Pre-Algebra Examples

$x^{4} - 13 x^{2} = - 36$

Step 1

Add $36$ to both sides of the equation.

$x^{4} - 13 x^{2} + 36 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

${(x^{2})}^{2} - 13 x^{2} + 36 = 0$

Step 2.2

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$u^{2} - 13 u + 36 = 0$

Step 2.3

Factor $u^{2} - 13 u + 36$ using the AC method.

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Step 2.3.1

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 $36$ and whose sum is $- 13$ .

$- 9, - 4$

Step 2.3.2

Write the factored form using these integers.

$(u - 9) (u - 4) = 0$

Step 2.4

Replace all occurrences of $u$ with $x^{2}$ .

$(x^{2} - 9) (x^{2} - 4) = 0$

Step 2.5

Rewrite $9$ as $3^{2}$ .

$(x^{2} - 3^{2}) (x^{2} - 4) = 0$

Step 2.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$(x + 3) (x - 3) (x^{2} - 4) = 0$

Step 2.7

Rewrite $4$ as $2^{2}$ .

$(x + 3) (x - 3) (x^{2} - 2^{2}) = 0$

Step 2.8

Factor.

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Step 2.8.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Step 2.8.2

Remove unnecessary parentheses.

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

Step 3

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

$x + 3 = 0$

$x - 3 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 4

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 6

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 8

The final solution is all the values that make $(x + 3) (x - 3) (x + 2) (x - 2) = 0$ true.

$x = - 3, 3, - 2, 2$