Find the Roots/Zeros Using the Rational Roots Test 11x^3+125x^2+45x+11

$11 x^{3} + 125 x^{2} + 45 x + 11$

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.

$p = \pm 1, \pm 11$

$q = \pm 1, \pm 11$

Step 2

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

$\pm 1, \pm \frac{1}{11}, \pm 11$

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.

$11 {(- 11)}^{3} + 125 {(- 11)}^{2} + 45 (- 11) + 11$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 11$ is a root of the polynomial.

Simplify each term.

Raise $- 11$ to the power of $3$ .

$11 \cdot - 1331 + 125 {(- 11)}^{2} + 45 (- 11) + 11$

Multiply $11$ by $- 1331$ .

$- 14641 + 125 {(- 11)}^{2} + 45 (- 11) + 11$

Raise $- 11$ to the power of $2$ .

$- 14641 + 125 \cdot 121 + 45 (- 11) + 11$

Multiply $125$ by $121$ .

$- 14641 + 15125 + 45 (- 11) + 11$

Multiply $45$ by $- 11$ .

$- 14641 + 15125 - 495 + 11$

Simplify by adding and subtracting.

Add $- 14641$ and $15125$ .

$484 - 495 + 11$

Subtract $495$ from $484$ .

$- 11 + 11$

Add $- 11$ and $11$ .

$0$

Step 5

Since $- 11$ is a known root, divide the polynomial by $x + 11$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{11 x^{3} + 125 x^{2} + 45 x + 11}{x + 11}$

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.

$- 11$	$11$	$125$	$45$	$11$
	\xa0			\xa0

The first number in the dividend $(11)$ is put into the first position of the result area (below the horizontal line).

$- 11$	$11$	$125$	$45$	$11$
	\xa0			\xa0
	$11$

Multiply the newest entry in the result $(11)$ by the divisor $(- 11)$ and place the result of $(- 121)$ under the next term in the dividend $(125)$ .

$- 11$	$11$	$125$	$45$	$11$
	\xa0	$- 121$		\xa0
	$11$

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

$- 11$	$11$	$125$	$45$	$11$
	\xa0	$- 121$		\xa0
	$11$	$4$

Multiply the newest entry in the result $(4)$ by the divisor $(- 11)$ and place the result of $(- 44)$ under the next term in the dividend $(45)$ .

$- 11$	$11$	$125$	$45$	$11$
	\xa0	$- 121$	$- 44$	\xa0
	$11$	$4$

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

$- 11$	$11$	$125$	$45$	$11$
	\xa0	$- 121$	$- 44$	\xa0
	$11$	$4$	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 11)$ and place the result of $(- 11)$ under the next term in the dividend $(11)$ .

$- 11$	$11$	$125$	$45$	$11$
	\xa0	$- 121$	$- 44$	$- 11$
	$11$	$4$	$1$

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

$- 11$	$11$	$125$	$45$	$11$
	\xa0	$- 121$	$- 44$	$- 11$
	$11$	$4$	$1$	$0$

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

$11 x^{2} + 4 x + 1$

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 $a = 11$ , $b = 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (11 \cdot 1)}}{2 \cdot 11}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 11 \cdot 1}}{2 \cdot 11}$

Multiply $- 4 \cdot 11 \cdot 1$ .

Multiply $- 4$ by $11$ .

$x = \frac{- 4 \pm \sqrt{16 - 44 \cdot 1}}{2 \cdot 11}$

Multiply $- 44$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 44}}{2 \cdot 11}$

Subtract $44$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 28}}{2 \cdot 11}$

Rewrite $- 28$ as $- 1 (28)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 28}}{2 \cdot 11}$

Rewrite $\sqrt{- 1 (28)}$ as $\sqrt{- 1} \cdot \sqrt{28}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{28}}{2 \cdot 11}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{28}}{2 \cdot 11}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4 (7)}}{2 \cdot 11}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 4 \pm i \cdot \sqrt{2^{2} \cdot 7}}{2 \cdot 11}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (2 \sqrt{7})}{2 \cdot 11}$

Move $2$ to the left of $i$ .

$x = \frac{- 4 \pm 2 i \sqrt{7}}{2 \cdot 11}$

Multiply $2$ by $11$ .

$x = \frac{- 4 \pm 2 i \sqrt{7}}{22}$

Simplify $\frac{- 4 \pm 2 i \sqrt{7}}{22}$ .

$x = \frac{- 2 \pm i \sqrt{7}}{11}$

The final answer is the combination of both solutions.

$x = - \frac{2 - i \sqrt{7}}{11}, - \frac{2 + i \sqrt{7}}{11}$

Step 8

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

$(x + 11) (x + \frac{2 - i \sqrt{7}}{11}) (x + \frac{2 + i \sqrt{7}}{11})$

Step 9

These are the roots (zeros) of the polynomial $11 x^{3} + 125 x^{2} + 45 x + 11$ .

$x = - 11, - \frac{2 - i \sqrt{7}}{11}, - \frac{2 + i \sqrt{7}}{11}$

Step 10

Find the Roots/Zeros Using the Rational Roots Test 11x^3+125x^2+45x+11

Related Questions

Support LCMGCF.com by disabling your adblocker.