Pre-Algebra Examples

x^{2} - 6 x + 8 = 0

$x^{2} - 6 x + 8 = 0$

Step 1

Factor

x^{2} - 6 x + 8

$x^{2} - 6 x + 8$ 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

8

$8$ and whose sum is

- 6

$- 6$ .

- 4, - 2

$- 4, - 2$

Step 1.2

Write the factored form using these integers.

(x - 4) (x - 2) = 0

$(x - 4) (x - 2) = 0$

(x - 4) (x - 2) = 0

$(x - 4) (x - 2) = 0$

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 - 4 = 0

$x - 4 = 0$

x - 2 = 0

$x - 2 = 0$

Step 3

Set

x - 4

$x - 4$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.1

Set

x - 4

$x - 4$ equal to

0

$0$ .

x - 4 = 0

$x - 4 = 0$

Step 3.2

Add

4

$4$ to both sides of the equation.

x = 4

$x = 4$

x = 4

$x = 4$

Step 4

Set

x - 2

$x - 2$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

x - 2

$x - 2$ equal to

0

$0$ .

x - 2 = 0

$x - 2 = 0$

Step 4.2

Add

2

$2$ to both sides of the equation.

x = 2

$x = 2$

x = 2

$x = 2$

Step 5

The final solution is all the values that make

(x - 4) (x - 2) = 0

$(x - 4) (x - 2) = 0$ true.

x = 4, 2

$x = 4, 2$