Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3+5x^2+50x+125

$f (x) = 2 x^{3} + 5 x^{2} + 50 x + 125$

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.

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

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 5, \pm \frac{5}{2}, \pm 25, \pm \frac{25}{2}, \pm 125, \pm \frac{125}{2}$

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.

$2 {(- \frac{5}{2})}^{3} + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - \frac{5}{2}$ is a root of the polynomial.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{5}{2}$ .

$2 ({(- 1)}^{3} {(\frac{5}{2})}^{3}) + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Apply the product rule to $\frac{5}{2}$ .

$2 ({(- 1)}^{3} \frac{5^{3}}{2^{3}}) + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

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

$2 (- \frac{5^{3}}{2^{3}}) + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Raise $5$ to the power of $3$ .

$2 (- \frac{125}{2^{3}}) + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

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

$2 (- \frac{125}{8}) + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{125}{8}$ into the numerator.

$2 (\frac{- 125}{8}) + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Factor $2$ out of $8$ .

$2 \frac{- 125}{2 (4)} + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Cancel the common factor.

$2 \frac{- 125}{2 \cdot 4} + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Rewrite the expression.

$\frac{- 125}{4} + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Move the negative in front of the fraction.

$- \frac{125}{4} + 5 {(- \frac{5}{2})}^{2} + 50 (- \frac{5}{2}) + 125$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{5}{2}$ .

$- \frac{125}{4} + 5 ({(- 1)}^{2} {(\frac{5}{2})}^{2}) + 50 (- \frac{5}{2}) + 125$

Apply the product rule to $\frac{5}{2}$ .

$- \frac{125}{4} + 5 ({(- 1)}^{2} \frac{5^{2}}{2^{2}}) + 50 (- \frac{5}{2}) + 125$

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

$- \frac{125}{4} + 5 (1 \frac{5^{2}}{2^{2}}) + 50 (- \frac{5}{2}) + 125$

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

$- \frac{125}{4} + 5 \frac{5^{2}}{2^{2}} + 50 (- \frac{5}{2}) + 125$

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

$- \frac{125}{4} + 5 \frac{25}{2^{2}} + 50 (- \frac{5}{2}) + 125$

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

$- \frac{125}{4} + 5 (\frac{25}{4}) + 50 (- \frac{5}{2}) + 125$

Multiply $5 (\frac{25}{4})$ .

Combine $5$ and $\frac{25}{4}$ .

$- \frac{125}{4} + \frac{5 \cdot 25}{4} + 50 (- \frac{5}{2}) + 125$

Multiply $5$ by $25$ .

$- \frac{125}{4} + \frac{125}{4} + 50 (- \frac{5}{2}) + 125$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5}{2}$ into the numerator.

$- \frac{125}{4} + \frac{125}{4} + 50 (\frac{- 5}{2}) + 125$

Factor $2$ out of $50$ .

$- \frac{125}{4} + \frac{125}{4} + 2 (25) \frac{- 5}{2} + 125$

Cancel the common factor.

$- \frac{125}{4} + \frac{125}{4} + 2 \cdot 25 \frac{- 5}{2} + 125$

Rewrite the expression.

$- \frac{125}{4} + \frac{125}{4} + 25 \cdot - 5 + 125$

Multiply $25$ by $- 5$ .

$- \frac{125}{4} + \frac{125}{4} - 125 + 125$

Combine fractions.

Combine the numerators over the common denominator.

$- 125 + 125 + \frac{- 125 + 125}{4}$

Add $- 125$ and $125$ .

$- 125 + 125 + \frac{0}{4}$

Find the common denominator.

Write $- 125$ as a fraction with denominator $1$ .

$\frac{- 125}{1} + 125 + \frac{0}{4}$

Multiply $\frac{- 125}{1}$ by $\frac{4}{4}$ .

$\frac{- 125}{1} \cdot \frac{4}{4} + 125 + \frac{0}{4}$

Multiply $\frac{- 125}{1}$ by $\frac{4}{4}$ .

$\frac{- 125 \cdot 4}{4} + 125 + \frac{0}{4}$

Write $125$ as a fraction with denominator $1$ .

$\frac{- 125 \cdot 4}{4} + \frac{125}{1} + \frac{0}{4}$

Multiply $\frac{125}{1}$ by $\frac{4}{4}$ .

$\frac{- 125 \cdot 4}{4} + \frac{125}{1} \cdot \frac{4}{4} + \frac{0}{4}$

Multiply $\frac{125}{1}$ by $\frac{4}{4}$ .

$\frac{- 125 \cdot 4}{4} + \frac{125 \cdot 4}{4} + \frac{0}{4}$

Combine the numerators over the common denominator.

$\frac{- 125 \cdot 4 + 125 \cdot 4}{4}$

Simplify each term.

Multiply $- 125$ by $4$ .

$\frac{- 500 + 125 \cdot 4}{4}$

Multiply $125$ by $4$ .

$\frac{- 500 + 500}{4}$

Simplify the expression.

Add $- 500$ and $500$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

Step 5

Since $- \frac{5}{2}$ is a known root, divide the polynomial by $x + \frac{5}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} + 5 x^{2} + 50 x + 125}{x + \frac{5}{2}}$

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.

$- \frac{5}{2}$	$2$	$5$	$50$	$125$
	\xa0			\xa0

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$- \frac{5}{2}$	$2$	$5$	$50$	$125$
	\xa0			\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(- \frac{5}{2})$ and place the result of $(- 5)$ under the next term in the dividend $(5)$ .

$- \frac{5}{2}$	$2$	$5$	$50$	$125$
	\xa0	$- 5$		\xa0
	$2$

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

$- \frac{5}{2}$	$2$	$5$	$50$	$125$
	\xa0	$- 5$		\xa0
	$2$	$0$

Multiply the newest entry in the result $(0)$ by the divisor $(- \frac{5}{2})$ and place the result of $(0)$ under the next term in the dividend $(50)$ .

$- \frac{5}{2}$	$2$	$5$	$50$	$125$
	\xa0	$- 5$	$0$	\xa0
	$2$	$0$

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

$- \frac{5}{2}$	$2$	$5$	$50$	$125$
	\xa0	$- 5$	$0$	\xa0
	$2$	$0$	$50$

Multiply the newest entry in the result $(50)$ by the divisor $(- \frac{5}{2})$ and place the result of $(- 125)$ under the next term in the dividend $(125)$ .

$- \frac{5}{2}$	$2$	$5$	$50$	$125$
	\xa0	$- 5$	$0$	$- 125$
	$2$	$0$	$50$

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

$- \frac{5}{2}$	$2$	$5$	$50$	$125$
	\xa0	$- 5$	$0$	$- 125$
	$2$	$0$	$50$	$0$

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

$2 x^{2} + 0 x + 50$

Simplify the quotient polynomial.

$2 x^{2} + 50$

Step 7

Factor $2$ out of $2 x^{2} + 50$ .

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

$(x + \frac{5}{2}) (2 (x^{2}) + 50)$

Factor $2$ out of $50$ .

$(x + \frac{5}{2}) (2 x^{2} + 2 \cdot 25)$

Factor $2$ out of $2 x^{2} + 2 \cdot 25$ .

$(x + \frac{5}{2}) (2 (x^{2} + 25))$

Step 8

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{3} + 5 x^{2}) + 50 x + 125 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 5$ .

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

Step 9

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

$2 x + 5 = 0$

$x^{2} + 25 = 0$

Step 10

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 11

Set $x^{2} + 25$ equal to $0$ and solve for $x$ .

Set $x^{2} + 25$ equal to $0$ .

$x^{2} + 25 = 0$

Solve $x^{2} + 25 = 0$ for $x$ .

Subtract $25$ from both sides of the equation.

$x^{2} = - 25$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 25}$

Simplify $\pm \sqrt{- 25}$ .

Rewrite $- 25$ as $- 1 (25)$ .

$x = \pm \sqrt{- 1 (25)}$

Rewrite $\sqrt{- 1 (25)}$ as $\sqrt{- 1} \cdot \sqrt{25}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{25}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{25}$

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

$x = \pm i \cdot \sqrt{5^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 5$

Move $5$ to the left of $i$ .

$x = \pm 5 i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 5 i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5 i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5 i, - 5 i$

Step 12

The final solution is all the values that make $(2 x + 5) (x^{2} + 25) = 0$ true.

$x = - \frac{5}{2}, 5 i, - 5 i$

Step 13

Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3+5x^2+50x+125

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