Find the x and y Intercepts f(x)=(x^2-7)(2x+1)^2(x+2)

$f (x) = (x^{2} - 7) {(2 x + 1)}^{2} (x + 2)$

Step 1

Find the x-intercepts.

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

$0 = (x^{2} - 7) {(2 x + 1)}^{2} (x + 2)$

Solve the equation.

Rewrite the equation as $(x^{2} - 7) {(2 x + 1)}^{2} (x + 2) = 0$ .

$(x^{2} - 7) {(2 x + 1)}^{2} (x + 2) = 0$

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

$x^{2} - 7 = 0$

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

$x + 2 = 0$

Set $x^{2} - 7$ equal to $0$ and solve for $x$ .

Set $x^{2} - 7$ equal to $0$ .

$x^{2} - 7 = 0$

Solve $x^{2} - 7 = 0$ for $x$ .

Add $7$ to both sides of the equation.

$x^{2} = 7$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{7}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{7}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{7}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{7}, - \sqrt{7}$

Set ${(2 x + 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(2 x + 1)}^{2}$ equal to $0$ .

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

Solve ${(2 x + 1)}^{2} = 0$ for $x$ .

Set the $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

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.

$x = - 2$

The final solution is all the values that make $(x^{2} - 7) {(2 x + 1)}^{2} (x + 2) = 0$ true.

$x = \sqrt{7}, - \sqrt{7}, - \frac{1}{2}, - 2$

x-intercept(s) in point form.

x-intercept(s): $(\sqrt{7}, 0), (- \sqrt{7}, 0), (- \frac{1}{2}, 0), (- 2, 0)$

Step 2

Find the y-intercepts.

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

$y = ({(0)}^{2} - 7) {(2 (0) + 1)}^{2} ((0) + 2)$

Solve the equation.

Remove parentheses.

$y = (0^{2} - 7) {(2 (0) + 1)}^{2} ((0) + 2)$

Remove parentheses.

$y = (0^{2} - 7) {(2 (0) + 1)}^{2} (0 + 2)$

Remove parentheses.

$y = ({(0)}^{2} - 7) {(2 (0) + 1)}^{2} ((0) + 2)$

Simplify $({(0)}^{2} - 7) {(2 (0) + 1)}^{2} ((0) + 2)$ .

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

$y = (0 - 7) {(2 (0) + 1)}^{2} ((0) + 2)$

Subtract $7$ from $0$ .

$y = - 7 {(2 (0) + 1)}^{2} ((0) + 2)$

Multiply $2$ by $0$ .

$y = - 7 {(0 + 1)}^{2} ((0) + 2)$

Add $0$ and $1$ .

$y = - 7 \cdot 1^{2} ((0) + 2)$

One to any power is one.

$y = - 7 \cdot 1 ((0) + 2)$

Multiply $- 7$ by $1$ .

$y = - 7 ((0) + 2)$

Add $0$ and $2$ .

$y = - 7 \cdot 2$

Multiply $- 7$ by $2$ .

$y = - 14$

y-intercept(s) in point form.

y-intercept(s): $(0, - 14)$

Step 3

List the intersections.

x-intercept(s): $(\sqrt{7}, 0), (- \sqrt{7}, 0), (- \frac{1}{2}, 0), (- 2, 0)$

y-intercept(s): $(0, - 14)$

Step 4

Find the x and y Intercepts f(x)=(x^2-7)(2x+1)^2(x+2)

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