Precalculus Examples

$x^{4} - 5 x^{3} = 25 x^{2} - 125 x$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $25 x^{2}$ from both sides of the equation.

$x^{4} - 5 x^{3} - 25 x^{2} = - 125 x$

Step 1.2

Add $125 x$ to both sides of the equation.

$x^{4} - 5 x^{3} - 25 x^{2} + 125 x = 0$

Step 2

Factor $x$ out of $x^{4} - 5 x^{3} - 25 x^{2} + 125 x$ .

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Step 2.1

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

$x \cdot x^{3} - 5 x^{3} - 25 x^{2} + 125 x = 0$

Step 2.2

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

$x \cdot x^{3} + x (- 5 x^{2}) - 25 x^{2} + 125 x = 0$

Step 2.3

Factor $x$ out of $- 25 x^{2}$ .

$x \cdot x^{3} + x (- 5 x^{2}) + x (- 25 x) + 125 x = 0$

Step 2.4

Factor $x$ out of $125 x$ .

$x \cdot x^{3} + x (- 5 x^{2}) + x (- 25 x) + x \cdot 125 = 0$

Step 2.5

Factor $x$ out of $x \cdot x^{3} + x (- 5 x^{2})$ .

$x (x^{3} - 5 x^{2}) + x (- 25 x) + x \cdot 125 = 0$

Step 2.6

Factor $x$ out of $x (x^{3} - 5 x^{2}) + x (- 25 x)$ .

$x (x^{3} - 5 x^{2} - 25 x) + x \cdot 125 = 0$

Step 2.7

Factor $x$ out of $x (x^{3} - 5 x^{2} - 25 x) + x \cdot 125$ .

$x (x^{3} - 5 x^{2} - 25 x + 125) = 0$

Step 3

Factor out the greatest common factor from each group.

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Step 3.1

Group the first two terms and the last two terms.

$x ((x^{3} - 5 x^{2}) - 25 x + 125) = 0$

Step 3.2

Factor out the greatest common factor (GCF) from each group.

$x (x^{2} (x - 5) - 25 (x - 5)) = 0$

Step 4

Factor the polynomial by factoring out the greatest common factor, $x - 5$ .

$x ((x - 5) (x^{2} - 25)) = 0$

Step 5

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

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

Step 6

Factor.

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Step 6.1

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 $b = 5$ .

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

Step 6.2

Remove unnecessary parentheses.

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

Step 7

Factor.

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Step 7.1

Combine exponents.

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Step 7.1.1

Raise $x - 5$ to the power of $1$ .

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

Step 7.1.2

Raise $x - 5$ to the power of $1$ .

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

Step 7.1.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 7.1.4

Add $1$ and $1$ .

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

Step 7.2

Remove unnecessary parentheses.

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

Step 8

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

$x = 0$

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

$x + 5 = 0$

Step 9

Set $x$ equal to $0$ .

$x = 0$

Step 10

Set ${(x - 5)}^{2}$ equal to $0$ and solve for $x$ .

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Step 10.1

Set ${(x - 5)}^{2}$ equal to $0$ .

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

Step 10.2

Solve ${(x - 5)}^{2} = 0$ for $x$ .

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Step 10.2.1

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

$x - 5 = 0$

Step 10.2.2

Add $5$ to both sides of the equation.

$x = 5$

Step 11

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

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Step 11.1

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

$x + 5 = 0$

Step 11.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 12

The final solution is all the values that make $x {(x - 5)}^{2} (x + 5) = 0$ true.

$x = 0, 5, - 5$