Pre-Algebra Examples

${(x - 8)}^{2} - 4 (x - 8) - 5 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = x - 8$ . Substitute $u$ for all occurrences of $x - 8$ .

$u^{2} - 4 u - 5 = 0$

Step 1.2

Factor $u^{2} - 4 u - 5$ using the AC method.

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Step 1.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Step 1.2.2

Write the factored form using these integers.

$(u - 5) (u + 1) = 0$

Step 1.3

Replace all occurrences of $u$ with $x - 8$ .

$(x - 8 - 5) (x - 8 + 1) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 8 - 5 = 0$

$x - 8 + 1 = 0$

Step 3

Set $x - 8 - 5$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x - 8 - 5$ equal to $0$ .

$x - 8 - 5 = 0$

Step 3.2

Solve $x - 8 - 5 = 0$ for $x$ .

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Step 3.2.1

Subtract $5$ from $- 8$ .

$x - 13 = 0$

Step 3.2.2

Add $13$ to both sides of the equation.

$x = 13$

Step 4

Set $x - 8 + 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x - 8 + 1$ equal to $0$ .

$x - 8 + 1 = 0$

Step 4.2

Solve $x - 8 + 1 = 0$ for $x$ .

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Step 4.2.1

Add $- 8$ and $1$ .

$x - 7 = 0$

Step 4.2.2

Add $7$ to both sides of the equation.

$x = 7$

Step 5

The final solution is all the values that make $(x - 8 - 5) (x - 8 + 1) = 0$ true.

$x = 13, 7$