Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+4x^2-35x-38

$f (x) = x^{3} + 4 x^{2} - 35 x - 38$

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.

$p = \pm 1, \pm 2, \pm 19, \pm 38$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 19, \pm 38$

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.

${(- 1)}^{3} + 4 {(- 1)}^{2} - 35 \cdot - 1 - 38$

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 $3$ .

$- 1 + 4 {(- 1)}^{2} - 35 \cdot - 1 - 38$

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

$- 1 + 4 \cdot 1 - 35 \cdot - 1 - 38$

Multiply $4$ by $1$ .

$- 1 + 4 - 35 \cdot - 1 - 38$

Multiply $- 35$ by $- 1$ .

$- 1 + 4 + 35 - 38$

Simplify by adding and subtracting.

Add $- 1$ and $4$ .

$3 + 35 - 38$

Add $3$ and $35$ .

$38 - 38$

Subtract $38$ from $38$ .

$0$

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.

$\frac{x^{3} + 4 x^{2} - 35 x - 38}{x + 1}$

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.

$- 1$	$1$	$4$	$- 35$	$- 38$
	\xa0			\xa0

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

$- 1$	$1$	$4$	$- 35$	$- 38$
	\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 $(4)$ .

$- 1$	$1$	$4$	$- 35$	$- 38$
	\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.

$- 1$	$1$	$4$	$- 35$	$- 38$
	\xa0	$- 1$		\xa0
	$1$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(- 1)$ and place the result of $(- 3)$ under the next term in the dividend $(- 35)$ .

$- 1$	$1$	$4$	$- 35$	$- 38$
	\xa0	$- 1$	$- 3$	\xa0
	$1$	$3$

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

$- 1$	$1$	$4$	$- 35$	$- 38$
	\xa0	$- 1$	$- 3$	\xa0
	$1$	$3$	$- 38$

Multiply the newest entry in the result $(- 38)$ by the divisor $(- 1)$ and place the result of $(38)$ under the next term in the dividend $(- 38)$ .

$- 1$	$1$	$4$	$- 35$	$- 38$
	\xa0	$- 1$	$- 3$	$38$
	$1$	$3$	$- 38$

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

$- 1$	$1$	$4$	$- 35$	$- 38$
	\xa0	$- 1$	$- 3$	$38$
	$1$	$3$	$- 38$	$0$

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

$1 x^{2} + 3 x - 38$

Simplify the quotient polynomial.

$x^{2} + 3 x - 38$

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 $a = 1$ , $b = 3$ , and $c = - 38$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 38)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 38}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 38$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 38}}{2 \cdot 1}$

Multiply $- 4$ by $- 38$ .

$x = \frac{- 3 \pm \sqrt{9 + 152}}{2 \cdot 1}$

Add $9$ and $152$ .

$x = \frac{- 3 \pm \sqrt{161}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{161}}{2}$

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

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 38}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 38$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 38}}{2 \cdot 1}$

Multiply $- 4$ by $- 38$ .

$x = \frac{- 3 \pm \sqrt{9 + 152}}{2 \cdot 1}$

Add $9$ and $152$ .

$x = \frac{- 3 \pm \sqrt{161}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{161}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{- 3 + \sqrt{161}}{2}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 + \sqrt{161}}{2}$

Factor $- 1$ out of $\sqrt{161}$ .

$x = \frac{- 1 \cdot 3 - 1 (- \sqrt{161})}{2}$

Factor $- 1$ out of $- 1 (3) - 1 (- \sqrt{161})$ .

$x = \frac{- 1 (3 - \sqrt{161})}{2}$

Move the negative in front of the fraction.

$x = - \frac{3 - \sqrt{161}}{2}$

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

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 38}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 38$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 38}}{2 \cdot 1}$

Multiply $- 4$ by $- 38$ .

$x = \frac{- 3 \pm \sqrt{9 + 152}}{2 \cdot 1}$

Add $9$ and $152$ .

$x = \frac{- 3 \pm \sqrt{161}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{161}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{- 3 - \sqrt{161}}{2}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 - \sqrt{161}}{2}$

Factor $- 1$ out of $- \sqrt{161}$ .

$x = \frac{- 1 \cdot 3 - (\sqrt{161})}{2}$

Factor $- 1$ out of $- 1 (3) - (\sqrt{161})$ .

$x = \frac{- 1 (3 + \sqrt{161})}{2}$

Move the negative in front of the fraction.

$x = - \frac{3 + \sqrt{161}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{3 - \sqrt{161}}{2}, - \frac{3 + \sqrt{161}}{2}$

Step 8

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

$(x + 1) (x + \frac{3 - \sqrt{161}}{2}) (x + \frac{3 + \sqrt{161}}{2})$

Step 9

These are the roots (zeros) of the polynomial $x^{3} + 4 x^{2} - 35 x - 38$ .

$x = - 1, - \frac{3 - \sqrt{161}}{2}, - \frac{3 + \sqrt{161}}{2}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = - 1, - \frac{3 - \sqrt{161}}{2}, - \frac{3 + \sqrt{161}}{2}$

Decimal Form:

$x = - 1, 4.84428877 \dots, - 7.84428877 \dots$

Step 11

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+4x^2-35x-38

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