Factor x^4-10x^2=-9

$x^{4} - 10 x^{2} = - 9$

Step 1

Add $9$ to both sides of the equation.

$x^{4} - 10 x^{2} + 9 = 0$

Step 2

Factor the left side.

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

${(x^{2})}^{2} - 10 x^{2} + 9 = 0$

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

$u^{2} - 10 u + 9 = 0$

Factor $u^{2} - 10 u + 9$ 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 $9$ and whose sum is $- 10$ .

$- 9, - 1$

Write the factored form using these integers.

$(u - 9) (u - 1) = 0$

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

$(x^{2} - 9) (x^{2} - 1) = 0$

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

$(x^{2} - 3^{2}) (x^{2} - 1) = 0$

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 $b = 3$ .

$(x + 3) (x - 3) (x^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$(x + 3) (x - 3) (x^{2} - 1^{2}) = 0$

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 $b = 1$ .

$(x + 3) (x - 3) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x + 3) (x - 3) (x + 1) (x - 1) = 0$

Factor x^4-10x^2=-9

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