Find the Roots/Zeros Using the Rational Roots Test x^3-17x^2+55x+25

$x^{3} - 17 x^{2} + 55 x + 25$

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.

$p = \pm 1, \pm 5, \pm 25$

$q = \pm 1$

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

$\pm 1, \pm 5, \pm 25$

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.

${(5)}^{3} - 17 {(5)}^{2} + 55 (5) + 25$

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 $3$ .

$125 - 17 {(5)}^{2} + 55 (5) + 25$

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

$125 - 17 \cdot 25 + 55 (5) + 25$

Multiply $- 17$ by $25$ .

$125 - 425 + 55 (5) + 25$

Multiply $55$ by $5$ .

$125 - 425 + 275 + 25$

Simplify by adding and subtracting.

Subtract $425$ from $125$ .

$- 300 + 275 + 25$

Add $- 300$ and $275$ .

$- 25 + 25$

Add $- 25$ and $25$ .

$0$

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.

$\frac{x^{3} - 17 x^{2} + 55 x + 25}{x - 5}$

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.

$5$	$1$	$- 17$	$55$	$25$
	\xa0			\xa0

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

$5$	$1$	$- 17$	$55$	$25$
	\xa0			\xa0
	$1$

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 $(- 17)$ .

$5$	$1$	$- 17$	$55$	$25$
	\xa0	$5$		\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.

$5$	$1$	$- 17$	$55$	$25$
	\xa0	$5$		\xa0
	$1$	$- 12$

Multiply the newest entry in the result $(- 12)$ by the divisor $(5)$ and place the result of $(- 60)$ under the next term in the dividend $(55)$ .

$5$	$1$	$- 17$	$55$	$25$
	\xa0	$5$	$- 60$	\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.

$5$	$1$	$- 17$	$55$	$25$
	\xa0	$5$	$- 60$	\xa0
	$1$	$- 12$	$- 5$

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 $(25)$ .

$5$	$1$	$- 17$	$55$	$25$
	\xa0	$5$	$- 60$	$- 25$
	$1$	$- 12$	$- 5$

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

$5$	$1$	$- 17$	$55$	$25$
	\xa0	$5$	$- 60$	$- 25$
	$1$	$- 12$	$- 5$	$0$

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

$1 x^{2} + - 12 x - 5$

Simplify the quotient polynomial.

$x^{2} - 12 x - 5$

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 $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot - 5)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $- 5$ .

$x = \frac{12 \pm \sqrt{144 + 20}}{2 \cdot 1}$

Add $144$ and $20$ .

$x = \frac{12 \pm \sqrt{164}}{2 \cdot 1}$

Rewrite $164$ as $2^{2} \cdot 41$ .

Factor $4$ out of $164$ .

$x = \frac{12 \pm \sqrt{4 (41)}}{2 \cdot 1}$

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

$x = \frac{12 \pm \sqrt{2^{2} \cdot 41}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{12 \pm 2 \sqrt{41}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{12 \pm 2 \sqrt{41}}{2}$

Simplify $\frac{12 \pm 2 \sqrt{41}}{2}$ .

$x = 6 \pm \sqrt{41}$

The final answer is the combination of both solutions.

$x = 6 + \sqrt{41}, 6 - \sqrt{41}$

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

$(x - 5) (x - 6 - \sqrt{41}) (x - 6 + \sqrt{41})$

These are the roots (zeros) of the polynomial $x^{3} - 17 x^{2} + 55 x + 25$ .

$x = 5, 6 + \sqrt{41}, 6 - \sqrt{41}$

The result can be shown in multiple forms.

Exact Form:

$x = 5, 6 + \sqrt{41}, 6 - \sqrt{41}$

Decimal Form:

$x = 5, 12.40312423 \dots, - 0.40312423 \dots$

Find the Roots/Zeros Using the Rational Roots Test x^3-17x^2+55x+25

Related Questions

Support LCMGCF.com by disabling your adblocker.