Pre-Algebra Examples

x^{2} + 5 x - 24 = 0

$x^{2} + 5 x - 24 = 0$

Step 1

Factor

x^{2} + 5 x - 24

$x^{2} + 5 x - 24$ using the AC method.

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Step 1.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 24

$- 24$ and whose sum is

5

$5$ .

- 3, 8

$- 3, 8$

Step 1.2

Write the factored form using these integers.

(x - 3) (x + 8) = 0

$(x - 3) (x + 8) = 0$

(x - 3) (x + 8) = 0

$(x - 3) (x + 8) = 0$

Step 2

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

x + 8 = 0

$x + 8 = 0$

Step 3

Set

x - 3

$x - 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.1

Set

x - 3

$x - 3$ equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

Step 3.2

Add

3

$3$ to both sides of the equation.

x = 3

$x = 3$

x = 3

$x = 3$

Step 4

Set

x + 8

$x + 8$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

x + 8

$x + 8$ equal to

0

$0$ .

x + 8 = 0

$x + 8 = 0$

Step 4.2

Subtract

8

$8$ from both sides of the equation.

x = - 8

$x = - 8$

x = - 8

$x = - 8$

Step 5

The final solution is all the values that make

(x - 3) (x + 8) = 0

$(x - 3) (x + 8) = 0$ true.

x = 3, - 8

$x = 3, - 8$