Find the x and y Intercepts x^3+x^2-4x-4

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

Write $x^{3} + x^{2} - 4 x - 4$ as an equation.

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

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

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

Solve the equation.

Rewrite the equation as $x^{3} + x^{2} - 4 x - 4 = 0$ .

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

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.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

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

Rewrite $4$ as $2^{2}$ .

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

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

$x + 1 = 0$

$x + 2 = 0$

$x - 2 = 0$

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 $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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.

$x = 2$

The final solution is all the values that make $(x + 1) (x + 2) (x - 2) = 0$ true.

$x = - 1, - 2, 2$

x-intercept(s) in point form.

x-intercept(s): $(- 1, 0), (- 2, 0), (2, 0)$

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = {(0)}^{3} + {(0)}^{2} - 4 \cdot 0 - 4$

Solve the equation.

Remove parentheses.

$y = 0^{3} + {(0)}^{2} - 4 \cdot 0 - 4$

Remove parentheses.

$y = 0^{3} + 0^{2} - 4 \cdot 0 - 4$

Remove parentheses.

$y = {(0)}^{3} + {(0)}^{2} - 4 \cdot 0 - 4$

Simplify ${(0)}^{3} + {(0)}^{2} - 4 \cdot 0 - 4$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 0 + {(0)}^{2} - 4 \cdot 0 - 4$

Raising $0$ to any positive power yields $0$ .

$y = 0 + 0 - 4 \cdot 0 - 4$

Multiply $- 4$ by $0$ .

$y = 0 + 0 + 0 - 4$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 - 4$

Add $0$ and $0$ .

$y = 0 - 4$

Subtract $4$ from $0$ .

$y = - 4$

y-intercept(s) in point form.

y-intercept(s): $(0, - 4)$

List the intersections.

x-intercept(s): $(- 1, 0), (- 2, 0), (2, 0)$

y-intercept(s): $(0, - 4)$

Find the x and y Intercepts x^3+x^2-4x-4

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