Solve by Factoring 8x^4-18x^3-5x^2=0

$8 x^{4} - 18 x^{3} - 5 x^{2} = 0$

Step 1

Factor $x^{2}$ out of $8 x^{4} - 18 x^{3} - 5 x^{2}$ .

Factor $x^{2}$ out of $8 x^{4}$ .

$x^{2} (8 x^{2}) - 18 x^{3} - 5 x^{2} = 0$

Factor $x^{2}$ out of $- 18 x^{3}$ .

$x^{2} (8 x^{2}) + x^{2} (- 18 x) - 5 x^{2} = 0$

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

$x^{2} (8 x^{2}) + x^{2} (- 18 x) + x^{2} \cdot - 5 = 0$

Factor $x^{2}$ out of $x^{2} (8 x^{2}) + x^{2} (- 18 x)$ .

$x^{2} (8 x^{2} - 18 x) + x^{2} \cdot - 5 = 0$

Factor $x^{2}$ out of $x^{2} (8 x^{2} - 18 x) + x^{2} \cdot - 5$ .

$x^{2} (8 x^{2} - 18 x - 5) = 0$

Step 2

Factor.

Factor by grouping.

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 $a \cdot c = 8 \cdot - 5 = - 40$ and whose sum is $b = - 18$ .

Factor $- 18$ out of $- 18 x$ .

$x^{2} (8 x^{2} - 18 x - 5) = 0$

Rewrite $- 18$ as $2$ plus $- 20$

$x^{2} (8 x^{2} + (2 - 20) x - 5) = 0$

Apply the distributive property.

$x^{2} (8 x^{2} + 2 x - 20 x - 5) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x^{2} ((8 x^{2} + 2 x) - 20 x - 5) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $4 x + 1$ .

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

Remove unnecessary parentheses.

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

Step 3

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

$x^{2} = 0$

$4 x + 1 = 0$

$2 x - 5 = 0$

Step 4

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 5

Set $4 x + 1$ equal to $0$ and solve for $x$ .

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Solve $4 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Divide each term in $4 x = - 1$ by $4$ and simplify.

Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Step 6

Set $2 x - 5$ equal to $0$ and solve for $x$ .

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Solve $2 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 7

The final solution is all the values that make $x^{2} (4 x + 1) (2 x - 5) = 0$ true.

$x = 0, - \frac{1}{4}, \frac{5}{2}$

Solve by Factoring 8x^4-18x^3-5x^2=0

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