Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+5x^2-139x-143

$f (x) = x^{3} + 5 x^{2} - 139 x - 143$

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.

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

$q = \pm 1$

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

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

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.

${(- 1)}^{3} + 5 {(- 1)}^{2} - 139 \cdot - 1 - 143$

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

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Simplify each term.

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

$- 1 + 5 {(- 1)}^{2} - 139 \cdot - 1 - 143$

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

$- 1 + 5 \cdot 1 - 139 \cdot - 1 - 143$

Multiply $5$ by $1$ .

$- 1 + 5 - 139 \cdot - 1 - 143$

Multiply $- 139$ by $- 1$ .

$- 1 + 5 + 139 - 143$

Simplify by adding and subtracting.

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Add $- 1$ and $5$ .

$4 + 139 - 143$

Add $4$ and $139$ .

$143 - 143$

Subtract $143$ from $143$ .

$0$

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.

$\frac{x^{3} + 5 x^{2} - 139 x - 143}{x + 1}$

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

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

$- 1$	$1$	$5$	$- 139$	$- 143$

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

$- 1$	$1$	$5$	$- 139$	$- 143$

	$1$

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 $(5)$ .

$- 1$	$1$	$5$	$- 139$	$- 143$
		$- 1$
	$1$

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

$- 1$	$1$	$5$	$- 139$	$- 143$
		$- 1$
	$1$	$4$

Multiply the newest entry in the result $(4)$ by the divisor $(- 1)$ and place the result of $(- 4)$ under the next term in the dividend $(- 139)$ .

$- 1$	$1$	$5$	$- 139$	$- 143$
		$- 1$	$- 4$
	$1$	$4$

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

$- 1$	$1$	$5$	$- 139$	$- 143$
		$- 1$	$- 4$
	$1$	$4$	$- 143$

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 $(- 143)$ .

$- 1$	$1$	$5$	$- 139$	$- 143$
		$- 1$	$- 4$	$143$
	$1$	$4$	$- 143$

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

$- 1$	$1$	$5$	$- 139$	$- 143$
		$- 1$	$- 4$	$143$
	$1$	$4$	$- 143$	$0$

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

$1 x^{2} + 4 x - 143$

Simplify the quotient polynomial.

$x^{2} + 4 x - 143$

Solve the equation to find any remaining roots.

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Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 4$ , and $c = - 143$ into the quadratic formula and solve for $x$ .

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

Simplify.

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Simplify the numerator.

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Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot (1 \cdot - 143)}}{2 \cdot 1}$

Multiply $- 143$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4$ by $- 143$ .

$x = \frac{- 4 \pm \sqrt{16 + 572}}{2 \cdot 1}$

Add $16$ and $572$ .

$x = \frac{- 4 \pm \sqrt{588}}{2 \cdot 1}$

Rewrite $588$ as $14^{2} \cdot 3$ .

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Factor $196$ out of $588$ .

$x = \frac{- 4 \pm \sqrt{196 (3)}}{2 \cdot 1}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{- 4 \pm \sqrt{14^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 14 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 14 \sqrt{3}}{2}$

Simplify $\frac{- 4 \pm 14 \sqrt{3}}{2}$ .

$x = - 2 \pm 7 \sqrt{3}$

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

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Simplify the numerator.

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Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot (1 \cdot - 143)}}{2 \cdot 1}$

Multiply $- 143$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4$ by $- 143$ .

$x = \frac{- 4 \pm \sqrt{16 + 572}}{2 \cdot 1}$

Add $16$ and $572$ .

$x = \frac{- 4 \pm \sqrt{588}}{2 \cdot 1}$

Rewrite $588$ as $14^{2} \cdot 3$ .

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Factor $196$ out of $588$ .

$x = \frac{- 4 \pm \sqrt{196 (3)}}{2 \cdot 1}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{- 4 \pm \sqrt{14^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 14 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 14 \sqrt{3}}{2}$

Simplify $\frac{- 4 \pm 14 \sqrt{3}}{2}$ .

$x = - 2 \pm 7 \sqrt{3}$

Change the $\pm$ to $+$ .

$x = - 2 + 7 \sqrt{3}$

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

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Simplify the numerator.

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Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot (1 \cdot - 143)}}{2 \cdot 1}$

Multiply $- 143$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 143}}{2 \cdot 1}$

Multiply $- 4$ by $- 143$ .

$x = \frac{- 4 \pm \sqrt{16 + 572}}{2 \cdot 1}$

Add $16$ and $572$ .

$x = \frac{- 4 \pm \sqrt{588}}{2 \cdot 1}$

Rewrite $588$ as $14^{2} \cdot 3$ .

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Factor $196$ out of $588$ .

$x = \frac{- 4 \pm \sqrt{196 (3)}}{2 \cdot 1}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{- 4 \pm \sqrt{14^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 14 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 14 \sqrt{3}}{2}$

Simplify $\frac{- 4 \pm 14 \sqrt{3}}{2}$ .

$x = - 2 \pm 7 \sqrt{3}$

Change the $\pm$ to $-$ .

$x = - 2 - 7 \sqrt{3}$

The final answer is the combination of both solutions.

$x = - 2 + 7 \sqrt{3}, - 2 - 7 \sqrt{3}$

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

$(x + 1) (x + 2 - 7 \sqrt{3}) (x + 2 + 7 \sqrt{3})$

These are the roots (zeros) of the polynomial $x^{3} + 5 x^{2} - 139 x - 143$ .

$x = - 1, - 2 + 7 \sqrt{3}, - 2 - 7 \sqrt{3}$

The result can be shown in multiple forms.

Exact Form:

$x = - 1, - 2 + 7 \sqrt{3}, - 2 - 7 \sqrt{3}$

Decimal Form:

$x = - 1, 10.12435565 \dots, - 14.12435565 \dots$

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