Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+6x^2-138x-143

$f (x) = x^{3} + 6 x^{2} - 138 x - 143$

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.

$p = \pm 1, \pm 11, \pm 13, \pm 143$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 11, \pm 13, \pm 143$

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.

${(- 1)}^{3} + 6 {(- 1)}^{2} - 138 \cdot - 1 - 143$

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 $3$ .

$- 1 + 6 {(- 1)}^{2} - 138 \cdot - 1 - 143$

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

$- 1 + 6 \cdot 1 - 138 \cdot - 1 - 143$

Multiply $6$ by $1$ .

$- 1 + 6 - 138 \cdot - 1 - 143$

Multiply $- 138$ by $- 1$ .

$- 1 + 6 + 138 - 143$

Simplify by adding and subtracting.

Add $- 1$ and $6$ .

$5 + 138 - 143$

Add $5$ and $138$ .

$143 - 143$

Subtract $143$ from $143$ .

$0$

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.

$\frac{x^{3} + 6 x^{2} - 138 x - 143}{x + 1}$

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.

$- 1$	$1$	$6$	$- 138$	$- 143$
	\xa0			\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$1$	$6$	$- 138$	$- 143$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 1)$ and place the result of $(- 1)$ under the next term in the dividend $(6)$ .

$- 1$	$1$	$6$	$- 138$	$- 143$
	\xa0	$- 1$		\xa0
	$1$

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

$- 1$	$1$	$6$	$- 138$	$- 143$
	\xa0	$- 1$		\xa0
	$1$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(- 1)$ and place the result of $(- 5)$ under the next term in the dividend $(- 138)$ .

$- 1$	$1$	$6$	$- 138$	$- 143$
	\xa0	$- 1$	$- 5$	\xa0
	$1$	$5$

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

$- 1$	$1$	$6$	$- 138$	$- 143$
	\xa0	$- 1$	$- 5$	\xa0
	$1$	$5$	$- 143$

Multiply the newest entry in the result $(- 143)$ by the divisor $(- 1)$ and place the result of $(143)$ under the next term in the dividend $(- 143)$ .

$- 1$	$1$	$6$	$- 138$	$- 143$
	\xa0	$- 1$	$- 5$	$143$
	$1$	$5$	$- 143$

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

$- 1$	$1$	$6$	$- 138$	$- 143$
	\xa0	$- 1$	$- 5$	$143$
	$1$	$5$	$- 143$	$0$

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

$1 x^{2} + 5 x - 143$

Simplify the quotient polynomial.

$x^{2} + 5 x - 143$

Step 7

Solve the equation to find any remaining roots.

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 5$ , and $c = - 143$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 143)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 143$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4$ by $- 143$ .

$x = \frac{- 5 \pm \sqrt{25 + 572}}{2 \cdot 1}$

Add $25$ and $572$ .

$x = \frac{- 5 \pm \sqrt{597}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm \sqrt{597}}{2}$

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

Simplify the numerator.

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

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 143$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4$ by $- 143$ .

$x = \frac{- 5 \pm \sqrt{25 + 572}}{2 \cdot 1}$

Add $25$ and $572$ .

$x = \frac{- 5 \pm \sqrt{597}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm \sqrt{597}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{- 5 + \sqrt{597}}{2}$

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

$x = \frac{- 1 \cdot 5 + \sqrt{597}}{2}$

Factor $- 1$ out of $\sqrt{597}$ .

$x = \frac{- 1 \cdot 5 - 1 (- \sqrt{597})}{2}$

Factor $- 1$ out of $- 1 (5) - 1 (- \sqrt{597})$ .

$x = \frac{- 1 (5 - \sqrt{597})}{2}$

Move the negative in front of the fraction.

$x = - \frac{5 - \sqrt{597}}{2}$

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

Simplify the numerator.

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

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 143$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4$ by $- 143$ .

$x = \frac{- 5 \pm \sqrt{25 + 572}}{2 \cdot 1}$

Add $25$ and $572$ .

$x = \frac{- 5 \pm \sqrt{597}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm \sqrt{597}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{- 5 - \sqrt{597}}{2}$

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

$x = \frac{- 1 \cdot 5 - \sqrt{597}}{2}$

Factor $- 1$ out of $- \sqrt{597}$ .

$x = \frac{- 1 \cdot 5 - (\sqrt{597})}{2}$

Factor $- 1$ out of $- 1 (5) - (\sqrt{597})$ .

$x = \frac{- 1 (5 + \sqrt{597})}{2}$

Move the negative in front of the fraction.

$x = - \frac{5 + \sqrt{597}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{5 - \sqrt{597}}{2}, - \frac{5 + \sqrt{597}}{2}$

Step 8

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

$(x + 1) (x + \frac{5 - \sqrt{597}}{2}) (x + \frac{5 + \sqrt{597}}{2})$

Step 9

These are the roots (zeros) of the polynomial $x^{3} + 6 x^{2} - 138 x - 143$ .

$x = - 1, - \frac{5 - \sqrt{597}}{2}, - \frac{5 + \sqrt{597}}{2}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = - 1, - \frac{5 - \sqrt{597}}{2}, - \frac{5 + \sqrt{597}}{2}$

Decimal Form:

$x = - 1, 9.71679172 \dots, - 14.71679172 \dots$

Step 11

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+6x^2-138x-143

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