Find the Roots (Zeros) f(x)=x^3+4x^2-x-4

$f (x) = x^{3} + 4 x^{2} - x - 4$

Step 1

Set $x^{3} + 4 x^{2} - x - 4$ equal to $0$ .

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

Step 2

Solve 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.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $x + 4$ .

$(x + 4) (x^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$(x + 4) (x^{2} - 1^{2}) = 0$

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 $b = 1$ .

$(x + 4) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x + 4) (x + 1) (x - 1) = 0$

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

$x + 4 = 0$

$x + 1 = 0$

$x - 1 = 0$

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.

$x = - 4$

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.

$x = - 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x + 4) (x + 1) (x - 1) = 0$ true.

$x = - 4, - 1, 1$

Step 3