Solve by Factoring 2x^3-5x^2+40x-100=0

$2 x^{3} - 5 x^{2} + 40 x - 100 = 0$

Step 1

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{3} - 5 x^{2}) + 40 x - 100 = 0$

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

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

Step 2

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

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

Step 3

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

$2 x - 5 = 0$

$x^{2} + 20 = 0$

Step 4

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 5

Set $x^{2} + 20$ equal to $0$ and solve for $x$ .

Set $x^{2} + 20$ equal to $0$ .

$x^{2} + 20 = 0$

Solve $x^{2} + 20 = 0$ for $x$ .

Subtract $20$ from both sides of the equation.

$x^{2} = - 20$

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

$x = \pm \sqrt{- 20}$

Simplify $\pm \sqrt{- 20}$ .

Rewrite $- 20$ as $- 1 (20)$ .

$x = \pm \sqrt{- 1 (20)}$

Rewrite $\sqrt{- 1 (20)}$ as $\sqrt{- 1} \cdot \sqrt{20}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{20}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{20}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \pm i \cdot \sqrt{4 (5)}$

Rewrite $4$ as $2^{2}$ .

$x = \pm i \cdot \sqrt{2^{2} \cdot 5}$

Pull terms out from under the radical.

$x = \pm i \cdot (2 \sqrt{5})$

Move $2$ to the left of $i$ .

$x = \pm 2 i \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2 i \sqrt{5}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2 i \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2 i \sqrt{5}, - 2 i \sqrt{5}$

Step 6

The final solution is all the values that make $(2 x - 5) (x^{2} + 20) = 0$ true.

$x = \frac{5}{2}, 2 i \sqrt{5}, - 2 i \sqrt{5}$

Solve by Factoring 2x^3-5x^2+40x-100=0

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