Find the Roots (Zeros) f(x)=x^3-5x^2-6x

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

Step 1

Set $x^{3} - 5 x^{2} - 6 x$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 5 x^{2} - 6 x$ .

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

$x \cdot x^{2} - 5 x^{2} - 6 x = 0$

Factor $x$ out of $- 5 x^{2}$ .

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

Factor $x$ out of $- 6 x$ .

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

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

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

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

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

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 $- 5$ .

$- 6, 1$

Write the factored form using these integers.

$x ((x - 6) (x + 1)) = 0$

Remove unnecessary parentheses.

$x (x - 6) (x + 1) = 0$

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

$x = 0$

$x - 6 = 0$

$x + 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $x (x - 6) (x + 1) = 0$ true.

$x = 0, 6, - 1$

Step 3

Find the Roots (Zeros) f(x)=x^3-5x^2-6x

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