Solve for x -x^6-3x^5+25x^4+75x^3=0

$- x^{6} - 3 x^{5} + 25 x^{4} + 75 x^{3} = 0$

Step 1

Factor the left side of the equation.

Factor $x^{3}$ out of $- x^{6} - 3 x^{5} + 25 x^{4} + 75 x^{3}$ .

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

$x^{3} (- x^{3}) - 3 x^{5} + 25 x^{4} + 75 x^{3} = 0$

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

$x^{3} (- x^{3}) + x^{3} (- 3 x^{2}) + 25 x^{4} + 75 x^{3} = 0$

Factor $x^{3}$ out of $25 x^{4}$ .

$x^{3} (- x^{3}) + x^{3} (- 3 x^{2}) + x^{3} (25 x) + 75 x^{3} = 0$

Factor $x^{3}$ out of $75 x^{3}$ .

$x^{3} (- x^{3}) + x^{3} (- 3 x^{2}) + x^{3} (25 x) + x^{3} \cdot 75 = 0$

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

$x^{3} (- x^{3} - 3 x^{2}) + x^{3} (25 x) + x^{3} \cdot 75 = 0$

Factor $x^{3}$ out of $x^{3} (- x^{3} - 3 x^{2}) + x^{3} (25 x)$ .

$x^{3} (- x^{3} - 3 x^{2} + 25 x) + x^{3} \cdot 75 = 0$

Factor $x^{3}$ out of $x^{3} (- x^{3} - 3 x^{2} + 25 x) + x^{3} \cdot 75$ .

$x^{3} (- x^{3} - 3 x^{2} + 25 x + 75) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x^{3} ((- x^{3} - 3 x^{2}) + 25 x + 75) = 0$

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

$x^{3} (x^{2} (- x - 3) - 25 (- x - 3)) = 0$

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

$x^{3} ((- x - 3) (x^{2} - 25)) = 0$

Rewrite $25$ as $5^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Step 2

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

$x^{3} = 0$

$- x - 3 = 0$

$x + 5 = 0$

$x - 5 = 0$

Step 3

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 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 $x$ .

Add $3$ to both sides of the equation.

$- x = 3$

Divide each term in $- x = 3$ by $- 1$ and simplify.

Divide each term in $- x = 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{3}{- 1}$

Simplify the right side.

Divide $3$ by $- 1$ .

$x = - 3$

Step 5

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 7

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

$x = 0, - 3, - 5, 5$

Solve for x -x^6-3x^5+25x^4+75x^3=0

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