Solve by Factoring x^4-17x^2+16=0

$x^{4} - 17 x^{2} + 16 = 0$

Step 1

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

${(x^{2})}^{2} - 17 x^{2} + 16 = 0$

Step 2

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

$u^{2} - 17 u + 16 = 0$

Step 3

Factor $u^{2} - 17 u + 16$ 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 $16$ and whose sum is $- 17$ .

$- 16, - 1$

Write the factored form using these integers.

$(u - 16) (u - 1) = 0$

Step 4

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

$(x^{2} - 16) (x^{2} - 1) = 0$

Step 5

Rewrite $16$ as $4^{2}$ .

$(x^{2} - 4^{2}) (x^{2} - 1) = 0$

Step 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 $b = 4$ .

$(x + 4) (x - 4) (x^{2} - 1) = 0$

Step 7

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

$(x + 4) (x - 4) (x^{2} - 1^{2}) = 0$

Step 8

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 $b = 1$ .

$(x + 4) (x - 4) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x + 4) (x - 4) (x + 1) (x - 1) = 0$

Step 9

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

$x + 4 = 0$

$x - 4 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 10

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 11

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 12

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 13

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 14

The final solution is all the values that make $(x + 4) (x - 4) (x + 1) (x - 1) = 0$ true.

$x = - 4, 4, - 1, 1$

Solve by Factoring x^4-17x^2+16=0

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