Find the x and y Intercepts f(x)=7x^2-5x+9x^3-5+2x^4

$f (x) = 7 x^{2} - 5 x + 9 x^{3} - 5 + 2 x^{4}$

Find the x-intercepts.

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

$0 = 7 x^{2} - 5 x + 9 x^{3} - 5 + 2 x^{4}$

Solve the equation.

Rewrite the equation as $7 x^{2} - 5 x + 9 x^{3} - 5 + 2 x^{4} = 0$ .

$7 x^{2} - 5 x + 9 x^{3} - 5 + 2 x^{4} = 0$

Factor the left side of the equation.

Regroup terms.

$- 5 x - 5 + 7 x^{2} + 9 x^{3} + 2 x^{4} = 0$

Factor $- 5$ out of $- 5 x - 5$ .

Factor $- 5$ out of $- 5 x$ .

$- 5 x - 5 + 7 x^{2} + 9 x^{3} + 2 x^{4} = 0$

Factor $- 5$ out of $- 5$ .

$- 5 x - 5 \cdot 1 + 7 x^{2} + 9 x^{3} + 2 x^{4} = 0$

Factor $- 5$ out of $- 5 (x) - 5 (1)$ .

$- 5 (x + 1) + 7 x^{2} + 9 x^{3} + 2 x^{4} = 0$

Factor $x^{2}$ out of $7 x^{2} + 9 x^{3} + 2 x^{4}$ .

Factor $x^{2}$ out of $7 x^{2}$ .

$- 5 (x + 1) + x^{2} \cdot 7 + 9 x^{3} + 2 x^{4} = 0$

Factor $x^{2}$ out of $9 x^{3}$ .

$- 5 (x + 1) + x^{2} \cdot 7 + x^{2} (9 x) + 2 x^{4} = 0$

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

$- 5 (x + 1) + x^{2} \cdot 7 + x^{2} (9 x) + x^{2} (2 x^{2}) = 0$

Factor $x^{2}$ out of $x^{2} \cdot 7 + x^{2} (9 x)$ .

$- 5 (x + 1) + x^{2} \cdot (7 + 9 x) + x^{2} (2 x^{2}) = 0$

Factor $x^{2}$ out of $x^{2} \cdot (7 + 9 x) + x^{2} (2 x^{2})$ .

$- 5 (x + 1) + x^{2} (7 + 9 x + 2 x^{2}) = 0$

Factor.

Factor by grouping.

Reorder terms.

$- 5 (x + 1) + x^{2} (2 x^{2} + 9 x + 7) = 0$

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 $a \cdot c = 2 \cdot 7 = 14$ and whose sum is $b = 9$ .

Factor $9$ out of $9 x$ .

$- 5 (x + 1) + x^{2} (2 x^{2} + 9 (x) + 7) = 0$

Rewrite $9$ as $2$ plus $7$

$- 5 (x + 1) + x^{2} (2 x^{2} + (2 + 7) x + 7) = 0$

Apply the distributive property.

$- 5 (x + 1) + x^{2} (2 x^{2} + 2 x + 7 x + 7) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- 5 (x + 1) + x^{2} ((2 x^{2} + 2 x) + 7 x + 7) = 0$

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

$- 5 (x + 1) + x^{2} (2 x (x + 1) + 7 (x + 1)) = 0$

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

$- 5 (x + 1) + x^{2} ((x + 1) (2 x + 7)) = 0$

Remove unnecessary parentheses.

$- 5 (x + 1) + x^{2} (x + 1) (2 x + 7) = 0$

Factor $x + 1$ out of $- 5 (x + 1) + x^{2} (x + 1) (2 x + 7)$ .

Factor $x + 1$ out of $- 5 (x + 1)$ .

$(x + 1) \cdot - 5 + x^{2} (x + 1) (2 x + 7) = 0$

Factor $x + 1$ out of $x^{2} (x + 1) (2 x + 7)$ .

$(x + 1) \cdot - 5 + (x + 1) (x^{2} (2 x + 7)) = 0$

Factor $x + 1$ out of $(x + 1) \cdot - 5 + (x + 1) (x^{2} (2 x + 7))$ .

$(x + 1) (- 5 + x^{2} (2 x + 7)) = 0$

Apply the distributive property.

$(x + 1) (- 5 + x^{2} (2 x) + x^{2} \cdot 7) = 0$

Rewrite using the commutative property of multiplication.

$(x + 1) (- 5 + 2 x^{2} x + x^{2} \cdot 7) = 0$

Move $7$ to the left of $x^{2}$ .

$(x + 1) (- 5 + 2 x^{2} x + 7 \cdot x^{2}) = 0$

Multiply $x^{2}$ by $x$ by adding the exponents.

Move $x$ .

$(x + 1) (- 5 + 2 (x \cdot x^{2}) + 7 \cdot x^{2}) = 0$

Multiply $x$ by $x^{2}$ .

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

$(x + 1) (- 5 + 2 (x \cdot x^{2}) + 7 \cdot x^{2}) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$(x + 1) (- 5 + 2 x^{1 + 2} + 7 \cdot x^{2}) = 0$

Add $1$ and $2$ .

$(x + 1) (- 5 + 2 x^{3} + 7 \cdot x^{2}) = 0$

$(x + 1) (- 5 + 2 x^{3} + 7 x^{2}) = 0$

Reorder terms.

$(x + 1) (2 x^{3} + 7 x^{2} - 5) = 0$

Factor.

Factor $2 x^{3} + 7 x^{2} - 5$ 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.

$p = \pm 1, \pm 5 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 5, \pm 2.5$

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.

$2 {(- 1)}^{3} + 7 {(- 1)}^{2} - 5$

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

$2 \cdot - 1 + 7 {(- 1)}^{2} - 5$

Multiply $2$ by $- 1$ .

$- 2 + 7 {(- 1)}^{2} - 5$

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

$- 2 + 7 \cdot 1 - 5$

Multiply $7$ by $1$ .

$- 2 + 7 - 5$

Add $- 2$ and $7$ .

$5 - 5$

Subtract $5$ from $5$ .

$0$

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.

$\frac{2 x^{3} + 7 x^{2} - 5}{x + 1}$

Divide $2 x^{3} + 7 x^{2} - 5$ 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 $0$ .

$x$

$1$

$2 x^{3}$

$7 x^{2}$

$0 x$

$5$

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $x$ .

					$2 x^{2}$
	$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			+	$2 x^{3}$	+	$2 x^{2}$

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

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$

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

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$

Divide the highest order term in the dividend $5 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	+	$5 x$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$5 x$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$
					+	$5 x^{2}$	+	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{2} + 5 x$

				$2 x^{2}$	+	$5 x$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$
					-	$5 x^{2}$	-	$5 x$

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

				$2 x^{2}$	+	$5 x$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$
					-	$5 x^{2}$	-	$5 x$
							-	$5 x$

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

				$2 x^{2}$	+	$5 x$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$
					-	$5 x^{2}$	-	$5 x$
							-	$5 x$	-	$5$

Divide the highest order term in the dividend $- 5 x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	+	$5 x$	-	$5$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$
					-	$5 x^{2}$	-	$5 x$
							-	$5 x$	-	$5$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$5 x$	-	$5$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$
					-	$5 x^{2}$	-	$5 x$
							-	$5 x$	-	$5$
							-	$5 x$	-	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x - 5$

				$2 x^{2}$	+	$5 x$	-	$5$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$
					-	$5 x^{2}$	-	$5 x$
							-	$5 x$	-	$5$
							+	$5 x$	+	$5$

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

				$2 x^{2}$	+	$5 x$	-	$5$
$x$	+	$1$		$2 x^{3}$	+	$7 x^{2}$	+	$0 x$	-	$5$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$5 x^{2}$	+	$0 x$
					-	$5 x^{2}$	-	$5 x$
							-	$5 x$	-	$5$
							+	$5 x$	+	$5$
										$0$

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

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

Write $2 x^{3} + 7 x^{2} - 5$ as a set of factors.

$(x + 1) ((x + 1) (2 x^{2} + 5 x - 5)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x + 1) (2 x^{2} + 5 x - 5) = 0$

Combine exponents.

Raise $x + 1$ to the power of $1$ .

$(x + 1) (x + 1) (2 x^{2} + 5 x - 5) = 0$

Raise $x + 1$ to the power of $1$ .

$(x + 1) (x + 1) (2 x^{2} + 5 x - 5) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(x + 1)}^{1 + 1} (2 x^{2} + 5 x - 5) = 0$

Add $1$ and $1$ .

${(x + 1)}^{2} (2 x^{2} + 5 x - 5) = 0$

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

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

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

Set ${(x + 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 1)}^{2}$ equal to $0$ .

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

Solve ${(x + 1)}^{2} = 0$ for $x$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $2 x^{2} + 5 x - 5$ equal to $0$ and solve for $x$ .

Set $2 x^{2} + 5 x - 5$ equal to $0$ .

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

Solve $2 x^{2} + 5 x - 5 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = 5$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (2 \cdot - 5)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 8 \cdot - 5}}{2 \cdot 2}$

Multiply $- 8$ by $- 5$ .

$x = \frac{- 5 \pm \sqrt{25 + 40}}{2 \cdot 2}$

Add $25$ and $40$ .

$x = \frac{- 5 \pm \sqrt{65}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 5 \pm \sqrt{65}}{4}$

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

Simplify the numerator.

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 8 \cdot - 5}}{2 \cdot 2}$

Multiply $- 8$ by $- 5$ .

$x = \frac{- 5 \pm \sqrt{25 + 40}}{2 \cdot 2}$

Add $25$ and $40$ .

$x = \frac{- 5 \pm \sqrt{65}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 5 \pm \sqrt{65}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{- 5 + \sqrt{65}}{4}$

Rewrite $- 5$ as $- 1 (5)$ .

$x = \frac{- 1 \cdot 5 + \sqrt{65}}{4}$

Factor $- 1$ out of $\sqrt{65}$ .

$x = \frac{- 1 \cdot 5 - 1 (- \sqrt{65})}{4}$

Factor $- 1$ out of $- 1 (5) - 1 (- \sqrt{65})$ .

$x = \frac{- 1 (5 - \sqrt{65})}{4}$

Move the negative in front of the fraction.

$x = - \frac{5 - \sqrt{65}}{4}$

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

Simplify the numerator.

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 8 \cdot - 5}}{2 \cdot 2}$

Multiply $- 8$ by $- 5$ .

$x = \frac{- 5 \pm \sqrt{25 + 40}}{2 \cdot 2}$

Add $25$ and $40$ .

$x = \frac{- 5 \pm \sqrt{65}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 5 \pm \sqrt{65}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{- 5 - \sqrt{65}}{4}$

Rewrite $- 5$ as $- 1 (5)$ .

$x = \frac{- 1 \cdot 5 - \sqrt{65}}{4}$

Factor $- 1$ out of $- \sqrt{65}$ .

$x = \frac{- 1 \cdot 5 - (\sqrt{65})}{4}$

Factor $- 1$ out of $- 1 (5) - (\sqrt{65})$ .

$x = \frac{- 1 (5 + \sqrt{65})}{4}$

Move the negative in front of the fraction.

$x = - \frac{5 + \sqrt{65}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{5 - \sqrt{65}}{4}, - \frac{5 + \sqrt{65}}{4}$

The final solution is all the values that make ${(x + 1)}^{2} (2 x^{2} + 5 x - 5) = 0$ true.

$x = - 1, - \frac{5 - \sqrt{65}}{4}, - \frac{5 + \sqrt{65}}{4}$

x-intercept(s) in point form.

x-intercept(s): $(- 1, 0), (- \frac{5 - \sqrt{65}}{4}, 0), (- \frac{5 + \sqrt{65}}{4}, 0)$

Find the y-intercepts.

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

$y = 7 {(0)}^{2} - 5 \cdot 0 + 9 {(0)}^{3} - 5 + 2 {(0)}^{4}$

Solve the equation.

Remove parentheses.

$y = 7 \cdot 0^{2} - 5 \cdot 0 + 9 {(0)}^{3} - 5 + 2 {(0)}^{4}$

Remove parentheses.

$y = 7 \cdot 0^{2} - 5 \cdot 0 + 9 \cdot 0^{3} - 5 + 2 {(0)}^{4}$

Remove parentheses.

$y = 7 \cdot 0^{2} - 5 \cdot 0 + 9 \cdot 0^{3} - 5 + 2 \cdot 0^{4}$

Remove parentheses.

$y = 7 {(0)}^{2} - 5 \cdot 0 + 9 {(0)}^{3} - 5 + 2 {(0)}^{4}$

Simplify $7 {(0)}^{2} - 5 \cdot 0 + 9 {(0)}^{3} - 5 + 2 {(0)}^{4}$ .

Simplify each term.

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

$y = 7 \cdot 0 - 5 \cdot 0 + 9 {(0)}^{3} - 5 + 2 {(0)}^{4}$

Multiply $7$ by $0$ .

$y = 0 - 5 \cdot 0 + 9 {(0)}^{3} - 5 + 2 {(0)}^{4}$

Multiply $- 5$ by $0$ .

$y = 0 + 0 + 9 {(0)}^{3} - 5 + 2 {(0)}^{4}$

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

$y = 0 + 0 + 9 \cdot 0 - 5 + 2 {(0)}^{4}$

Multiply $9$ by $0$ .

$y = 0 + 0 + 0 - 5 + 2 {(0)}^{4}$

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

$y = 0 + 0 + 0 - 5 + 2 \cdot 0$

Multiply $2$ by $0$ .

$y = 0 + 0 + 0 - 5 + 0$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 - 5 + 0$

Add $0$ and $0$ .

$y = 0 - 5 + 0$

Subtract $5$ from $0$ .

$y = - 5 + 0$

Add $- 5$ and $0$ .

$y = - 5$

y-intercept(s) in point form.

y-intercept(s): $(0, - 5)$

List the intersections.

x-intercept(s): $(- 1, 0), (- \frac{5 - \sqrt{65}}{4}, 0), (- \frac{5 + \sqrt{65}}{4}, 0)$

y-intercept(s): $(0, - 5)$

Find the x and y Intercepts f(x)=7x^2-5x+9x^3-5+2x^4

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