Find the Roots (Zeros) P(x)=-x^3+x^2+20x

$P (x) = - x^{3} + x^{2} + 20 x$

Step 1

Set $- x^{3} + x^{2} + 20 x$ equal to $0$ .

$- x^{3} + x^{2} + 20 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $- x$ out of $- x^{3} + x^{2} + 20 x$ .

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

$- x \cdot x^{2} + x^{2} + 20 x = 0$

Factor $- x$ out of $x^{2}$ .

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

Factor $- x$ out of $20 x$ .

$- x \cdot x^{2} - x (- x) - x \cdot - 20 = 0$

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

$- x (x^{2} - x) - x \cdot - 20 = 0$

Factor $- x$ out of $- x (x^{2} - x) - x (- 20)$ .

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

Factor.

Factor $x^{2} - x - 20$ 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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Write the factored form using these integers.

$- x ((x - 5) (x + 4)) = 0$

Remove unnecessary parentheses.

$- x (x - 5) (x + 4) = 0$

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

$x = 0$

$x - 5 = 0$

$x + 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = 5$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $- x (x - 5) (x + 4) = 0$ true.

$x = 0, 5, - 4$

Step 3

Find the Roots (Zeros) P(x)=-x^3+x^2+20x

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