Find the Roots/Zeros Using the Rational Roots Test x^3-11x^2+33x+45

$x^{3} - 11 x^{2} + 33 x + 45$

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.

$p = \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45$

$q = \pm 1$

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

$\pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45$

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.

${(- 1)}^{3} - 11 {(- 1)}^{2} + 33 (- 1) + 45$

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 $3$ .

$- 1 - 11 {(- 1)}^{2} + 33 (- 1) + 45$

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

$- 1 - 11 \cdot 1 + 33 (- 1) + 45$

Multiply $- 11$ by $1$ .

$- 1 - 11 + 33 (- 1) + 45$

Multiply $33$ by $- 1$ .

$- 1 - 11 - 33 + 45$

Simplify by adding and subtracting.

Subtract $11$ from $- 1$ .

$- 12 - 33 + 45$

Subtract $33$ from $- 12$ .

$- 45 + 45$

Add $- 45$ and $45$ .

$0$

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.

$\frac{x^{3} - 11 x^{2} + 33 x + 45}{x + 1}$

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.

$- 1$	$1$	$- 11$	$33$	$45$
	\xa0			\xa0

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

$- 1$	$1$	$- 11$	$33$	$45$
	\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 $(- 11)$ .

$- 1$	$1$	$- 11$	$33$	$45$
	\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.

$- 1$	$1$	$- 11$	$33$	$45$
	\xa0	$- 1$		\xa0
	$1$	$- 12$

Multiply the newest entry in the result $(- 12)$ by the divisor $(- 1)$ and place the result of $(12)$ under the next term in the dividend $(33)$ .

$- 1$	$1$	$- 11$	$33$	$45$
	\xa0	$- 1$	$12$	\xa0
	$1$	$- 12$

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

$- 1$	$1$	$- 11$	$33$	$45$
	\xa0	$- 1$	$12$	\xa0
	$1$	$- 12$	$45$

Multiply the newest entry in the result $(45)$ by the divisor $(- 1)$ and place the result of $(- 45)$ under the next term in the dividend $(45)$ .

$- 1$	$1$	$- 11$	$33$	$45$
	\xa0	$- 1$	$12$	$- 45$
	$1$	$- 12$	$45$

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

$- 1$	$1$	$- 11$	$33$	$45$
	\xa0	$- 1$	$12$	$- 45$
	$1$	$- 12$	$45$	$0$

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

$1 x^{2} + - 12 x + 45$

Simplify the quotient polynomial.

$x^{2} - 12 x + 45$

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 $a = 1$ , $b = - 12$ , and $c = 45$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot 45)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot 45}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 45}}{2 \cdot 1}$

Multiply $- 4$ by $45$ .

$x = \frac{12 \pm \sqrt{144 - 180}}{2 \cdot 1}$

Subtract $180$ from $144$ .

$x = \frac{12 \pm \sqrt{- 36}}{2 \cdot 1}$

Rewrite $- 36$ as $- 1 (36)$ .

$x = \frac{12 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \frac{12 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

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

$x = \frac{12 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{12 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

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

$x = \frac{12 \pm i \cdot 6}{2 \cdot 1}$

Move $6$ to the left of $i$ .

$x = \frac{12 \pm 6 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{12 \pm 6 i}{2}$

Simplify $\frac{12 \pm 6 i}{2}$ .

$x = 6 \pm 3 i$

The final answer is the combination of both solutions.

$x = 6 + 3 i, 6 - 3 i$

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

$(x + 1) (x - 6 - 3 i) (x - 6 + 3 i)$

These are the roots (zeros) of the polynomial $x^{3} - 11 x^{2} + 33 x + 45$ .

$x = - 1, 6 + 3 i, 6 - 3 i$

Find the Roots/Zeros Using the Rational Roots Test x^3-11x^2+33x+45

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