Find the x and y Intercepts (2x^3+x^2-8x-4)/(x^2-3x+2)

$\frac{2 x^{3} + x^{2} - 8 x - 4}{x^{2} - 3 x + 2}$

Step 1

Write $\frac{2 x^{3} + x^{2} - 8 x - 4}{x^{2} - 3 x + 2}$ as an equation.

$y = \frac{2 x^{3} + x^{2} - 8 x - 4}{x^{2} - 3 x + 2}$

Step 2

Find the x-intercepts.

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

$0 = \frac{2 x^{3} + x^{2} - 8 x - 4}{x^{2} - 3 x + 2}$

Solve the equation.

Set the numerator equal to zero.

$2 x^{3} + x^{2} - 8 x - 4 = 0$

Solve the equation for $x$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{3} + x^{2}) - 8 x - 4 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

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

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

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$(2 x + 1) ((x + 2) (x - 2)) = 0$

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 $0$ .

$2 x + 1 = 0$

$x + 2 = 0$

$x - 2 = 0$

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ 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.

$x = - 2$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

Exclude the solutions that do not make $0 = \frac{2 x^{3} + x^{2} - 8 x - 4}{x^{2} - 3 x + 2}$ true.

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

x-intercept(s) in point form.

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

Step 3

Find the y-intercepts.

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

$y = \frac{2 {(0)}^{3} + {(0)}^{2} - 8 \cdot 0 - 4}{{(0)}^{2} - 3 \cdot 0 + 2}$

Solve the equation.

Remove parentheses.

$y = \frac{2 \cdot 0^{3} + {(0)}^{2} - 8 \cdot 0 - 4}{{(0)}^{2} - 3 \cdot 0 + 2}$

Remove parentheses.

$y = \frac{2 \cdot 0^{3} + 0^{2} - 8 \cdot 0 - 4}{{(0)}^{2} - 3 \cdot 0 + 2}$

Remove parentheses.

$y = \frac{2 \cdot 0^{3} + 0^{2} - 8 \cdot 0 - 4}{0^{2} - 3 \cdot 0 + 2}$

Remove parentheses.

$y = \frac{2 {(0)}^{3} + {(0)}^{2} - 8 \cdot 0 - 4}{{(0)}^{2} - 3 \cdot 0 + 2}$

Simplify $\frac{2 {(0)}^{3} + {(0)}^{2} - 8 \cdot 0 - 4}{{(0)}^{2} - 3 \cdot 0 + 2}$ .

Simplify the numerator.

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

$y = \frac{2 \cdot 0 + 0^{2} - 8 \cdot 0 - 4}{0^{2} - 3 \cdot 0 + 2}$

Multiply $2$ by $0$ .

$y = \frac{0 + 0^{2} - 8 \cdot 0 - 4}{0^{2} - 3 \cdot 0 + 2}$

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

$y = \frac{0 + 0 - 8 \cdot 0 - 4}{0^{2} - 3 \cdot 0 + 2}$

Multiply $- 8$ by $0$ .

$y = \frac{0 + 0 + 0 - 4}{0^{2} - 3 \cdot 0 + 2}$

Add $0$ and $0$ .

$y = \frac{0 + 0 - 4}{0^{2} - 3 \cdot 0 + 2}$

Add $0$ and $0$ .

$y = \frac{0 - 4}{0^{2} - 3 \cdot 0 + 2}$

Subtract $4$ from $0$ .

$y = \frac{- 4}{0^{2} - 3 \cdot 0 + 2}$

Simplify the denominator.

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

$y = \frac{- 4}{0 - 3 \cdot 0 + 2}$

Multiply $- 3$ by $0$ .

$y = \frac{- 4}{0 + 0 + 2}$

Add $0$ and $0$ .

$y = \frac{- 4}{0 + 2}$

Add $0$ and $2$ .

$y = \frac{- 4}{2}$

Divide $- 4$ by $2$ .

$y = - 2$

y-intercept(s) in point form.

y-intercept(s): $(0, - 2)$

Step 4

List the intersections.

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

y-intercept(s): $(0, - 2)$

Step 5

Find the x and y Intercepts (2x^3+x^2-8x-4)/(x^2-3x+2)

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