Solve quadratic equations by factoring

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Difficulty: Intermediate
Estimated time: 28 minutes
Domains: Advanced Math
Techniques: FactoringZero-product-propertySelected-rootsRepeated-expression

What you’ll learn

Spot a quadratic equation you can solve by factoring.
Graph both sides instead when the factors are hard to see and a decimal or answer choices you can tell apart are all you need.
Move every term to one side so the other side is $0$ .
Use the zero-product property to find every real solution.
Use words like positive, negative, greater or sum to pick or combine solutions before you answer.

Prerequisites

Why this matters on the SAT

Pick the faster way to solve

An SAT quadratic question might ask for one solution, every solution, a particular one, like the positive one, or a number built from the solutions, like their sum. So before you start any algebra, ask which way gets you there with less work: a graph, or factors you can already see.

SAT example

To the nearest tenth, what is the positive solution to the equation

7x^2-9x=4?

A
$-0.3$
B
$0.3$
C
$1.6$
D
$2.3$

Solution to the example

The question wants a decimal, rounded to the nearest tenth. And you can't spot factors for this quadratic quickly. In fact, it doesn't factor with whole numbers at all. So graph it. Put each side of the original equation on its own line:

\begin{aligned} y&=7x^2-9x,\\[1.4em] y&=4. \end{aligned}

The graphs cross at $x\approx-0.349$ and $x\approx1.635$ . The positive solution rounds to $1.6$ , so the answer is C.

The graph skipped a hunt for factors that would have come up empty, and it handed you the decimal directly. When the factors are easy to see, it goes the other way, and factoring is faster.

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Click both crossing points, keep the positive x-value, and round it to the nearest tenth.

Choose graphing or factoring

Two things decide it: what the question wants back, and whether you can see the factors.

Graph both sides when no factors jump out and a decimal is enough, like "to the nearest tenth," or answer choices you can tell apart on the graph.
Factor first when you can see the factors, like the common factor $2x$ in $2x^2-12x$ , a difference of squares like $x^2-25$ , or a trinomial that splits easily. That matters most when the answer has to be exact, like a fraction.
Use an exact method like the quadratic formula when you can't see any factors and the question wants an exact answer with a square root, like $3+\sqrt2$ , or asks how many real solutions there are.

When you graph, do what the opening example did and keep the equation as it's written. Put the left side on one line and the right side on another, then click every point where the graphs cross. Each crossing's $x$ -value is one solution. If one side is already $0$ , you can graph the other side alone and click every $x$ -intercept instead.

For more calculator practice, see Solve one-variable equations and Read points of interest from a graph.

Check your understanding:

Would you graph or factor $x^2-81=0$ ? What are its solutions?

Get zero on one side, then factor

Factoring solves an equation because of one fact, the zero-product property: if two numbers multiply to $0$ , at least one of them is $0$ .

ab=0\quad\Longrightarrow\quad a=0\ \text{or}\ b=0.

That's because two nonzero numbers can't multiply to $0$ . So once a product equals $0$ , you can split it into two small equations:

\begin{aligned} (x-4)(x+6)&=0,\\[1.4em] x-4=0 \quad&\text{or}\quad x+6=0,\\[1.4em] x=4 \quad&\text{or}\quad x=-6. \end{aligned}

This only works with $0$ on the other side. If $(x-4)(x+6)=10$ , neither factor has to be $0$ , since plenty of pairs of nonzero numbers multiply to $10$ . So when you factor, the first move is always to get $0$ on one side.

Common mistake:

Two slips happen here. One is setting the factors to $0$ while the other side is still some other number. Move every term to one side first. The other is reading the answer straight off a factor: $x+6$ doesn’t give $x=6$ . Solve the whole equation $x+6=0$ , and you get $x=-6$ .

Keep a zero root

Sometimes the first thing to factor out is a common factor:

\begin{aligned} 2x^2-12x&=0,\\[1.4em] 2x(x-6)&=0. \end{aligned}

The factors with $x$ in them give $x=0$ or $x=6$ . It's tempting to divide both sides by $x$ first, but that throws away $x=0$ , which is a real solution. So factor out $x$ instead of dividing by it.

Keep every distinct root

A pattern you recognize makes the rest quick. Here's a difference of squares:

\begin{aligned} 9x^2-25&=0,\\[1.4em] (3x-5)(3x+5)&=0,\\[1.4em] x=\frac53 \quad&\text{or}\quad x=-\frac53. \end{aligned}

If a factor repeats, as in $(x+4)^2=0$ , both copies give $x=-4$ . That's one distinct solution, not two, so you list it once.

Find every solution first. Only then use words like positive, negative, greater or sum to pick one or combine them.

Example: Find an exact root and use it

Worked example

The positive solution to the equation

6x^2+x=12

can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

Step 1

Pick factoring, then get zero on one side

The answer has to be an exact fraction, and this quadratic factors, so factoring is the way in. Subtract $12$ from both sides:

6x^2+x-12=0.

Step 2

Split the middle term and group

Multiply the first and last numbers: $ac=(6)(-12)=-72$ . Now look for two numbers that multiply to $-72$ and add to $1$ , the number in front of $x$ . The pair $9$ and $-8$ works. Split $x$ into $9x-8x$ and factor each pair of terms:

\begin{aligned} 6x^2+x-12 &=6x^2+9x-8x-12\\[1.4em] &=3x(2x+3)-4(2x+3)\\[1.4em] &=(3x-4)(2x+3). \end{aligned}

Both groups share $(2x+3)$ , so it comes out as a factor.

Step 3

Solve both factor equations

Set each factor equal to $0$ and solve:

\begin{aligned} 3x-4=0 &\quad\Longrightarrow\quad x=\frac43,\\[1.4em] 2x+3=0 &\quad\Longrightarrow\quad x=-\frac32. \end{aligned}

Step 4

Answer what the question asks

The positive solution is $\frac43$ . Since $4$ and $3$ share no common factor, $m=4$ and $n=3$ . So

\boxed{m+n=7}.

Common mistake:

Stopping at the root. Finding $\frac43$ feels like the finish line, but the question asks for a number built from it. Once you have the right solution, do the last step too. Here the answer is $m+n=7$ , not $\frac{m}{n}$ .

Treat a repeated expression as one letter

Look at this equation:

(x+2)^2-7(x+2)+10=0.

The same expression, $x+2$ , shows up twice. Instead of expanding, give the whole expression a temporary name. Let $u=x+2$ . Then the equation turns into a plain quadratic:

\begin{aligned} u^2-7u+10&=0,\\[1.4em] (u-5)(u-2)&=0, \end{aligned}

so $u=5$ or $u=2$ . You're not done yet, because the question is about $x$ . Put $x+2$ back in for $u$ in both cases:

\begin{aligned} x+2=5 &\quad\Longrightarrow\quad x=3,\\[1.4em] x+2=2 &\quad\Longrightarrow\quad x=0. \end{aligned}

The letter $u$ keeps the factoring tidy, but your answers are always values of $x$ .

Practice problems

Before you solve each one, decide: graph it, or factor it?

Read decimal solutions from a graph

Practice problem

To the nearest hundredth, what are all solutions to the equation

11x^2+7x=3

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Graph both sides, click both crossing points, and round their x-values.

Use a difference of squares

Practice problem

The equation

9t^2=64

has two real solutions. What is the negative solution?

Your answer

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Factor this one. Afterward, you can graph both sides to check the negative solution.

Rename a repeated expression

Practice problem

The equation

3(2x-1)^2-14(2x-1)-5=0

has two real solutions. What is the sum of the solutions?

Your answer

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Factor with a temporary letter to get the exact sum. Afterward, you can graph both original sides to check the two x-values before adding.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Graph both original sides when you can't see the factors and decimals or answer choices are enough.
To factor, get $0$ on one side, factor completely, and set each factor with $x$ in it equal to $0$ .
Keep $x=0$ when $x$ is a factor, and list a repeated root once unless the question asks about repeats.
Find every solution before you use words like positive, negative, greater or least.
Keep exact answers exact with algebra, then do the final calculation the question asks for, like $m+n$ .

Next lesson

Use the quadratic formula and discriminant

Solve quadratics that don’t factor easily and find how many real solutions they have.

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Practice

Practice this lesson

165 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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