Find the Roots/Zeros Using the Rational Roots Test x^3-2x^2-25x+50=0

$x^{3} - 2 x^{2} - 25 x + 50 = 0$

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.

$p = \pm 1, \pm 2, \pm 5, \pm 10, \pm 25, \pm 50$

$q = \pm 1$

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

$\pm 1, \pm 2, \pm 5, \pm 10, \pm 25, \pm 50$

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.

${(2)}^{3} - 2 {(2)}^{2} - 25 \cdot 2 + 50$

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

Simplify each term.

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

$8 - 2 {(2)}^{2} - 25 \cdot 2 + 50$

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

$8 - 2 \cdot 4 - 25 \cdot 2 + 50$

Multiply $- 2$ by $4$ .

$8 - 8 - 25 \cdot 2 + 50$

Multiply $- 25$ by $2$ .

$8 - 8 - 50 + 50$

Simplify by adding and subtracting.

Subtract $8$ from $8$ .

$0 - 50 + 50$

Subtract $50$ from $0$ .

$- 50 + 50$

Add $- 50$ and $50$ .

$0$

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 2 x^{2} - 25 x + 50}{x - 2}$

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.

$2$	$1$	$- 2$	$- 25$	$50$
	\xa0			\xa0

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

$2$	$1$	$- 2$	$- 25$	$50$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(2)$ and place the result of $(2)$ under the next term in the dividend $(- 2)$ .

$2$	$1$	$- 2$	$- 25$	$50$
	\xa0	$2$		\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.

$2$	$1$	$- 2$	$- 25$	$50$
	\xa0	$2$		\xa0
	$1$	$0$

Multiply the newest entry in the result $(0)$ by the divisor $(2)$ and place the result of $(0)$ under the next term in the dividend $(- 25)$ .

$2$	$1$	$- 2$	$- 25$	$50$
	\xa0	$2$	$0$	\xa0
	$1$	$0$

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

$2$	$1$	$- 2$	$- 25$	$50$
	\xa0	$2$	$0$	\xa0
	$1$	$0$	$- 25$

Multiply the newest entry in the result $(- 25)$ by the divisor $(2)$ and place the result of $(- 50)$ under the next term in the dividend $(50)$ .

$2$	$1$	$- 2$	$- 25$	$50$
	\xa0	$2$	$0$	$- 50$
	$1$	$0$	$- 25$

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

$2$	$1$	$- 2$	$- 25$	$50$
	\xa0	$2$	$0$	$- 50$
	$1$	$0$	$- 25$	$0$

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

$1 x^{2} + 0 x - 25$

Simplify the quotient polynomial.

$x^{2} - 25$

Rewrite $25$ as $5^{2}$ .

$(x - 2) x^{2} - 5^{2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$(x - 2) (x + 5) (x - 5)$

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} - 2 x^{2}) - 25 x + 50 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x - 2) - 25 (x - 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 2$ .

$(x - 2) (x^{2} - 25) = 0$

Rewrite $25$ as $5^{2}$ .

$(x - 2) (x^{2} - 5^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$(x - 2) ((x + 5) (x - 5)) = 0$

Remove unnecessary parentheses.

$(x - 2) (x + 5) (x - 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 2 = 0$

$x + 5 = 0$

$x - 5 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $(x - 2) (x + 5) (x - 5) = 0$ true.

$x = 2, - 5, 5$

Find the Roots/Zeros Using the Rational Roots Test x^3-2x^2-25x+50=0

Related Questions

Support LCMGCF.com by disabling your adblocker.