Find the x and y Intercepts 7x^4-32x^3+6x^2+72x+27

$7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27$

Step 1

Write $7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27$ as an equation.

$y = 7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27$

Step 2

Find the x-intercepts.

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

$0 = 7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27$

Solve the equation.

Rewrite the equation as $7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27 = 0$ .

$7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27 = 0$

Factor the left side of the equation.

Factor $7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27$ 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.

$p = \pm 1, \pm 27, \pm 3, \pm 9 q = \pm 1, \pm 7$

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

$\pm 1, \pm 0.\overline{142857}, \pm 27, \pm 3.\overline{857142}, \pm 3, \pm 0.\overline{428571}, \pm 9, \pm 1.\overline{285714}$

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.

$7 {(- 1)}^{4} - 32 {(- 1)}^{3} + 6 {(- 1)}^{2} + 72 \cdot - 1 + 27$

Raise $- 1$ to the power of $4$ .

$7 \cdot 1 - 32 {(- 1)}^{3} + 6 {(- 1)}^{2} + 72 \cdot - 1 + 27$

Multiply $7$ by $1$ .

$7 - 32 {(- 1)}^{3} + 6 {(- 1)}^{2} + 72 \cdot - 1 + 27$

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

$7 - 32 \cdot - 1 + 6 {(- 1)}^{2} + 72 \cdot - 1 + 27$

Multiply $- 32$ by $- 1$ .

$7 + 32 + 6 {(- 1)}^{2} + 72 \cdot - 1 + 27$

Add $7$ and $32$ .

$39 + 6 {(- 1)}^{2} + 72 \cdot - 1 + 27$

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

$39 + 6 \cdot 1 + 72 \cdot - 1 + 27$

Multiply $6$ by $1$ .

$39 + 6 + 72 \cdot - 1 + 27$

Add $39$ and $6$ .

$45 + 72 \cdot - 1 + 27$

Multiply $72$ by $- 1$ .

$45 - 72 + 27$

Subtract $72$ from $45$ .

$- 27 + 27$

Add $- 27$ and $27$ .

$0$

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.

$\frac{7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27}{x + 1}$

Divide $7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27$ 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 $0$ .

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

Divide the highest order term in the dividend $7 x^{4}$ by the highest order term in divisor $x$ .

$7 x^{3}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

Multiply the new quotient term by the divisor.

$7 x^{3}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{4} + 7 x^{3}$

$7 x^{3}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

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

$7 x^{3}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

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

$7 x^{3}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

Divide the highest order term in the dividend $- 39 x^{3}$ by the highest order term in divisor $x$ .

$7 x^{3}$

$39 x^{2}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

Multiply the new quotient term by the divisor.

$7 x^{3}$

$39 x^{2}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- 39 x^{3} - 39 x^{2}$

$7 x^{3}$

$39 x^{2}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

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

$7 x^{3}$

$39 x^{2}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

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

$7 x^{3}$

$39 x^{2}$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

Divide the highest order term in the dividend $45 x^{2}$ by the highest order term in divisor $x$ .

$7 x^{3}$

$39 x^{2}$

$45 x$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

Multiply the new quotient term by the divisor.

$7 x^{3}$

$39 x^{2}$

$45 x$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

$45 x^{2}$

$45 x$

The expression needs to be subtracted from the dividend, so change all the signs in $45 x^{2} + 45 x$

$7 x^{3}$

$39 x^{2}$

$45 x$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

$45 x^{2}$

$45 x$

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

$7 x^{3}$

$39 x^{2}$

$45 x$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

$45 x^{2}$

$45 x$

$27 x$

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

$7 x^{3}$

$39 x^{2}$

$45 x$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

$45 x^{2}$

$45 x$

$27 x$

$27$

Divide the highest order term in the dividend $27 x$ by the highest order term in divisor $x$ .

$7 x^{3}$

$39 x^{2}$

$45 x$

$27$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

$45 x^{2}$

$45 x$

$27 x$

$27$

Multiply the new quotient term by the divisor.

$7 x^{3}$

$39 x^{2}$

$45 x$

$27$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

$45 x^{2}$

$45 x$

$27 x$

$27$

$27 x$

$27$

The expression needs to be subtracted from the dividend, so change all the signs in $27 x + 27$

$7 x^{3}$

$39 x^{2}$

$45 x$

$27$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

$45 x^{2}$

$45 x$

$27 x$

$27$

$27 x$

$27$

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

$7 x^{3}$

$39 x^{2}$

$45 x$

$27$

$x$

$1$

$7 x^{4}$

$32 x^{3}$

$6 x^{2}$

$72 x$

$27$

$7 x^{4}$

$7 x^{3}$

$39 x^{3}$

$6 x^{2}$

$39 x^{3}$

$39 x^{2}$

$45 x^{2}$

$72 x$

$45 x^{2}$

$45 x$

$27 x$

$27$

$27 x$

$27$

$0$

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

$7 x^{3} - 39 x^{2} + 45 x + 27$

Write $7 x^{4} - 32 x^{3} + 6 x^{2} + 72 x + 27$ as a set of factors.

$(x + 1) (7 x^{3} - 39 x^{2} + 45 x + 27) = 0$

Factor $7 x^{3} - 39 x^{2} + 45 x + 27$ 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.

$p = \pm 1, \pm 27, \pm 3, \pm 9 q = \pm 1, \pm 7$

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

$\pm 1, \pm 0.\overline{142857}, \pm 27, \pm 3.\overline{857142}, \pm 3, \pm 0.\overline{428571}, \pm 9, \pm 1.\overline{285714}$

Substitute $- 0.\overline{428571}$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.\overline{428571}$ is a root of the polynomial.

Substitute $- 0.\overline{428571}$ into the polynomial.

$7 {(- 0.\overline{428571})}^{3} - 39 {(- 0.\overline{428571})}^{2} + 45 \cdot - 0.\overline{428571} + 27$

Raise $- 0.\overline{428571}$ to the power of $3$ .

$7 \cdot - 0.0787172 - 39 {(- 0.\overline{428571})}^{2} + 45 \cdot - 0.\overline{428571} + 27$

Multiply $7$ by $- 0.0787172$ .

$- 0.5510204 - 39 {(- 0.\overline{428571})}^{2} + 45 \cdot - 0.\overline{428571} + 27$

Raise $- 0.\overline{428571}$ to the power of $2$ .

$- 0.5510204 - 39 \cdot 0.18367346 + 45 \cdot - 0.\overline{428571} + 27$

Multiply $- 39$ by $0.18367346$ .

$- 0.5510204 - 7.1632653 + 45 \cdot - 0.\overline{428571} + 27$

Subtract $7.1632653$ from $- 0.5510204$ .

$- 7.71428571 + 45 \cdot - 0.\overline{428571} + 27$

Multiply $45$ by $- 0.\overline{428571}$ .

$- 7.71428571 - 19.\overline{285714} + 27$

Subtract $19.\overline{285714}$ from $- 7.71428571$ .

$- 27 + 27$

Add $- 27$ and $27$ .

$0$

Since $- 0.\overline{428571}$ is a known root, divide the polynomial by $7 x + 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{7 x^{3} - 39 x^{2} + 45 x + 27}{7 x + 3}$

Divide $7 x^{3} - 39 x^{2} + 45 x + 27$ by $7 x + 3$ .

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

$7 x$

$3$

$7 x^{3}$

$39 x^{2}$

$45 x$

$27$

Divide the highest order term in the dividend $7 x^{3}$ by the highest order term in divisor $7 x$ .

					$x^{2}$
	$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$

Multiply the new quotient term by the divisor.

				$x^{2}$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			+	$7 x^{3}$	+	$3 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{3} + 3 x^{2}$

				$x^{2}$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$

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

				$x^{2}$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$

Divide the highest order term in the dividend $- 42 x^{2}$ by the highest order term in divisor $7 x$ .

				$x^{2}$	-	$6 x$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$
					-	$42 x^{2}$	-	$18 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 42 x^{2} - 18 x$

				$x^{2}$	-	$6 x$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$
					+	$42 x^{2}$	+	$18 x$

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

				$x^{2}$	-	$6 x$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$
					+	$42 x^{2}$	+	$18 x$
							+	$63 x$

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

				$x^{2}$	-	$6 x$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$
					+	$42 x^{2}$	+	$18 x$
							+	$63 x$	+	$27$

Divide the highest order term in the dividend $63 x$ by the highest order term in divisor $7 x$ .

				$x^{2}$	-	$6 x$	+	$9$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$
					+	$42 x^{2}$	+	$18 x$
							+	$63 x$	+	$27$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$9$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$
					+	$42 x^{2}$	+	$18 x$
							+	$63 x$	+	$27$
							+	$63 x$	+	$27$

The expression needs to be subtracted from the dividend, so change all the signs in $63 x + 27$

				$x^{2}$	-	$6 x$	+	$9$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$
					+	$42 x^{2}$	+	$18 x$
							+	$63 x$	+	$27$
							-	$63 x$	-	$27$

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

				$x^{2}$	-	$6 x$	+	$9$
$7 x$	+	$3$		$7 x^{3}$	-	$39 x^{2}$	+	$45 x$	+	$27$
			-	$7 x^{3}$	-	$3 x^{2}$
					-	$42 x^{2}$	+	$45 x$
					+	$42 x^{2}$	+	$18 x$
							+	$63 x$	+	$27$
							-	$63 x$	-	$27$
										$0$

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

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

Write $7 x^{3} - 39 x^{2} + 45 x + 27$ as a set of factors.

$(x + 1) ((7 x + 3) (x^{2} - 6 x + 9)) = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$(x + 1) ((7 x + 3) (x^{2} - 6 x + 3^{2})) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$(x + 1) ((7 x + 3) (x^{2} - 2 \cdot x \cdot 3 + 3^{2})) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$(x + 1) ((7 x + 3) {(x - 3)}^{2}) = 0$

Remove unnecessary parentheses.

$(x + 1) (7 x + 3) {(x - 3)}^{2} = 0$

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

$x + 1 = 0$

$7 x + 3 = 0$

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

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $7 x + 3$ equal to $0$ and solve for $x$ .

Set $7 x + 3$ equal to $0$ .

$7 x + 3 = 0$

Solve $7 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$7 x = - 3$

Divide each term in $7 x = - 3$ by $7$ and simplify.

Divide each term in $7 x = - 3$ by $7$ .

$\frac{7 x}{7} = \frac{- 3}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 3}{7}$

Divide $x$ by $1$ .

$x = \frac{- 3}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{7}$

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 $(x + 1) (7 x + 3) {(x - 3)}^{2} = 0$ true.

$x = - 1, - \frac{3}{7}, 3$

x-intercept(s) in point form.

x-intercept(s): $(- 1, 0), (- \frac{3}{7}, 0), (3, 0)$

Step 3

Find the y-intercepts.

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

$y = 7 {(0)}^{4} - 32 {(0)}^{3} + 6 {(0)}^{2} + 72 (0) + 27$

Solve the equation.

Remove parentheses.

$y = 7 \cdot 0^{4} - 32 {(0)}^{3} + 6 {(0)}^{2} + 72 (0) + 27$

Remove parentheses.

$y = 7 \cdot 0^{4} - 32 \cdot 0^{3} + 6 {(0)}^{2} + 72 (0) + 27$

Remove parentheses.

$y = 7 \cdot 0^{4} - 32 \cdot 0^{3} + 6 \cdot 0^{2} + 72 (0) + 27$

Remove parentheses.

$y = 7 {(0)}^{4} - 32 {(0)}^{3} + 6 {(0)}^{2} + 72 (0) + 27$

Simplify $7 {(0)}^{4} - 32 {(0)}^{3} + 6 {(0)}^{2} + 72 (0) + 27$ .

Simplify each term.

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

$y = 7 \cdot 0 - 32 {(0)}^{3} + 6 {(0)}^{2} + 72 (0) + 27$

Multiply $7$ by $0$ .

$y = 0 - 32 {(0)}^{3} + 6 {(0)}^{2} + 72 (0) + 27$

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

$y = 0 - 32 \cdot 0 + 6 {(0)}^{2} + 72 (0) + 27$

Multiply $- 32$ by $0$ .

$y = 0 + 0 + 6 {(0)}^{2} + 72 (0) + 27$

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

$y = 0 + 0 + 6 \cdot 0 + 72 (0) + 27$

Multiply $6$ by $0$ .

$y = 0 + 0 + 0 + 72 (0) + 27$

Multiply $72$ by $0$ .

$y = 0 + 0 + 0 + 0 + 27$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 0 + 0 + 27$

Add $0$ and $0$ .

$y = 0 + 0 + 27$

Add $0$ and $0$ .

$y = 0 + 27$

Add $0$ and $27$ .

$y = 27$

y-intercept(s) in point form.

y-intercept(s): $(0, 27)$

Step 4

List the intersections.

x-intercept(s): $(- 1, 0), (- \frac{3}{7}, 0), (3, 0)$

y-intercept(s): $(0, 27)$

Step 5

Find the x and y Intercepts 7x^4-32x^3+6x^2+72x+27

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