Solve for x x^5+4x^3=5x^4-2x^3

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

Step 1

Move all terms containing $x$ to the left side of the equation.

Subtract $5 x^{4}$ from both sides of the equation.

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

Add $2 x^{3}$ to both sides of the equation.

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

Add $4 x^{3}$ and $2 x^{3}$ .

$x^{5} - 5 x^{4} + 6 x^{3} = 0$

Step 2

Factor the left side of the equation.

Factor $x^{3}$ out of $x^{5} - 5 x^{4} + 6 x^{3}$ .

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

$x^{3} x^{2} - 5 x^{4} + 6 x^{3} = 0$

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

$x^{3} x^{2} + x^{3} (- 5 x) + 6 x^{3} = 0$

Factor $x^{3}$ out of $6 x^{3}$ .

$x^{3} x^{2} + x^{3} (- 5 x) + x^{3} \cdot 6 = 0$

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

$x^{3} (x^{2} - 5 x) + x^{3} \cdot 6 = 0$

Factor $x^{3}$ out of $x^{3} (x^{2} - 5 x) + x^{3} \cdot 6$ .

$x^{3} (x^{2} - 5 x + 6) = 0$

Factor.

Factor $x^{2} - 5 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

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

$x^{3} = 0$

$x - 3 = 0$

$x - 2 = 0$

Step 4

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

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

$x^{3} = 0$

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

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

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

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

$x = 0$

Step 5

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 6

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 7

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

$x = 0, 3, 2$

Solve for x x^5+4x^3=5x^4-2x^3

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