Find the x and y Intercepts -x^3-5x^2-3x+9

$- x^{3} - 5 x^{2} - 3 x + 9$

Step 1

Write $- x^{3} - 5 x^{2} - 3 x + 9$ as an equation.

$y = - x^{3} - 5 x^{2} - 3 x + 9$

Step 2

Find the x-intercepts.

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

$0 = - x^{3} - 5 x^{2} - 3 x + 9$

Solve the equation.

Rewrite the equation as $- x^{3} - 5 x^{2} - 3 x + 9 = 0$ .

$- x^{3} - 5 x^{2} - 3 x + 9 = 0$

Factor the left side of the equation.

Factor $- x^{3} - 5 x^{2} - 3 x + 9$ 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.

$p = \pm 1, \pm 9, \pm 3 q = \pm 1$

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

$\pm 1, \pm 9, \pm 3$

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.

$- 1^{3} - 5 \cdot 1^{2} - 3 \cdot 1 + 9$

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

$- 1 \cdot 1 - 5 \cdot 1^{2} - 3 \cdot 1 + 9$

Multiply $- 1$ by $1$ .

$- 1 - 5 \cdot 1^{2} - 3 \cdot 1 + 9$

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

$- 1 - 5 \cdot 1 - 3 \cdot 1 + 9$

Multiply $- 5$ by $1$ .

$- 1 - 5 - 3 \cdot 1 + 9$

Subtract $5$ from $- 1$ .

$- 6 - 3 \cdot 1 + 9$

Multiply $- 3$ by $1$ .

$- 6 - 3 + 9$

Subtract $3$ from $- 6$ .

$- 9 + 9$

Add $- 9$ and $9$ .

$0$

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.

$\frac{- x^{3} - 5 x^{2} - 3 x + 9}{x - 1}$

Divide $- x^{3} - 5 x^{2} - 3 x + 9$ 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 $0$ .

$x$

$1$

$x^{3}$

$5 x^{2}$

$3 x$

$9$

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

				-	$x^{2}$
	$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			-	$x^{3}$	+	$x^{2}$

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

			-	$x^{2}$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$

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

			-	$x^{2}$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$

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

			-	$x^{2}$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x$ .

			-	$x^{2}$	-	$6 x$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$6 x$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$
					-	$6 x^{2}$	+	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} + 6 x$

			-	$x^{2}$	-	$6 x$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$
					+	$6 x^{2}$	-	$6 x$

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

			-	$x^{2}$	-	$6 x$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$
					+	$6 x^{2}$	-	$6 x$
							-	$9 x$

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

			-	$x^{2}$	-	$6 x$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$
					+	$6 x^{2}$	-	$6 x$
							-	$9 x$	+	$9$

Divide the highest order term in the dividend $- 9 x$ by the highest order term in divisor $x$ .

			-	$x^{2}$	-	$6 x$	-	$9$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$
					+	$6 x^{2}$	-	$6 x$
							-	$9 x$	+	$9$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$6 x$	-	$9$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$
					+	$6 x^{2}$	-	$6 x$
							-	$9 x$	+	$9$
							-	$9 x$	+	$9$

The expression needs to be subtracted from the dividend, so change all the signs in $- 9 x + 9$

			-	$x^{2}$	-	$6 x$	-	$9$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$
					+	$6 x^{2}$	-	$6 x$
							-	$9 x$	+	$9$
							+	$9 x$	-	$9$

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

			-	$x^{2}$	-	$6 x$	-	$9$
$x$	-	$1$	-	$x^{3}$	-	$5 x^{2}$	-	$3 x$	+	$9$
			+	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$3 x$
					+	$6 x^{2}$	-	$6 x$
							-	$9 x$	+	$9$
							+	$9 x$	-	$9$
										$0$

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

$- x^{2} - 6 x - 9$

Write $- x^{3} - 5 x^{2} - 3 x + 9$ as a set of factors.

$(x - 1) (- x^{2} - 6 x - 9) = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 9 = 9$ and whose sum is $b = - 6$ .

Factor $- 6$ out of $- 6 x$ .

$(x - 1) (- x^{2} - 6 x - 9) = 0$

Rewrite $- 6$ as $- 3$ plus $- 3$

$(x - 1) (- x^{2} + (- 3 - 3) x - 9) = 0$

Apply the distributive property.

$(x - 1) (- x^{2} - 3 x - 3 x - 9) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) ((- x^{2} - 3 x) - 3 x - 9) = 0$

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

$(x - 1) (x (- x - 3) + 3 (- x - 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $- x - 3$ .

$(x - 1) ((- x - 3) (x + 3)) = 0$

Remove unnecessary parentheses.

$(x - 1) (- x - 3) (x + 3) = 0$

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

$x - 1 = 0$

$- x - 3 = 0$

$x + 3 = 0$

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 $- x - 3$ equal to $0$ and solve for $x$ .

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

$- x - 3 = 0$

Solve $- x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$- x = 3$

Divide each term in $- x = 3$ by $- 1$ and simplify.

Divide each term in $- x = 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{3}{- 1}$

Simplify the right side.

Divide $3$ by $- 1$ .

$x = - 3$

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

$x = 1, - 3$

x-intercept(s) in point form.

x-intercept(s): $(1, 0), (- 3, 0)$

Step 3

Find the y-intercepts.

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

$y = - {(0)}^{3} - 5 {(0)}^{2} - 3 \cdot 0 + 9$

Solve the equation.

Remove parentheses.

$y = - 0^{3} - 5 {(0)}^{2} - 3 \cdot 0 + 9$

Remove parentheses.

$y = - 0^{3} - 5 \cdot 0^{2} - 3 \cdot 0 + 9$

Remove parentheses.

$y = - {(0)}^{3} - 5 {(0)}^{2} - 3 \cdot 0 + 9$

Simplify $- {(0)}^{3} - 5 {(0)}^{2} - 3 \cdot 0 + 9$ .

Simplify each term.

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

$y = - 0 - 5 {(0)}^{2} - 3 \cdot 0 + 9$

Multiply $- 1$ by $0$ .

$y = 0 - 5 {(0)}^{2} - 3 \cdot 0 + 9$

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

$y = 0 - 5 \cdot 0 - 3 \cdot 0 + 9$

Multiply $- 5$ by $0$ .

$y = 0 + 0 - 3 \cdot 0 + 9$

Multiply $- 3$ by $0$ .

$y = 0 + 0 + 0 + 9$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 0 + 9$

Add $0$ and $0$ .

$y = 0 + 9$

Add $0$ and $9$ .

$y = 9$

y-intercept(s) in point form.

y-intercept(s): $(0, 9)$

Step 4

List the intersections.

x-intercept(s): $(1, 0), (- 3, 0)$

y-intercept(s): $(0, 9)$

Step 5

Find the x and y Intercepts -x^3-5x^2-3x+9

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