Algebra Examples

$x^{4} - 26 x^{2} + 25 = 0$

Step 1

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

${(x^{2})}^{2} - 26 x^{2} + 25 = 0$

Step 2

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

$u^{2} - 26 u + 25 = 0$

Step 3

Factor $u^{2} - 26 u + 25$ using the AC method.

Tap for more steps...

Step 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 $25$ and whose sum is $- 26$ .

$- 25, - 1$

Step 3.2

Write the factored form using these integers.

$(u - 25) (u - 1) = 0$

Step 4

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

$(x^{2} - 25) (x^{2} - 1) = 0$

Step 5

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

$(x^{2} - 5^{2}) (x^{2} - 1) = 0$

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 $b = 5$ .

$(x + 5) (x - 5) (x^{2} - 1) = 0$

Step 7

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

$(x + 5) (x - 5) (x^{2} - 1^{2}) = 0$

Step 8

Factor.

Tap for more steps...

Step 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 $b = 1$ .

$(x + 5) (x - 5) ((x + 1) (x - 1)) = 0$

Step 8.2

Remove unnecessary parentheses.

$(x + 5) (x - 5) (x + 1) (x - 1) = 0$

Step 9

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

$x + 5 = 0$

$x - 5 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 10

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

Tap for more steps...

Step 10.1

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

$x + 5 = 0$

Step 10.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 11

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

Tap for more steps...

Step 11.1

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

$x - 5 = 0$

Step 11.2

Add $5$ to both sides of the equation.

$x = 5$

Step 12

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

Tap for more steps...

Step 12.1

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

$x + 1 = 0$

Step 12.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 13

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

Tap for more steps...

Step 13.1

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

$x - 1 = 0$

Step 13.2

Add $1$ to both sides of the equation.

$x = 1$

Step 14

The final solution is all the values that make $(x + 5) (x - 5) (x + 1) (x - 1) = 0$ true.

$x = - 5, 5, - 1, 1$