Solve by Factoring x^4-29x^2=-100

$x^{4} - 29 x^{2} = - 100$

Step 1

Add $100$ to both sides of the equation.

$x^{4} - 29 x^{2} + 100 = 0$

Step 2

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

${(x^{2})}^{2} - 29 x^{2} + 100 = 0$

Step 3

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

$u^{2} - 29 u + 100 = 0$

Step 4

Factor $u^{2} - 29 u + 100$ using the AC method.

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 $100$ and whose sum is $- 29$ .

$- 25, - 4$

Write the factored form using these integers.

$(u - 25) (u - 4) = 0$

Step 5

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

$(x^{2} - 25) (x^{2} - 4) = 0$

Step 6

Rewrite $25$ as $5^{2}$ .

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

Step 7

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 $b = 5$ .

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

Step 8

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

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

Step 9

Factor.

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 $b = 2$ .

$(x + 5) (x - 5) ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x + 5) (x - 5) (x + 2) (x - 2) = 0$

Step 10

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

$x + 5 = 0$

$x - 5 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 11

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 12

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 13

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 14

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 15

The final solution is all the values that make $(x + 5) (x - 5) (x + 2) (x - 2) = 0$ true.

$x = - 5, 5, - 2, 2$

Solve by Factoring x^4-29x^2=-100

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