Find the Roots/Zeros Using the Rational Roots Test 5x^3+28x^2+16x+5

$5 x^{3} + 28 x^{2} + 16 x + 5$

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.

$p = \pm 1, \pm 5$

$q = \pm 1, \pm 5$

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

$\pm 1, \pm \frac{1}{5}, \pm 5$

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.

$5 {(- 5)}^{3} + 28 {(- 5)}^{2} + 16 (- 5) + 5$

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 5$ is a root of the polynomial.

Simplify each term.

Raise $- 5$ to the power of $3$ .

$5 \cdot - 125 + 28 {(- 5)}^{2} + 16 (- 5) + 5$

Multiply $5$ by $- 125$ .

$- 625 + 28 {(- 5)}^{2} + 16 (- 5) + 5$

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

$- 625 + 28 \cdot 25 + 16 (- 5) + 5$

Multiply $28$ by $25$ .

$- 625 + 700 + 16 (- 5) + 5$

Multiply $16$ by $- 5$ .

$- 625 + 700 - 80 + 5$

Simplify by adding and subtracting.

Add $- 625$ and $700$ .

$75 - 80 + 5$

Subtract $80$ from $75$ .

$- 5 + 5$

Add $- 5$ and $5$ .

$0$

Since $- 5$ is a known root, divide the polynomial by $x + 5$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{5 x^{3} + 28 x^{2} + 16 x + 5}{x + 5}$

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.

$- 5$	$5$	$28$	$16$	$5$
	\xa0			\xa0

The first number in the dividend $(5)$ is put into the first position of the result area (below the horizontal line).

$- 5$	$5$	$28$	$16$	$5$
	\xa0			\xa0
	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(- 5)$ and place the result of $(- 25)$ under the next term in the dividend $(28)$ .

$- 5$	$5$	$28$	$16$	$5$
	\xa0	$- 25$		\xa0
	$5$

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

$- 5$	$5$	$28$	$16$	$5$
	\xa0	$- 25$		\xa0
	$5$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(- 5)$ and place the result of $(- 15)$ under the next term in the dividend $(16)$ .

$- 5$	$5$	$28$	$16$	$5$
	\xa0	$- 25$	$- 15$	\xa0
	$5$	$3$

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

$- 5$	$5$	$28$	$16$	$5$
	\xa0	$- 25$	$- 15$	\xa0
	$5$	$3$	$1$

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

$- 5$	$5$	$28$	$16$	$5$
	\xa0	$- 25$	$- 15$	$- 5$
	$5$	$3$	$1$

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

$- 5$	$5$	$28$	$16$	$5$
	\xa0	$- 25$	$- 15$	$- 5$
	$5$	$3$	$1$	$0$

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

$5 x^{2} + 3 x + 1$

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 $a = 5$ , $b = 3$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (5 \cdot 1)}}{2 \cdot 5}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 5 \cdot 1}}{2 \cdot 5}$

Multiply $- 4$ by $5$ .

$x = \frac{- 3 \pm \sqrt{9 - 20 \cdot 1}}{2 \cdot 5}$

Multiply $- 20$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 20}}{2 \cdot 5}$

Subtract $20$ from $9$ .

$x = \frac{- 3 \pm \sqrt{- 11}}{2 \cdot 5}$

Rewrite $- 11$ as $- 1 (11)$ .

$x = \frac{- 3 \pm \sqrt{- 1 \cdot 11}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (11)}$ as $\sqrt{- 1} \cdot \sqrt{11}$ .

$x = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{11}}{2 \cdot 5}$

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

$x = \frac{- 3 \pm i \sqrt{11}}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 3 \pm i \sqrt{11}}{10}$

The final answer is the combination of both solutions.

$x = - \frac{3 - i \sqrt{11}}{10}, - \frac{3 + i \sqrt{11}}{10}$

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

$(x + 5) (x + \frac{3 - i \sqrt{11}}{10}) (x + \frac{3 + i \sqrt{11}}{10})$

These are the roots (zeros) of the polynomial $5 x^{3} + 28 x^{2} + 16 x + 5$ .

$x = - 5, - \frac{3 - i \sqrt{11}}{10}, - \frac{3 + i \sqrt{11}}{10}$

Find the Roots/Zeros Using the Rational Roots Test 5x^3+28x^2+16x+5

Related Questions

Support LCMGCF.com by disabling your adblocker.