Solve by Factoring 2x^4-26x^2+72=0

$2 x^{4} - 26 x^{2} + 72 = 0$

Step 1

Factor $2$ out of $2 x^{4} - 26 x^{2} + 72$ .

Factor $2$ out of $2 x^{4}$ .

$2 (x^{4}) - 26 x^{2} + 72 = 0$

Factor $2$ out of $- 26 x^{2}$ .

$2 (x^{4}) + 2 (- 13 x^{2}) + 72 = 0$

Factor $2$ out of $72$ .

$2 x^{4} + 2 (- 13 x^{2}) + 2 \cdot 36 = 0$

Factor $2$ out of $2 x^{4} + 2 (- 13 x^{2})$ .

$2 (x^{4} - 13 x^{2}) + 2 \cdot 36 = 0$

Factor $2$ out of $2 (x^{4} - 13 x^{2}) + 2 \cdot 36$ .

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

Step 2

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

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

Step 3

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

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

Step 4

Factor $u^{2} - 13 u + 36$ 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 $36$ and whose sum is $- 13$ .

$- 9, - 4$

Write the factored form using these integers.

$2 ((u - 9) (u - 4)) = 0$

Step 5

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

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

Step 6

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

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

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 $b = 3$ .

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

Step 8

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

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Step 10

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

$x + 3 = 0$

$x - 3 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 11

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 12

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

Solve by Factoring 2x^4-26x^2+72=0

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