Algebra Examples

$P (x) = - 5 x^{3} - 4 x^{2} + x$

Step 1

Set $- 5 x^{3} - 4 x^{2} + x$ equal to $0$ .

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

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

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

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Step 2.1.1.1

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

$- x (5 x^{2}) - 4 x^{2} + x = 0$

Step 2.1.1.2

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

$- x (5 x^{2}) - x (4 x) + x = 0$

Step 2.1.1.3

Rewrite $x$ as $1 x$ .

$- x (5 x^{2}) - x (4 x) + 1 x = 0$

Step 2.1.1.4

Factor $- x$ out of $1 x$ .

$- x (5 x^{2}) - x (4 x) - x \cdot - 1 = 0$

Step 2.1.1.5

Factor $- x$ out of $- x (5 x^{2}) - x (4 x)$ .

$- x (5 x^{2} + 4 x) - x \cdot - 1 = 0$

Step 2.1.1.6

Factor $- x$ out of $- x (5 x^{2} + 4 x) - x (- 1)$ .

$- x (5 x^{2} + 4 x - 1) = 0$

Step 2.1.2

Factor.

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Step 2.1.2.1

Factor by grouping.

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Step 2.1.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot - 1 = - 5$ and whose sum is $b = 4$ .

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Step 2.1.2.1.1.1

Factor $4$ out of $4 x$ .

$- x (5 x^{2} + 4 (x) - 1) = 0$

Step 2.1.2.1.1.2

Rewrite $4$ as $- 1$ plus $5$

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

Step 2.1.2.1.1.3

Apply the distributive property.

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

Step 2.1.2.1.2

Factor out the greatest common factor from each group.

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Step 2.1.2.1.2.1

Group the first two terms and the last two terms.

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

Step 2.1.2.1.2.2

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

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

Step 2.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $5 x - 1$ .

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

Step 2.1.2.2

Remove unnecessary parentheses.

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

Step 2.2

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

$x = 0$

$5 x - 1 = 0$

$x + 1 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set $5 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Step 2.4.2

Solve $5 x - 1 = 0$ for $x$ .

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Step 2.4.2.1

Add $1$ to both sides of the equation.

$5 x = 1$

Step 2.4.2.2

Divide each term in $5 x = 1$ by $5$ and simplify.

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Step 2.4.2.2.1

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $5$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{5}$

Step 2.5

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

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Step 2.5.1

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

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

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

$x = 0, \frac{1}{5}, - 1$

Step 3