Find the x and y Intercepts f(x)=2x^3+2x^2-24x

$f (x) = 2 x^{3} + 2 x^{2} - 24 x$

Step 1

Find the x-intercepts.

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

$0 = 2 x^{3} + 2 x^{2} - 24 x$

Solve the equation.

Rewrite the equation as $2 x^{3} + 2 x^{2} - 24 x = 0$ .

$2 x^{3} + 2 x^{2} - 24 x = 0$

Factor the left side of the equation.

Factor $2 x$ out of $2 x^{3} + 2 x^{2} - 24 x$ .

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

$2 x (x^{2}) + 2 x^{2} - 24 x = 0$

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

$2 x (x^{2}) + 2 x (x) - 24 x = 0$

Factor $2 x$ out of $- 24 x$ .

$2 x (x^{2}) + 2 x (x) + 2 x (- 12) = 0$

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

$2 x (x^{2} + x) + 2 x (- 12) = 0$

Factor $2 x$ out of $2 x (x^{2} + x) + 2 x (- 12)$ .

$2 x (x^{2} + x - 12) = 0$

Factor.

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$2 x ((x - 3) (x + 4)) = 0$

Remove unnecessary parentheses.

$2 x (x - 3) (x + 4) = 0$

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

$x = 0$

$x - 3 = 0$

$x + 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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 $2 x (x - 3) (x + 4) = 0$ true.

$x = 0, 3, - 4$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (3, 0), (- 4, 0)$

Step 2

Find the y-intercepts.

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

$y = 2 {(0)}^{3} + 2 {(0)}^{2} - 24 \cdot 0$

Solve the equation.

Remove parentheses.

$y = 2 \cdot 0^{3} + 2 {(0)}^{2} - 24 \cdot 0$

Remove parentheses.

$y = 2 \cdot 0^{3} + 2 \cdot 0^{2} - 24 \cdot 0$

Remove parentheses.

$y = 2 {(0)}^{3} + 2 {(0)}^{2} - 24 \cdot 0$

Simplify $2 {(0)}^{3} + 2 {(0)}^{2} - 24 \cdot 0$ .

Simplify each term.

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

$y = 2 \cdot 0 + 2 {(0)}^{2} - 24 \cdot 0$

Multiply $2$ by $0$ .

$y = 0 + 2 {(0)}^{2} - 24 \cdot 0$

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

$y = 0 + 2 \cdot 0 - 24 \cdot 0$

Multiply $2$ by $0$ .

$y = 0 + 0 - 24 \cdot 0$

Multiply $- 24$ by $0$ .

$y = 0 + 0 + 0$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

x-intercept(s): $(0, 0), (3, 0), (- 4, 0)$

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

Step 4

Find the x and y Intercepts f(x)=2x^3+2x^2-24x

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