Find the Roots (Zeros) p(x)=(2x^4-5x^3+10x-25)(x^3+5)

$p (x) = (2 x^{4} - 5 x^{3} + 10 x - 25) (x^{3} + 5)$

Step 1

Set $(2 x^{4} - 5 x^{3} + 10 x - 25) (x^{3} + 5)$ equal to $0$ .

$(2 x^{4} - 5 x^{3} + 10 x - 25) (x^{3} + 5) = 0$

Step 2

Solve for $x$ .

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

$2 x^{4} - 5 x^{3} + 10 x - 25 = 0$

$x^{3} + 5 = 0$

Set $2 x^{4} - 5 x^{3} + 10 x - 25$ equal to $0$ and solve for $x$ .

Set $2 x^{4} - 5 x^{3} + 10 x - 25$ equal to $0$ .

$2 x^{4} - 5 x^{3} + 10 x - 25 = 0$

Solve $2 x^{4} - 5 x^{3} + 10 x - 25 = 0$ for $x$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{4} - 5 x^{3}) + 10 x - 25 = 0$

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

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

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

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

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

$2 x - 5 = 0$

$x^{3} + 5 = 0$

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 $1$ .

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

Set $x^{3} + 5$ equal to $0$ and solve for $x$ .

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

$x^{3} + 5 = 0$

Solve $x^{3} + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$x^{3} = - 5$

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

$x = \sqrt[3]{- 5}$

Simplify $\sqrt[3]{- 5}$ .

Rewrite $- 5$ as ${(- 1)}^{3} \cdot 5$ .

Rewrite $- 5$ as $- 1 (5)$ .

$x = \sqrt[3]{- 1 (5)}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$x = \sqrt[3]{{(- 1)}^{3} \cdot 5}$

Pull terms out from under the radical.

$x = - 1 \sqrt[3]{5}$

Rewrite $- 1 \sqrt[3]{5}$ as $- \sqrt[3]{5}$ .

$x = - \sqrt[3]{5}$

The final solution is all the values that make $(2 x - 5) (x^{3} + 5) = 0$ true.

$x = \frac{5}{2}, - \sqrt[3]{5}$

Set $x^{3} + 5$ equal to $0$ and solve for $x$ .

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

$x^{3} + 5 = 0$

Solve $x^{3} + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$x^{3} = - 5$

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

$x = \sqrt[3]{- 5}$

Simplify $\sqrt[3]{- 5}$ .

Rewrite $- 5$ as ${(- 1)}^{3} \cdot 5$ .

Rewrite $- 5$ as $- 1 (5)$ .

$x = \sqrt[3]{- 1 (5)}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$x = \sqrt[3]{{(- 1)}^{3} \cdot 5}$

Pull terms out from under the radical.

$x = - 1 \sqrt[3]{5}$

Rewrite $- 1 \sqrt[3]{5}$ as $- \sqrt[3]{5}$ .

$x = - \sqrt[3]{5}$

The final solution is all the values that make $(2 x^{4} - 5 x^{3} + 10 x - 25) (x^{3} + 5) = 0$ true.

$x = \frac{5}{2}, - \sqrt[3]{5}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{2}, - \sqrt[3]{5}$

Decimal Form:

$x = 2.5, - 1.70997594 \dots$

Mixed Number Form:

$x = 2 \frac{1}{2}, - \sqrt[3]{5}$

Step 4

Find the Roots (Zeros) p(x)=(2x^4-5x^3+10x-25)(x^3+5)

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