Algebra Examples

$4 x^{5} - 8 x^{3} + 4 x = 0$

Step 1

Factor $4 x$ out of $4 x^{5} - 8 x^{3} + 4 x$ .

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Step 1.1

Factor $4 x$ out of $4 x^{5}$ .

$4 x (x^{4}) - 8 x^{3} + 4 x = 0$

Step 1.2

Factor $4 x$ out of $- 8 x^{3}$ .

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

Step 1.3

Factor $4 x$ out of $4 x$ .

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

Step 1.4

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

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

Step 1.5

Factor $4 x$ out of $4 x (x^{4} - 2 x^{2}) + 4 x (1)$ .

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

Step 2

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

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

Step 3

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

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

Step 4

Factor using the perfect square rule.

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Step 4.1

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

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

Step 4.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Step 4.3

Rewrite the polynomial.

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

Step 4.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 1$ .

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

Step 5

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

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

Step 6

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

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

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 $b = 1$ .

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

Step 8

Factor.

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Step 8.1

Apply the product rule to $(x + 1) (x - 1)$ .

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

Step 8.2

Remove unnecessary parentheses.

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

Step 9

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

$x = 0$

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

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

Step 10

Set $x$ equal to $0$ .

$x = 0$

Step 11

Set ${(x + 1)}^{2}$ equal to $0$ and solve for $x$ .

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Step 11.1

Set ${(x + 1)}^{2}$ equal to $0$ .

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

Step 11.2

Solve ${(x + 1)}^{2} = 0$ for $x$ .

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Step 11.2.1

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

$x + 1 = 0$

Step 11.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 12

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

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Step 12.1

Set ${(x - 1)}^{2}$ equal to $0$ .

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

Step 12.2

Solve ${(x - 1)}^{2} = 0$ for $x$ .

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Step 12.2.1

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

$x - 1 = 0$

Step 12.2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 13

The final solution is all the values that make $4 x {(x + 1)}^{2} {(x - 1)}^{2} = 0$ true.

$x = 0, - 1, 1$