Math · College algebra · Worked example

Solve a quadratic equation by factoring

Solve $x^{2} - 5 x + 6 = 0$ , then connect the answer to a graph.

Identify what you are solving

The goal is every real value of $x$ that makes the expression zero. The $x^{2}$ coefficient is $1$ , so start by looking for two numbers with product $6$ and sum $- 5$ .

x^{2} - 5 x + 6 = 0

Choose and check the factors

The numbers $- 2$ and $- 3$ work. Their product is positive and their sum is $- 5$ . Multiply back once to check the middle term before using the factorization.

(x - 2) (x - 3) = x^{2} - 3 x - 2 x + 6 = x^{2} - 5 x + 6

Rewrite the equation

Replace the polynomial with its factored form. The equation is written differently, but it has the same solutions.

\begin{array}{l} x^{2} - 5 x + 6 = 0 \\ (x - 2) (x - 3) = 0 \end{array}

Keep both possibilities

For the product to be zero, either $x - 2$ is zero or $x - 3$ is zero. Solve each small equation, and drop neither one.

x - 2 = 0 or x - 3 = 0

Why not divide by

x - 2

Dividing by $x - 2$ would assume $x$ is not $2$ . That would throw away one of the solutions you are trying to find. Setting each factor equal to zero keeps both possibilities.

Substitute each answer into the original equation

Check $x = 2$ and $x = 3$ in the original expression. Both give zero, so both are solutions. The substitutions also check the signs independently.

\begin{aligned} 2^{2} - 5 (2) + 6 & = 4 - 10 + 6 = 0 \\ 3^{2} - 5 (3) + 6 & = 9 - 15 + 6 = 0 \end{aligned}

Predict the graph before opening it

The curve should cross the horizontal axis at $2$ and $3$ . At $x = 2.5$ its height is $- 0.25$ , so it should dip below the axis between the roots. Open the graph to compare those predictions with the curve.

(2.5 - 2) (2.5 - 3) = (0.5) (- 0.5) = - 0.25

Result

The solutions are $x = 2$ and $x = 3$ .

Your turn

1. One like the example

Solve $x^{2} - 7 x + 12 = 0$ by factoring.

Show the answer and explanation

$x = 3$ or $x = 4$ .

The numbers $- 3$ and $- 4$ have product $12$ and sum $- 7$ , so $x^{2} - 7 x + 12$ $= (x - 3) (x - 4)$ . Each factor gives one solution. Check: $9 - 21 + 12 = 0$ and $16 - 28 + 12 = 0$ .

\begin{array}{l} x^{2} - 7 x + 12 = 0 \\ (x - 3) (x - 4) = 0 \\ x = 3 or x = 4 \end{array}

2. A variation

Solve $x^{2} + 2 x - 15 = 0$ . The constant is negative this time.

Show the answer and explanation

$x = - 5$ or $x = 3$ .

A negative product means the two numbers have opposite signs. The pair $5$ and $- 3$ has product $- 15$ and sum $2$ , so the factors are $(x + 5) (x - 3)$ . Check: $25 - 10 - 15 = 0$ and $9 + 6 - 15 = 0$ .

\begin{array}{l} x^{2} + 2 x - 15 = 0 \\ (x + 5) (x - 3) = 0 \\ x = - 5 or x = 3 \end{array}

3. Somewhere new

Solve $x^{3} = 9 x$ . Do not divide by $x$ .

Show the answer and explanation

$x = 0$ , $x = 3$ or $x = - 3$ .

Subtract $9 x$ to get $x^{3} - 9 x = 0$ . The common factor $x$ comes out first, leaving $x (x^{2} - 9) = 0$ , and $x^{2} - 9$ is a difference of squares. $x (x - 3) (x + 3) = 0$ has three factors, so there are three solutions. Dividing by $x$ would have lost $x = 0$ .

\begin{array}{l} x^{3} = 9 x \\ x^{3} - 9 x = 0 \\ x (x - 3) (x + 3) = 0 \\ x = 0 or x = 3 or x = - 3 \end{array}

Keep exploring

Open Graph and change the $6$ to $4$ . The curve then crosses the x-axis at $1$ and $4$ , the zeros of $(x - 1) (x - 4)$ .

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 6 key terms Check the factoring steps Explore the graph Open worked example on a board Factoring a quadratic in Math Reference

Your existing work stays on this device. Examples open as editable copies.

Identify what you are solving

Choose and check the factors

Rewrite the equation

Keep both possibilities

Substitute each answer into the original equation

Predict the graph before opening it

Result

Your turn

1. One like the example

x=3x=3 or x=4x=4.

2. A variation

x=−5x=−5 or x=3x=3.

3. Somewhere new

x=0x=0, x=3x=3 or x=−3x=−3.

Keep exploring

Study it, then try it

$x = 3$ or $x = 4$ .

$x = - 5$ or $x = 3$ .

$x = 0$ , $x = 3$ or $x = - 3$ .