^2 + bx + c = 0$ where $a \neq 0$.
If $AB = 0$, then $A = 0$ or $B = 0$ (or both).
- Write in standard form (everything on one side, $= 0$).
- Factor the quadratic expression.
- Set each factor equal to zero and solve.
$x^2 - 5x + 6 = 0$
$(x - 2)(x - 3) = 0 \;\Rightarrow\; x = 2$ or $x = 3$.
$2x^2 + 7x = -3$
- Standard form: $2x^2 + 7x + 3 = 0$
- Factor: $(2x + 1)(x + 3) = 0$
- $x = -\dfrac{1}{2}$ or $x = -3$
$x^2 = 9x$
$x^2 - 9x = 0 \;\Rightarrow\; x(x - 9) = 0 \;\Rightarrow\; x = 0$ or $x = 9$.
Never divide both sides by $x$ — you'll lose the solution $x = 0$! Always factor instead.
$3x^2 - 12 = 0$
$3(x^2 - 4) = 0 \;\Rightarrow\; 3(x+2)(x-2) = 0 \;\Rightarrow\; x = -2$ or $x = 2$.
$6x^2 - x - 2 = 0$
AC: $6(-2) = -12$. Numbers $-4, 3$. $6x^2 - 4x + 3x - 2 = 0$.
$2x(3x-2) + 1(3x-2) = 0 \;\Rightarrow\; (3x-2)(2x+1) = 0$.
$x = \dfrac{2}{3}$ or $x = -\dfrac{1}{2}$.
$(x - 4)(x + 1) = 6$
Expand: $x^2 - 3x - 4 = 6 \;\Rightarrow\; x^2 - 3x - 10 = 0$.
$(x - 5)(x + 2) = 0 \;\Rightarrow\; x = 5$ or $x = -2$.
Practice Problems
Show Answer Key
1. $x = 2$ or $x = -5$
2. $x = 3$ or $x = -3$
3. $x = 2$ or $x = -\dfrac{3}{2}$
4. $x = 0$ or $x = 7$
5. $x = -3$ (double root)
6. $x = \dfrac{5}{2}$ or $x = -\dfrac{5}{2}$
7. $x = -\dfrac{4}{3}$ or $x = -2$
8. $x = 3$ or $x = 5$
9. $x^2 - x - 20 = 0$; $x = 5$ or $x = -4$
10. $x = 0$ or $x = 4$
11. $x = 7$ or $x = -8$
12. $x = \dfrac{1}{2}$ or $x = -\dfrac{4}{3}$