Find the x and y Intercepts f(x)=4(4x+3)(x+1)^2(x^2-3)

$f (x) = 4 (4 x + 3) {(x + 1)}^{2} (x^{2} - 3)$

Find the x-intercepts.

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

$0 = 4 (4 x + 3) {(x + 1)}^{2} (x^{2} - 3)$

Solve the equation.

Rewrite the equation as $4 (4 x + 3) {(x + 1)}^{2} (x^{2} - 3) = 0$ .

$4 (4 x + 3) {(x + 1)}^{2} (x^{2} - 3) = 0$

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

$4 x + 3 = 0$

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

$x^{2} - 3 = 0$

Set $4 x + 3$ equal to $0$ and solve for $x$ .

Set $4 x + 3$ equal to $0$ .

$4 x + 3 = 0$

Solve $4 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$4 x = - 3$

Divide each term in $4 x = - 3$ by $4$ and simplify.

Divide each term in $4 x = - 3$ by $4$ .

$\frac{4 x}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 3}{4}$

Divide $x$ by $1$ .

$x = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{4}$

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

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

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

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

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

$x^{2} - 3 = 0$

Solve $x^{2} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

The final solution is all the values that make $4 (4 x + 3) {(x + 1)}^{2} (x^{2} - 3) = 0$ true.

$x = - \frac{3}{4}, - 1, \sqrt{3}, - \sqrt{3}$

x-intercept(s) in point form.

x-intercept(s): $(- \frac{3}{4}, 0), (- 1, 0), (\sqrt{3}, 0), (- \sqrt{3}, 0)$

Find the y-intercepts.

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

$y = 4 (4 (0) + 3) {((0) + 1)}^{2} ({(0)}^{2} - 3)$

Solve the equation.

Remove parentheses.

$y = 4 (4 (0) + 3) {(0 + 1)}^{2} ({(0)}^{2} - 3)$

Remove parentheses.

$y = 4 (4 (0) + 3) {(0 + 1)}^{2} (0^{2} - 3)$

Remove parentheses.

$y = 4 (4 (0) + 3) {((0) + 1)}^{2} ({(0)}^{2} - 3)$

Simplify $4 (4 (0) + 3) {((0) + 1)}^{2} ({(0)}^{2} - 3)$ .

Multiply $4$ by $0$ .

$y = 4 (0 + 3) {((0) + 1)}^{2} ({(0)}^{2} - 3)$

Add $0$ and $3$ .

$y = 4 \cdot 3 {((0) + 1)}^{2} ({(0)}^{2} - 3)$

Multiply $4$ by $3$ .

$y = 12 {((0) + 1)}^{2} ({(0)}^{2} - 3)$

Add $0$ and $1$ .

$y = 12 \cdot 1^{2} ({(0)}^{2} - 3)$

One to any power is one.

$y = 12 \cdot 1 ({(0)}^{2} - 3)$

Multiply $12$ by $1$ .

$y = 12 ({(0)}^{2} - 3)$

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

$y = 12 (0 - 3)$

Subtract $3$ from $0$ .

$y = 12 \cdot - 3$

Multiply $12$ by $- 3$ .

$y = - 36$

y-intercept(s) in point form.

y-intercept(s): $(0, - 36)$

List the intersections.

x-intercept(s): $(- \frac{3}{4}, 0), (- 1, 0), (\sqrt{3}, 0), (- \sqrt{3}, 0)$

y-intercept(s): $(0, - 36)$

Find the x and y Intercepts f(x)=4(4x+3)(x+1)^2(x^2-3)

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