Find the Roots (Zeros) (3x^4-8x^3+15x-40)(3x-8)^2

$(3 x^{4} - 8 x^{3} + 15 x - 40) {(3 x - 8)}^{2}$

Step 1

Set $(3 x^{4} - 8 x^{3} + 15 x - 40) {(3 x - 8)}^{2}$ equal to $0$ .

$(3 x^{4} - 8 x^{3} + 15 x - 40) {(3 x - 8)}^{2} = 0$

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 $0$ .

$3 x^{4} - 8 x^{3} + 15 x - 40 = 0$

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

Set $3 x^{4} - 8 x^{3} + 15 x - 40$ equal to $0$ and solve for $x$ .

Set $3 x^{4} - 8 x^{3} + 15 x - 40$ equal to $0$ .

$3 x^{4} - 8 x^{3} + 15 x - 40 = 0$

Solve $3 x^{4} - 8 x^{3} + 15 x - 40 = 0$ 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.

$(3 x^{4} - 8 x^{3}) + 15 x - 40 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 8$ .

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

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

$3 x - 8 = 0$

$x^{3} + 5 = 0$

Set $3 x - 8$ equal to $0$ and solve for $x$ .

Set $3 x - 8$ equal to $0$ .

$3 x - 8 = 0$

Solve $3 x - 8 = 0$ for $x$ .

Add $8$ to both sides of the equation.

$3 x = 8$

Divide each term in $3 x = 8$ by $3$ and simplify.

Divide each term in $3 x = 8$ by $3$ .

$\frac{3 x}{3} = \frac{8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{8}{3}$

Divide $x$ by $1$ .

$x = \frac{8}{3}$

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 $(3 x - 8) (x^{3} + 5) = 0$ true.

$x = \frac{8}{3}, - \sqrt[3]{5}$

Set ${(3 x - 8)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(3 x - 8)}^{2}$ equal to $0$ .

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

Solve ${(3 x - 8)}^{2} = 0$ for $x$ .

Set the $3 x - 8$ equal to $0$ .

$3 x - 8 = 0$

Solve for $x$ .

Add $8$ to both sides of the equation.

$3 x = 8$

Divide each term in $3 x = 8$ by $3$ and simplify.

Divide each term in $3 x = 8$ by $3$ .

$\frac{3 x}{3} = \frac{8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{8}{3}$

Divide $x$ by $1$ .

$x = \frac{8}{3}$

The final solution is all the values that make $(3 x^{4} - 8 x^{3} + 15 x - 40) {(3 x - 8)}^{2} = 0$ true.

$x = \frac{8}{3}, - \sqrt[3]{5}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{8}{3}, - \sqrt[3]{5}$

Decimal Form:

$x = 2.\overline{6}, - 1.70997594 \dots$

Mixed Number Form:

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

Step 4

Find the Roots (Zeros) (3x^4-8x^3+15x-40)(3x-8)^2

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