Math · College algebra · Worked example

Factor a quadratic by splitting the middle term

Solve $6 x^{2} + x - 2 = 0$ by factoring.

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

Look for a common factor

$6$ , $1$ and $- 2$ share no factor bigger than $1$ , so there is nothing to take out. The $x^{2}$ coefficient is $6$ , not $1$ , so the product and sum rule will not work directly. Split the middle term instead.

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

Find the pair for $a c$

Multiply $a$ by $c$ to get $6 \times (- 2) = - 12$ . Look for two numbers with product $- 12$ and sum $1$ , the coefficient of $x$ . The product is negative, so the signs differ. The pair $4$ and $- 3$ works.

4 \cdot (- 3) = - 12, 4 + (- 3) = 1

Split the middle term

Write $x$ as $4 x - 3 x$ . The expression has the same value, but now it has four terms that pair up.

6 x^{2} + 4 x - 3 x - 2 = 0

Factor each pair

The first pair has the common factor $2 x$ , leaving $3 x + 2$ . Take $- 1$ out of the second pair so that the same bracket appears, since $- 3 x - 2 = - (3 x + 2)$ .

2 x (3 x + 2) - (3 x + 2) = 0

Take out the shared bracket

Both parts contain $3 x + 2$ . Taking it out leaves $2 x$ from the first part and $- 1$ from the second.

(3 x + 2) (2 x - 1) = 0

Set each factor to zero

$3 x + 2 = 0$ gives $x = - 2 / 3$ , and $2 x - 1 = 0$ gives $x = 1 / 2$ . Write both answers on one row, joined by “or”.

x = - \frac{2}{3} or x = \frac{1}{2}

Check both answers

At $x = 1 / 2$ , $6 (1 / 4) + 1 / 2 - 2$ $= 3 / 2 + 1 / 2 - 2$ $= 0$ . At $x = - 2 / 3$ , $6 (4 / 9) - 2 / 3 - 2$ $= 8 / 3 - 2 / 3 - 2$ $= 0$ . Multiplying the factors back also gives $6 x^{2} - 3 x + 4 x - 2$ $= 6 x^{2} + x - 2$ .

\begin{aligned} 6 (\frac{1}{4}) + \frac{1}{2} - 2 & = 0 \\ 6 (\frac{4}{9}) - \frac{2}{3} - 2 & = 0 \end{aligned}

Result

$x = - 2 / 3$ or $x = 1 / 2$ , from $6 x^{2} + x - 2$ $= (3 x + 2) (2 x - 1)$ .

x = - \frac{2}{3} or x = \frac{1}{2}

Your turn

1. One like the example

Solve $2 x^{2} + 7 x + 3 = 0$ by splitting the middle term.

Show the answer and explanation

$x = - 1 / 2$ or $x = - 3$ .

$a c = 2 \times 3 = 6$ , and the pair $6$ and $1$ has sum $7$ . Then $2 x^{2} + 6 x + x + 3$ $= 2 x (x + 3) + (x + 3)$ $= (2 x + 1) (x + 3)$ . Each factor gives one solution.

\begin{array}{l} 2 x^{2} + 7 x + 3 = 0 \\ 2 x^{2} + 6 x + x + 3 = 0 \\ (2 x + 1) (x + 3) = 0 \\ x = - \frac{1}{2} or x = - 3 \end{array}

2. A variation

Solve $3 x^{2} - 12 = 0$ . Look for a common factor first.

Show the answer and explanation

$x = 2$ or $x = - 2$ .

The common factor $3$ comes out, leaving $3 (x^{2} - 4) = 0$ . $x^{2} - 4$ is a difference of squares, so $3 (x - 2) (x + 2) = 0$ . The factor $3$ is never zero, and the other two give $x = 2$ and $x = - 2$ .

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

3. Somewhere new

A ball is thrown up from a $15 m$ roof. Its height after $t$ seconds is $h = - 5 t^{2} + 10 t + 15$ meters. When does it hit the ground?

Show the answer and explanation

After $3 seconds$ .

The ground is $h = 0$ . The common factor $- 5$ comes out, leaving $- 5 (t^{2} - 2 t - 3) = 0$ , and $t^{2} - 2 t - 3$ $= (t - 3) (t + 1)$ . The solutions are $t = 3$ and $t = - 1$ . A time before the throw makes no sense here, so the ball lands at $t = 3 s$ . In a Math box, write both roots on the last row and reject $t = - 1$ in words. A last row of $t = 3$ alone reads ✓, with a note that $t = - 1$ also solves the equation. Write $t > 0$ on a row above, and it reads ✓ with no note.

\begin{array}{l} - 5 t^{2} + 10 t + 15 = 0 \\ - 5 (t^{2} - 2 t - 3) = 0 \\ - 5 (t - 3) (t + 1) = 0 \\ t = 3 or t = - 1 \end{array}

Keep exploring

Open the steps in Math and change the last row to $x = 2 / 3$ or $x = 1 / 2$ . The Checker marks it ✗, because $2 / 3$ does not solve the equation.

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Study the 6 key terms Check the steps in Math See the x-intercepts in 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.

Look for a common factor

Find the pair for a⁢cac

Split the middle term

Factor each pair

Take out the shared bracket

Set each factor to zero

Check both answers

Result

Your turn

1. One like the example

x=−1⁢/2x=−1/2 or x=−3x=−3.

2. A variation

x=2x=2 or x=−2x=−2.

3. Somewhere new

After 3 seconds3 seconds.

Keep exploring

Study it, then try it

Find the pair for $a c$

$x = - 1 / 2$ or $x = - 3$ .

$x = 2$ or $x = - 2$ .

After $3 seconds$ .