Find the Roots/Zeros Using the Rational Roots Test f(x)=7x^4-58x^3+60x^2+250x+125

$f (x) = 7 x^{4} - 58 x^{3} + 60 x^{2} + 250 x + 125$

Step 1

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.

$p = \pm 1, \pm 5, \pm 25, \pm 125$

$q = \pm 1, \pm 7$

Step 2

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

$\pm 1, \pm \frac{1}{7}, \pm 5, \pm \frac{5}{7}, \pm 25, \pm \frac{25}{7}, \pm 125, \pm \frac{125}{7}$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

$7 {(- 1)}^{4} - 58 {(- 1)}^{3} + 60 {(- 1)}^{2} + 250 (- 1) + 125$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

Simplify each term.

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

$7 \cdot 1 - 58 {(- 1)}^{3} + 60 {(- 1)}^{2} + 250 (- 1) + 125$

Multiply $7$ by $1$ .

$7 - 58 {(- 1)}^{3} + 60 {(- 1)}^{2} + 250 (- 1) + 125$

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

$7 - 58 \cdot - 1 + 60 {(- 1)}^{2} + 250 (- 1) + 125$

Multiply $- 58$ by $- 1$ .

$7 + 58 + 60 {(- 1)}^{2} + 250 (- 1) + 125$

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

$7 + 58 + 60 \cdot 1 + 250 (- 1) + 125$

Multiply $60$ by $1$ .

$7 + 58 + 60 + 250 (- 1) + 125$

Multiply $250$ by $- 1$ .

$7 + 58 + 60 - 250 + 125$

Simplify by adding and subtracting.

Add $7$ and $58$ .

$65 + 60 - 250 + 125$

Add $65$ and $60$ .

$125 - 250 + 125$

Subtract $250$ from $125$ .

$- 125 + 125$

Add $- 125$ and $125$ .

$0$

Step 5

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.

$\frac{7 x^{4} - 58 x^{3} + 60 x^{2} + 250 x + 125}{x + 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Place the numbers representing the divisor and the dividend into a division-like configuration.

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0				\xa0

The first number in the dividend $(7)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0				\xa0
	$7$

Multiply the newest entry in the result $(7)$ by the divisor $(- 1)$ and place the result of $(- 7)$ under the next term in the dividend $(- 58)$ .

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0	$- 7$			\xa0
	$7$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0	$- 7$			\xa0
	$7$	$- 65$

Multiply the newest entry in the result $(- 65)$ by the divisor $(- 1)$ and place the result of $(65)$ under the next term in the dividend $(60)$ .

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0	$- 7$	$65$		\xa0
	$7$	$- 65$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0	$- 7$	$65$		\xa0
	$7$	$- 65$	$125$

Multiply the newest entry in the result $(125)$ by the divisor $(- 1)$ and place the result of $(- 125)$ under the next term in the dividend $(250)$ .

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0	$- 7$	$65$	$- 125$	\xa0
	$7$	$- 65$	$125$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0	$- 7$	$65$	$- 125$	\xa0
	$7$	$- 65$	$125$	$125$

Multiply the newest entry in the result $(125)$ by the divisor $(- 1)$ and place the result of $(- 125)$ under the next term in the dividend $(125)$ .

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0	$- 7$	$65$	$- 125$	$- 125$
	$7$	$- 65$	$125$	$125$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$7$	$- 58$	$60$	$250$	$125$
	\xa0	$- 7$	$65$	$- 125$	$- 125$
	$7$	$- 65$	$125$	$125$	$0$

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$7 x^{3} + - 65 x^{2} + (125) x + 125$

Simplify the quotient polynomial.

$7 x^{3} - 65 x^{2} + 125 x + 125$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $7 x^{3} - 65 x^{2} + 125 x + 125$ 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.

$p = \pm 1, \pm 125, \pm 5, \pm 25 q = \pm 1, \pm 7$

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

$\pm 1, \pm 0.\overline{142857}, \pm 125, \pm 17.\overline{857142}, \pm 5, \pm 0.\overline{714285}, \pm 25, \pm 3.\overline{571428}$

Substitute $- 0.\overline{714285}$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.\overline{714285}$ is a root of the polynomial.

Substitute $- 0.\overline{714285}$ into the polynomial.

$7 {(- 0.\overline{714285})}^{3} - 65 {(- 0.\overline{714285})}^{2} + 125 \cdot - 0.\overline{714285} + 125$

Raise $- 0.\overline{714285}$ to the power of $3$ .

$7 \cdot - 0.36443148 - 65 {(- 0.\overline{714285})}^{2} + 125 \cdot - 0.\overline{714285} + 125$

Multiply $7$ by $- 0.36443148$ .

$- 2.5510204 - 65 {(- 0.\overline{714285})}^{2} + 125 \cdot - 0.\overline{714285} + 125$

Raise $- 0.\overline{714285}$ to the power of $2$ .

$- 2.5510204 - 65 \cdot 0.51020408 + 125 \cdot - 0.\overline{714285} + 125$

Multiply $- 65$ by $0.51020408$ .

$- 2.5510204 - 33.1632653 + 125 \cdot - 0.\overline{714285} + 125$

Subtract $33.1632653$ from $- 2.5510204$ .

$- 35.71428571 + 125 \cdot - 0.\overline{714285} + 125$

Multiply $125$ by $- 0.\overline{714285}$ .

$- 35.71428571 - 89.\overline{285714} + 125$

Subtract $89.\overline{285714}$ from $- 35.71428571$ .

$- 125 + 125$

Add $- 125$ and $125$ .

$0$

Since $- 0.\overline{714285}$ is a known root, divide the polynomial by $7 x + 5$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{7 x^{3} - 65 x^{2} + 125 x + 125}{7 x + 5}$

Divide $7 x^{3} - 65 x^{2} + 125 x + 125$ by $7 x + 5$ .

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

$7 x$

$5$

$7 x^{3}$

$65 x^{2}$

$125 x$

$125$

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

					$x^{2}$
	$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$

Multiply the new quotient term by the divisor.

				$x^{2}$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			+	$7 x^{3}$	+	$5 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{3} + 5 x^{2}$

				$x^{2}$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$

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

				$x^{2}$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$

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

				$x^{2}$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$

Divide the highest order term in the dividend $- 70 x^{2}$ by the highest order term in divisor $7 x$ .

				$x^{2}$	-	$10 x$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$10 x$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$
					-	$70 x^{2}$	-	$50 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 70 x^{2} - 50 x$

				$x^{2}$	-	$10 x$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$
					+	$70 x^{2}$	+	$50 x$

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

				$x^{2}$	-	$10 x$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$
					+	$70 x^{2}$	+	$50 x$
							+	$175 x$

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

				$x^{2}$	-	$10 x$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$
					+	$70 x^{2}$	+	$50 x$
							+	$175 x$	+	$125$

Divide the highest order term in the dividend $175 x$ by the highest order term in divisor $7 x$ .

				$x^{2}$	-	$10 x$	+	$25$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$
					+	$70 x^{2}$	+	$50 x$
							+	$175 x$	+	$125$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$10 x$	+	$25$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$
					+	$70 x^{2}$	+	$50 x$
							+	$175 x$	+	$125$
							+	$175 x$	+	$125$

The expression needs to be subtracted from the dividend, so change all the signs in $175 x + 125$

				$x^{2}$	-	$10 x$	+	$25$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$
					+	$70 x^{2}$	+	$50 x$
							+	$175 x$	+	$125$
							-	$175 x$	-	$125$

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

				$x^{2}$	-	$10 x$	+	$25$
$7 x$	+	$5$		$7 x^{3}$	-	$65 x^{2}$	+	$125 x$	+	$125$
			-	$7 x^{3}$	-	$5 x^{2}$
					-	$70 x^{2}$	+	$125 x$
					+	$70 x^{2}$	+	$50 x$
							+	$175 x$	+	$125$
							-	$175 x$	-	$125$
										$0$

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

$x^{2} - 10 x + 25$

Write $7 x^{3} - 65 x^{2} + 125 x + 125$ as a set of factors.

$(7 x + 5) (x^{2} - 10 x + 25) = 0$

Factor using the perfect square rule.

Rewrite $25$ as $5^{2}$ .

$(7 x + 5) (x^{2} - 10 x + 5^{2}) = 0$

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

$10 x = 2 \cdot x \cdot 5$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 5$ .

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

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

$7 x + 5 = 0$

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

Set $7 x + 5$ equal to $0$ and solve for $x$ .

Set $7 x + 5$ equal to $0$ .

$7 x + 5 = 0$

Solve $7 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$7 x = - 5$

Divide each term in $7 x = - 5$ by $7$ and simplify.

Divide each term in $7 x = - 5$ by $7$ .

$\frac{7 x}{7} = \frac{- 5}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 5}{7}$

Divide $x$ by $1$ .

$x = \frac{- 5}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{7}$

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

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

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

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

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $(7 x + 5) {(x - 5)}^{2} = 0$ true.

$x = - \frac{5}{7}, 5$

Step 8

The polynomial can be written as a set of linear factors.

$(x + 1) (x + \frac{5}{7}) (x - 5)$

Step 9

These are the roots (zeros) of the polynomial $7 x^{4} - 58 x^{3} + 60 x^{2} + 250 x + 125$ .

$x = - 1, - \frac{5}{7}, 5$

Step 10

Find the Roots/Zeros Using the Rational Roots Test f(x)=7x^4-58x^3+60x^2+250x+125

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