Find the Roots (Zeros) f(x)=x^3+3x^2-6x-8

$f (x) = x^{3} + 3 x^{2} - 6 x - 8$

Step 1

Set $x^{3} + 3 x^{2} - 6 x - 8$ equal to $0$ .

$x^{3} + 3 x^{2} - 6 x - 8 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$x^{3} - 8 + 3 x^{2} - 6 x = 0$

Rewrite $8$ as $2^{3}$ .

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

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 2$ .

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

Simplify.

Move $2$ to the left of $x$ .

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

Raise $2$ to the power of $2$ .

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

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

Factor $3 x$ out of $3 x^{2}$ .

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

Factor $3 x$ out of $- 6 x$ .

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

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

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

Factor $x - 2$ out of $(x - 2) (x^{2} + 2 x + 4) + 3 x (x - 2)$ .

Factor $x - 2$ out of $3 x (x - 2)$ .

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

Factor $x - 2$ out of $(x - 2) (x^{2} + 2 x + 4) + (x - 2) (3 x)$ .

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

Add $2 x$ and $3 x$ .

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

Factor.

Factor $x^{2} + 5 x + 4$ 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 $4$ and whose sum is $5$ .

$1, 4$

Write the factored form using these integers.

$(x - 2) ((x + 1) (x + 4)) = 0$

Remove unnecessary parentheses.

$(x - 2) (x + 1) (x + 4) = 0$

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

$x - 2 = 0$

$x + 1 = 0$

$x + 4 = 0$

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.

$x = 2$

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 $x + 4$ 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 $(x - 2) (x + 1) (x + 4) = 0$ true.

$x = 2, - 1, - 4$

Step 3

Find the Roots (Zeros) f(x)=x^3+3x^2-6x-8

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