Find the x and y Intercepts f(x)=2x^3+8x^2-8x-32

$f (x) = 2 x^{3} + 8 x^{2} - 8 x - 32$

Step 1

Find the x-intercepts.

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

$0 = 2 x^{3} + 8 x^{2} - 8 x - 32$

Solve the equation.

Rewrite the equation as $2 x^{3} + 8 x^{2} - 8 x - 32 = 0$ .

$2 x^{3} + 8 x^{2} - 8 x - 32 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{3} + 8 x^{2} - 8 x - 32$ .

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

$2 (x^{3}) + 8 x^{2} - 8 x - 32 = 0$

Factor $2$ out of $8 x^{2}$ .

$2 (x^{3}) + 2 (4 x^{2}) - 8 x - 32 = 0$

Factor $2$ out of $- 8 x$ .

$2 (x^{3}) + 2 (4 x^{2}) + 2 (- 4 x) - 32 = 0$

Factor $2$ out of $- 32$ .

$2 x^{3} + 2 (4 x^{2}) + 2 (- 4 x) + 2 \cdot - 16 = 0$

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

$2 (x^{3} + 4 x^{2}) + 2 (- 4 x) + 2 \cdot - 16 = 0$

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

$2 (x^{3} + 4 x^{2} - 4 x) + 2 \cdot - 16 = 0$

Factor $2$ out of $2 (x^{3} + 4 x^{2} - 4 x) + 2 \cdot - 16$ .

$2 (x^{3} + 4 x^{2} - 4 x - 16) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 ((x^{3} + 4 x^{2}) - 4 x - 16) = 0$

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

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

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

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

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

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

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

$x + 4 = 0$

$x + 2 = 0$

$x - 2 = 0$

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

$x = - 4, - 2, 2$

x-intercept(s) in point form.

x-intercept(s): $(- 4, 0), (- 2, 0), (2, 0)$

Step 2

Find the y-intercepts.

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

$y = 2 {(0)}^{3} + 8 {(0)}^{2} - 8 \cdot 0 - 32$

Solve the equation.

Remove parentheses.

$y = 2 \cdot 0^{3} + 8 {(0)}^{2} - 8 \cdot 0 - 32$

Remove parentheses.

$y = 2 \cdot 0^{3} + 8 \cdot 0^{2} - 8 \cdot 0 - 32$

Remove parentheses.

$y = 2 {(0)}^{3} + 8 {(0)}^{2} - 8 \cdot 0 - 32$

Simplify $2 {(0)}^{3} + 8 {(0)}^{2} - 8 \cdot 0 - 32$ .

Simplify each term.

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

$y = 2 \cdot 0 + 8 {(0)}^{2} - 8 \cdot 0 - 32$

Multiply $2$ by $0$ .

$y = 0 + 8 {(0)}^{2} - 8 \cdot 0 - 32$

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

$y = 0 + 8 \cdot 0 - 8 \cdot 0 - 32$

Multiply $8$ by $0$ .

$y = 0 + 0 - 8 \cdot 0 - 32$

Multiply $- 8$ by $0$ .

$y = 0 + 0 + 0 - 32$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 - 32$

Add $0$ and $0$ .

$y = 0 - 32$

Subtract $32$ from $0$ .

$y = - 32$

y-intercept(s) in point form.

y-intercept(s): $(0, - 32)$

Step 3

List the intersections.

x-intercept(s): $(- 4, 0), (- 2, 0), (2, 0)$

y-intercept(s): $(0, - 32)$

Step 4

Find the x and y Intercepts f(x)=2x^3+8x^2-8x-32

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