Find the x and y Intercepts x^3+2x^2-15x-36

$x^{3} + 2 x^{2} - 15 x - 36$

Step 1

Write $x^{3} + 2 x^{2} - 15 x - 36$ as an equation.

$y = x^{3} + 2 x^{2} - 15 x - 36$

Step 2

Find the x-intercepts.

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

$0 = x^{3} + 2 x^{2} - 15 x - 36$

Solve the equation.

Rewrite the equation as $x^{3} + 2 x^{2} - 15 x - 36 = 0$ .

$x^{3} + 2 x^{2} - 15 x - 36 = 0$

Factor the left side of the equation.

Factor $x^{3} + 2 x^{2} - 15 x - 36$ 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.

$p = \pm 1, \pm 36, \pm 2, \pm 18, \pm 3, \pm 12, \pm 4, \pm 9, \pm 6 q = \pm 1$

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

$\pm 1, \pm 36, \pm 2, \pm 18, \pm 3, \pm 12, \pm 4, \pm 9, \pm 6$

Substitute $- 3$ and simplify the expression. In this case, the expression is equal to $0$ so $- 3$ is a root of the polynomial.

Substitute $- 3$ into the polynomial.

${(- 3)}^{3} + 2 {(- 3)}^{2} - 15 \cdot - 3 - 36$

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

$- 27 + 2 {(- 3)}^{2} - 15 \cdot - 3 - 36$

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

$- 27 + 2 \cdot 9 - 15 \cdot - 3 - 36$

Multiply $2$ by $9$ .

$- 27 + 18 - 15 \cdot - 3 - 36$

Add $- 27$ and $18$ .

$- 9 - 15 \cdot - 3 - 36$

Multiply $- 15$ by $- 3$ .

$- 9 + 45 - 36$

Add $- 9$ and $45$ .

$36 - 36$

Subtract $36$ from $36$ .

$0$

Since $- 3$ is a known root, divide the polynomial by $x + 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} + 2 x^{2} - 15 x - 36}{x + 3}$

Divide $x^{3} + 2 x^{2} - 15 x - 36$ by $x + 3$ .

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

$x$

$3$

$x^{3}$

$2 x^{2}$

$15 x$

$36$

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

					$x^{2}$
	$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			+	$x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$

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

				$x^{2}$	-	$x$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$
					-	$x^{2}$	-	$3 x$

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

				$x^{2}$	-	$x$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$
					+	$x^{2}$	+	$3 x$

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

				$x^{2}$	-	$x$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$
					+	$x^{2}$	+	$3 x$
							-	$12 x$

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

				$x^{2}$	-	$x$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$
					+	$x^{2}$	+	$3 x$
							-	$12 x$	-	$36$

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$x$	-	$12$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$
					+	$x^{2}$	+	$3 x$
							-	$12 x$	-	$36$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$12$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$
					+	$x^{2}$	+	$3 x$
							-	$12 x$	-	$36$
							-	$12 x$	-	$36$

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 x - 36$

				$x^{2}$	-	$x$	-	$12$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$
					+	$x^{2}$	+	$3 x$
							-	$12 x$	-	$36$
							+	$12 x$	+	$36$

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

				$x^{2}$	-	$x$	-	$12$
$x$	+	$3$		$x^{3}$	+	$2 x^{2}$	-	$15 x$	-	$36$
			-	$x^{3}$	-	$3 x^{2}$
					-	$x^{2}$	-	$15 x$
					+	$x^{2}$	+	$3 x$
							-	$12 x$	-	$36$
							+	$12 x$	+	$36$
										$0$

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

$x^{2} - x - 12$

Write $x^{3} + 2 x^{2} - 15 x - 36$ as a set of factors.

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

Factor $x^{2} - x - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$(x + 3) ((x - 4) (x + 3)) = 0$

Remove unnecessary parentheses.

$(x + 3) (x - 4) (x + 3) = 0$

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

$x + 3 = 0$

$x - 4 = 0$

$x + 3 = 0$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

$x = - 3, 4$

x-intercept(s) in point form.

x-intercept(s): $(- 3, 0), (4, 0)$

Step 3

Find the y-intercepts.

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

$y = {(0)}^{3} + 2 {(0)}^{2} - 15 \cdot 0 - 36$

Solve the equation.

Remove parentheses.

$y = 0^{3} + 2 {(0)}^{2} - 15 \cdot 0 - 36$

Remove parentheses.

$y = 0^{3} + 2 \cdot 0^{2} - 15 \cdot 0 - 36$

Remove parentheses.

$y = {(0)}^{3} + 2 {(0)}^{2} - 15 \cdot 0 - 36$

Simplify ${(0)}^{3} + 2 {(0)}^{2} - 15 \cdot 0 - 36$ .

Simplify each term.

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

$y = 0 + 2 {(0)}^{2} - 15 \cdot 0 - 36$

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

$y = 0 + 2 \cdot 0 - 15 \cdot 0 - 36$

Multiply $2$ by $0$ .

$y = 0 + 0 - 15 \cdot 0 - 36$

Multiply $- 15$ by $0$ .

$y = 0 + 0 + 0 - 36$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 - 36$

Add $0$ and $0$ .

$y = 0 - 36$

Subtract $36$ from $0$ .

$y = - 36$

y-intercept(s) in point form.

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

Step 4

List the intersections.

x-intercept(s): $(- 3, 0), (4, 0)$

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

Step 5

Find the x and y Intercepts x^3+2x^2-15x-36

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