Find the x and y Intercepts f(x)=6x+2x^3-4x^2-4

$f (x) = 6 x + 2 x^{3} - 4 x^{2} - 4$

Find the x-intercepts.

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

$0 = 6 x + 2 x^{3} - 4 x^{2} - 4$

Solve the equation.

Rewrite the equation as $6 x + 2 x^{3} - 4 x^{2} - 4 = 0$ .

$6 x + 2 x^{3} - 4 x^{2} - 4 = 0$

Factor the left side of the equation.

Factor $2$ out of $6 x + 2 x^{3} - 4 x^{2} - 4$ .

Factor $2$ out of $6 x$ .

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

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

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

Factor $2$ out of $- 4 x^{2}$ .

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

Factor $2$ out of $- 4$ .

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

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

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

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

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

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

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

Reorder terms.

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

Factor.

Factor $x^{3} - 2 x^{2} + 3 x - 2$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 2 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 2$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$1^{3} - 2 \cdot 1^{2} + 3 \cdot 1 - 2$

Raise $1$ to the power of $3$ .

$1 - 2 \cdot 1^{2} + 3 \cdot 1 - 2$

Raise $1$ to the power of $2$ .

$1 - 2 \cdot 1 + 3 \cdot 1 - 2$

Multiply $- 2$ by $1$ .

$1 - 2 + 3 \cdot 1 - 2$

Subtract $2$ from $1$ .

$- 1 + 3 \cdot 1 - 2$

Multiply $3$ by $1$ .

$- 1 + 3 - 2$

Add $- 1$ and $3$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 2 x^{2} + 3 x - 2}{x - 1}$

Divide $x^{3} - 2 x^{2} + 3 x - 2$ by $x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{3}$

$2 x^{2}$

$3 x$

$2$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			+	$x^{3}$	-	$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - x^{2}$

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$

Divide the highest order term in the dividend $- x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$
					-	$x^{2}$	+	$x$

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{2} + x$

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$
					+	$x^{2}$	-	$x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$
					+	$x^{2}$	-	$x$
							+	$2 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$
					+	$x^{2}$	-	$x$
							+	$2 x$	-	$2$

Divide the highest order term in the dividend $2 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$x$	+	$2$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$
					+	$x^{2}$	-	$x$
							+	$2 x$	-	$2$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	+	$2$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$
					+	$x^{2}$	-	$x$
							+	$2 x$	-	$2$
							+	$2 x$	-	$2$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x - 2$

				$x^{2}$	-	$x$	+	$2$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$
					+	$x^{2}$	-	$x$
							+	$2 x$	-	$2$
							-	$2 x$	+	$2$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$x$	+	$2$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	+	$3 x$	-	$2$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	+	$3 x$
					+	$x^{2}$	-	$x$
							+	$2 x$	-	$2$
							-	$2 x$	+	$2$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - x + 2$

Write $x^{3} - 2 x^{2} + 3 x - 2$ as a set of factors.

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

Remove unnecessary parentheses.

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

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

$x - 1 = 0$

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

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

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

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

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 1$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Subtract $8$ from $1$ .

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

Rewrite $- 7$ as $- 1 (7)$ .

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

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

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

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

$x = \frac{1 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{7}}{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Subtract $8$ from $1$ .

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

Rewrite $- 7$ as $- 1 (7)$ .

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

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

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

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

$x = \frac{1 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{7}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{1 + i \sqrt{7}}{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Subtract $8$ from $1$ .

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

Rewrite $- 7$ as $- 1 (7)$ .

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

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

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

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

$x = \frac{1 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{7}}{2}$

Change the $\pm$ to $-$ .

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

The final answer is the combination of both solutions.

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

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

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

x-intercept(s) in point form.

x-intercept(s): $(1, 0)$

Find the y-intercepts.

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

$y = 6 (0) + 2 {(0)}^{3} - 4 {(0)}^{2} - 4$

Solve the equation.

Remove parentheses.

$y = 6 (0) + 2 \cdot 0^{3} - 4 {(0)}^{2} - 4$

Remove parentheses.

$y = 6 (0) + 2 \cdot 0^{3} - 4 \cdot 0^{2} - 4$

Remove parentheses.

$y = 6 (0) + 2 {(0)}^{3} - 4 {(0)}^{2} - 4$

Simplify $6 (0) + 2 {(0)}^{3} - 4 {(0)}^{2} - 4$ .

Simplify each term.

Multiply $6$ by $0$ .

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

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

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

Multiply $2$ by $0$ .

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

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

$y = 0 + 0 - 4 \cdot 0 - 4$

Multiply $- 4$ by $0$ .

$y = 0 + 0 + 0 - 4$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 - 4$

Add $0$ and $0$ .

$y = 0 - 4$

Subtract $4$ from $0$ .

$y = - 4$

y-intercept(s) in point form.

y-intercept(s): $(0, - 4)$

List the intersections.

x-intercept(s): $(1, 0)$

y-intercept(s): $(0, - 4)$

Find the x and y Intercepts f(x)=6x+2x^3-4x^2-4

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