Algebra Examples

$4 x^{5} - 40 x^{3} + 36 x = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $4 x$ out of $4 x^{5} - 40 x^{3} + 36 x$ .

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Step 1.1.1

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

$4 x (x^{4}) - 40 x^{3} + 36 x = 0$

Step 1.1.2

Factor $4 x$ out of $- 40 x^{3}$ .

$4 x (x^{4}) + 4 x (- 10 x^{2}) + 36 x = 0$

Step 1.1.3

Factor $4 x$ out of $36 x$ .

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

Step 1.1.4

Factor $4 x$ out of $4 x (x^{4}) + 4 x (- 10 x^{2})$ .

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

Step 1.1.5

Factor $4 x$ out of $4 x (x^{4} - 10 x^{2}) + 4 x (9)$ .

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

Factor $u^{2} - 10 u + 9$ using the AC method.

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Step 1.4.1

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 $9$ and whose sum is $- 10$ .

$- 9, - 1$

Step 1.4.2

Write the factored form using these integers.

$4 x ((u - 9) (u - 1)) = 0$

Step 1.5

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

$4 x ((x^{2} - 9) (x^{2} - 1)) = 0$

Step 1.6

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

$4 x ((x^{2} - 3^{2}) (x^{2} - 1)) = 0$

Step 1.7

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 $b = 3$ .

$4 x ((x + 3) (x - 3) (x^{2} - 1)) = 0$

Step 1.8

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

$4 x ((x + 3) (x - 3) (x^{2} - 1^{2})) = 0$

Step 1.9

Factor.

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Step 1.9.1

Factor.

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Step 1.9.1.1

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 $b = 1$ .

$4 x ((x + 3) (x - 3) ((x + 1) (x - 1))) = 0$

Step 1.9.1.2

Remove unnecessary parentheses.

$4 x ((x + 3) (x - 3) (x + 1) (x - 1)) = 0$

Step 1.9.2

Remove unnecessary parentheses.

$4 x (x + 3) (x - 3) (x + 1) (x - 1) = 0$

Step 2

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

$x = 0$

$x + 3 = 0$

$x - 3 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 3

Set $x$ equal to $0$ .

$x = 0$

Step 4

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

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Step 4.1

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

$x + 3 = 0$

Step 4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

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

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Step 5.1

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

$x - 3 = 0$

Step 5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 6

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

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Step 6.1

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

$x + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

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

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Step 7.1

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

$x - 1 = 0$

Step 7.2

Add $1$ to both sides of the equation.

$x = 1$

Step 8

The final solution is all the values that make $4 x (x + 3) (x - 3) (x + 1) (x - 1) = 0$ true.

$x = 0, - 3, 3, - 1, 1$