Solving Quadratic Equations by Factoring

A quadratic equation asks: which values of $x$ make $ax^2 + bx + c$ equal to zero (or to some other number)? You can’t just “get $x$ by itself” the way you do with a linear equation, because $x$ shows up squared and on its own. Instead, you factor, and let a simple fact about zero do the work.

Key ideas

Quadratic equations and roots

A quadratic equation can be written in the form

$$ax^2 + bx + c = 0, \qquad a \ne 0$$

The solutions are called the roots of the equation. A quadratic equation can have two roots, one root, or no real roots.

The zero product property

If two numbers multiply to give $0$, then at least one of them must be $0$:

$$\text{if } AB = 0, \text{ then } A = 0 \text{ or } B = 0$$

So once the equation looks like $(x - 2)(x + 4) = 0$, you know that $x - 2 = 0$ or $x + 4 = 0$.

This only works with $0$. If $AB = 6$, then $A$ and $B$ could be $2$ and $3$, or $1$ and $6$, or $12$ and $\tfrac{1}{2}$… you can’t conclude anything about either factor. That’s why the first step is always to get $0$ on one side.

The steps

1. Rearrange so that one side is $0$: $ax^2 + bx + c = 0$. Expand brackets first if there are any.
2. Factor the other side completely. Look for a common factor first, then factor the trinomial or special case.
3. Set each factor equal to $0$ and solve.
4. Check each root by substituting it into the original equation.

Double roots

Sometimes both factors are the same, as in $(x - 3)^2 = 0$. Then both give $x = 3$, so there’s only one root. It’s called a double root. On the graph, the parabola just touches the $x$-axis at its vertex instead of crossing it.

Roots and x-intercepts

The roots of $ax^2 + bx + c = 0$ are exactly the $x$-intercepts (zeros) of the relation $y = ax^2 + bx + c$, because the $x$-intercepts are where $y = 0$. So solving the equation tells you where the graph crosses the $x$-axis, and a graph (on paper or with graphing technology like Desmos) lets you check your roots.

Two roots means the graph crosses the $x$-axis twice; a double root means it touches the axis once.

Worked examples

Example 1: Two roots

Solve $x^2 + 2x - 8 = 0$.

Solution. The equation already has $0$ on the right. Find two numbers that multiply to $-8$ and add to $2$: they’re $4$ and $-2$.

$$\begin{aligned} x^2 + 2x - 8 &= 0 \\ (x + 4)(x - 2) &= 0 \end{aligned}$$

So $x + 4 = 0$ or $x - 2 = 0$, which gives $x = -4$ or $x = 2$.

Check: $(-4)^2 + 2(-4) - 8 = 16 - 8 - 8 = 0$ ✓ and $2^2 + 2(2) - 8 = 4 + 4 - 8 = 0$ ✓

These are the $x$-intercepts in the left graph above.

Example 2: Rearranging first

Solve $2x^2 = 5x + 3$.

Solution. Move everything to the left side so the right side is $0$:

$$2x^2 - 5x - 3 = 0$$

Factor by looking for two numbers that multiply to $2(-3) = -6$ and add to $-5$: they’re $-6$ and $1$.

$$\begin{aligned} 2x^2 - 6x + x - 3 &= 0 \\ 2x(x - 3) + 1(x - 3) &= 0 \\ (2x + 1)(x - 3) &= 0 \end{aligned}$$

So $2x + 1 = 0$ or $x - 3 = 0$, which gives $x = -\tfrac{1}{2}$ or $x = 3$.

Check in the original equation. For $x = 3$: the left side is $2(9) = 18$ and the right side is $5(3) + 3 = 18$. ✓ For $x = -\tfrac{1}{2}$: the left side is $2\left(\tfrac{1}{4}\right) = \tfrac{1}{2}$ and the right side is $5\left(-\tfrac{1}{2}\right) + 3 = \tfrac{1}{2}$. ✓

Example 3: A double root

Solve $x^2 + 9 = 6x$, and describe what the graph of $y = x^2 - 6x + 9$ looks like near its root.

Solution. Rearrange:

$$x^2 - 6x + 9 = 0$$

This is a perfect square trinomial:

$$(x - 3)^2 = 0$$

Both factors give $x - 3 = 0$, so the only root is $x = 3$, a double root.

Check: the left side is $3^2 + 9 = 18$ and the right side is $6(3) = 18$. ✓

The graph of $y = x^2 - 6x + 9$ has its vertex at $(3, 0)$, so it touches the $x$-axis at $3$ without crossing it (the right graph above).

Example 4: Expanding before you factor

Solve $(x + 2)(x - 3) = 14$.

Solution. It’s tempting to set each bracket equal to $14$, but the zero product property only works with $0$. Expand, then rearrange:

$$\begin{aligned} (x + 2)(x - 3) &= 14 \\ x^2 - 3x + 2x - 6 &= 14 \\ x^2 - x - 6 - 14 &= 0 \\ x^2 - x - 20 &= 0 \\ (x - 5)(x + 4) &= 0 \end{aligned}$$

So $x = 5$ or $x = -4$.

Check: $(5 + 2)(5 - 3) = 7 \times 2 = 14$ ✓ and $(-4 + 2)(-4 - 3) = (-2)(-7) = 14$ ✓

Common mistakes

Using the zero product property when the other side isn’t $0$. From $(x + 2)(x - 3) = 14$ you can’t say $x + 2 = 14$. Expand and rearrange to get $0$ on one side first.

Dividing both sides by $x$. In $3x^2 = 12x$, dividing by $x$ gives $x = 4$ and loses the root $x = 0$. Instead, rearrange and factor: $3x^2 - 12x = 0$, so $3x(x - 4) = 0$, which gives $x = 0$ or $x = 4$.

Getting the sign of a root wrong. $(x + 4)(x - 2) = 0$ gives $x = -4$ and $x = 2$. Solve each factor $= 0$ rather than just copying the numbers.

Forgetting to solve a factor like $2x + 1 = 0$. The root is $-\tfrac{1}{2}$, not $-1$.

Listing a double root twice, or treating a common factor as a root. $(x - 3)^2 = 0$ has one root, $3$. And in $2(x - 7)(x + 3) = 0$, the $2$ can never be $0$, so it doesn’t give a root: the roots are just $7$ and $-3$.

Practice

1. (Warm-up) Solve each equation.

• (a) $(x - 7)(x + 1) = 0$
• (b) $x(2x - 5) = 0$

Solution

(a) $x - 7 = 0$ or $x + 1 = 0$, so $x = 7$ or $x = -1$.

(b) $x = 0$ or $2x - 5 = 0$, so $x = 0$ or $x = \tfrac{5}{2}$.

2. (Warm-up) Solve $x^2 - 9x + 20 = 0$.

Solution

Two numbers that multiply to $20$ and add to $-9$: $-4$ and $-5$.

$$(x - 4)(x - 5) = 0$$

So $x = 4$ or $x = 5$.

3. (Warm-up) Solve $x^2 - 49 = 0$.

Solution

This is a difference of squares:

$$(x - 7)(x + 7) = 0$$

So $x = 7$ or $x = -7$.

4. (Core) Solve $5x^2 = 20x$.

Solution

Don’t divide by $x$. Rearrange and take out the common factor:

$$\begin{aligned} 5x^2 - 20x &= 0 \\ 5x(x - 4) &= 0 \end{aligned}$$

So $x = 0$ or $x = 4$.

Check: $5(0)^2 = 0 = 20(0)$ ✓ and $5(16) = 80 = 20(4)$ ✓

5. (Core) Solve $3x^2 + 10x - 8 = 0$.

Solution

Two numbers that multiply to $3(-8) = -24$ and add to $10$: $12$ and $-2$.

$$\begin{aligned} 3x^2 + 12x - 2x - 8 &= 0 \\ 3x(x + 4) - 2(x + 4) &= 0 \\ (3x - 2)(x + 4) &= 0 \end{aligned}$$

So $x = \tfrac{2}{3}$ or $x = -4$.

Check $x = -4$: $3(16) + 10(-4) - 8 = 48 - 40 - 8 = 0$ ✓

6. (Core) Solve $x^2 + 25 = 10x$. What does your answer tell you about the graph of $y = x^2 - 10x + 25$?

Solution

$$\begin{aligned} x^2 - 10x + 25 &= 0 \\ (x - 5)^2 &= 0 \end{aligned}$$

The only root is $x = 5$, a double root. The graph of $y = x^2 - 10x + 25$ touches the $x$-axis at $(5, 0)$, which is its vertex.

7. (Core) Solve $2x^2 - 8x - 42 = 0$.

Solution

Take out the common factor $2$ first:

$$\begin{aligned} 2(x^2 - 4x - 21) &= 0 \\ 2(x - 7)(x + 3) &= 0 \end{aligned}$$

The factor $2$ can’t be $0$, so $x - 7 = 0$ or $x + 3 = 0$: $x = 7$ or $x = -3$.

8. (Core) Solve $(x + 1)(x + 4) = 10$.

Solution

$$\begin{aligned} x^2 + 4x + x + 4 &= 10 \\ x^2 + 5x - 6 &= 0 \\ (x + 6)(x - 1) &= 0 \end{aligned}$$

So $x = -6$ or $x = 1$.

Check: $(1 + 1)(1 + 4) = 2 \times 5 = 10$ ✓ and $(-6 + 1)(-6 + 4) = (-5)(-2) = 10$ ✓

9. (Challenge) One root of $x^2 + kx - 12 = 0$ is $x = 3$. Find $k$, then find the other root.

Solution

Substitute $x = 3$:

$$9 + 3k - 12 = 0 \quad\Rightarrow\quad 3k = 3 \quad\Rightarrow\quad k = 1$$

The equation is $x^2 + x - 12 = 0$, which factors as $(x + 4)(x - 3) = 0$. The other root is $x = -4$.

10. (Challenge) Maya solves $x(x - 3) = 2(x - 3)$ by dividing both sides by $(x - 3)$, and gets $x = 2$. Solve the equation correctly, and explain what went wrong.

Solution

Expand and rearrange:

$$\begin{aligned} x^2 - 3x &= 2x - 6 \\ x^2 - 5x + 6 &= 0 \\ (x - 2)(x - 3) &= 0 \end{aligned}$$

So $x = 2$ or $x = 3$.

Check $x = 3$: both sides are $0$, since $3(0) = 2(0)$. ✓ Maya lost this root because when $x = 3$, she was dividing by $x - 3 = 0$, and you can never divide by zero. Never divide both sides by an expression that contains $x$; rearrange and factor instead.