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

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

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

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

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

$x = 0$

$x - 1 = 0$

$x^{2} + 4 = 0$

$x - 1 = 0$

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

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

Set $x$ equal to $0$ .

$x = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4$ equal to $0$ .

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

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

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{4}$

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

$x = \pm i \cdot \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

Set ${(4 x + 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(4 x + 3)}^{2}$ equal to $0$ .

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

Solve ${(4 x + 3)}^{2} = 0$ for $x$ .

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

$4 x + 3 = 0$

Solve 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 ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

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

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

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 0, 1, 2 i, - 2 i, - \frac{3}{4}, 3$

x-intercept(s) in point form.

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

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

Subtract $1$ from $0$ .

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

Multiply $0$ by $- 1$ .

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

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

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

Add $0$ and $4$ .

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

Multiply $0$ by $4$ .

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

Subtract $1$ from $0$ .

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

Multiply $0$ by $- 1$ .

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

Multiply $4$ by $0$ .

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

Add $0$ and $3$ .

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

Raise $3$ to the power of $2$ .

$y = 0 \cdot 9 {((0) - 3)}^{2}$

Multiply $0$ by $9$ .

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

Subtract $3$ from $0$ .

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

Raise $- 3$ to the power of $2$ .

$y = 0 \cdot 9$

Multiply $0$ by $9$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

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

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

Step 4

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

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