Find the Roots/Zeros Using the Rational Roots Test w^4-w^3-11w^2-w-12

$w^{4} - w^{3} - 11 w^{2} - w - 12$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

$q = \pm 1$

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

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.

${(- 3)}^{4} - {(- 3)}^{3} - 11 {(- 3)}^{2} - (- 3) - 12$

Simplify the expression. In this case, the expression is equal to $0$ so $w = - 3$ is a root of the polynomial.

Simplify each term.

Raise $- 3$ to the power of $4$ .

$81 - {(- 3)}^{3} - 11 {(- 3)}^{2} - (- 3) - 12$

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

$81 - - 27 - 11 {(- 3)}^{2} - (- 3) - 12$

Multiply $- 1$ by $- 27$ .

$81 + 27 - 11 {(- 3)}^{2} - (- 3) - 12$

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

$81 + 27 - 11 \cdot 9 - (- 3) - 12$

Multiply $- 11$ by $9$ .

$81 + 27 - 99 - (- 3) - 12$

Multiply $- 1$ by $- 3$ .

$81 + 27 - 99 + 3 - 12$

Simplify by adding and subtracting.

Add $81$ and $27$ .

$108 - 99 + 3 - 12$

Subtract $99$ from $108$ .

$9 + 3 - 12$

Add $9$ and $3$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

Since $- 3$ is a known root, divide the polynomial by $w + 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{w^{4} - w^{3} - 11 w^{2} - w - 12}{w + 3}$

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.

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0				\xa0

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

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0				\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 3)$ and place the result of $(- 3)$ under the next term in the dividend $(- 1)$ .

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0	$- 3$			\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.

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0	$- 3$			\xa0
	$1$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(- 3)$ and place the result of $(12)$ under the next term in the dividend $(- 11)$ .

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0	$- 3$	$12$		\xa0
	$1$	$- 4$

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

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0	$- 3$	$12$		\xa0
	$1$	$- 4$	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 3)$ and place the result of $(- 3)$ under the next term in the dividend $(- 1)$ .

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0	$- 3$	$12$	$- 3$	\xa0
	$1$	$- 4$	$1$

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

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0	$- 3$	$12$	$- 3$	\xa0
	$1$	$- 4$	$1$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(- 3)$ and place the result of $(12)$ under the next term in the dividend $(- 12)$ .

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0	$- 3$	$12$	$- 3$	$12$
	$1$	$- 4$	$1$	$- 4$

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

$- 3$	$1$	$- 1$	$- 11$	$- 1$	$- 12$
	\xa0	$- 3$	$12$	$- 3$	$12$
	$1$	$- 4$	$1$	$- 4$	$0$

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

$1 w^{3} + - 4 w^{2} + (1) w - 4$

Simplify the quotient polynomial.

$w^{3} - 4 w^{2} + w - 4$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(w + 3) (w^{3} - 4 w^{2}) + w - 4$

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

$(w + 3) + w^{2} (w - 4) + 1 (w - 4)$

$(w + 3) w^{2} (w - 4) + 1 (w - 4)$

Factor the polynomial by factoring out the greatest common factor, $w - 4$ .

$(w + 3) (w - 4) (w^{2} + 1)$

Factor the left side of the equation.

Regroup terms.

$- w^{3} - w + w^{4} - 11 w^{2} - 12 = 0$

Factor $- w$ out of $- w^{3} - w$ .

Factor $- w$ out of $- w^{3}$ .

$- w \cdot w^{2} - w + w^{4} - 11 w^{2} - 12 = 0$

Factor $- w$ out of $- w$ .

$- w \cdot w^{2} - w \cdot 1 + w^{4} - 11 w^{2} - 12 = 0$

Factor $- w$ out of $- w (w^{2}) - w (1)$ .

$- w (w^{2} + 1) + w^{4} - 11 w^{2} - 12 = 0$

Rewrite $w^{4}$ as ${(w^{2})}^{2}$ .

$- w (w^{2} + 1) + {(w^{2})}^{2} - 11 w^{2} - 12 = 0$

Let $u = w^{2}$ . Substitute $u$ for all occurrences of $w^{2}$ .

$- w (w^{2} + 1) + u^{2} - 11 u - 12 = 0$

Factor $u^{2} - 11 u - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 11$ .

$- 12, 1$

Write the factored form using these integers.

$- w (w^{2} + 1) + (u - 12) (u + 1) = 0$

Replace all occurrences of $u$ with $w^{2}$ .

$- w (w^{2} + 1) + (w^{2} - 12) (w^{2} + 1) = 0$

Factor $w^{2} + 1$ out of $- w (w^{2} + 1) + (w^{2} - 12) (w^{2} + 1)$ .

Factor $w^{2} + 1$ out of $- w (w^{2} + 1)$ .

$(w^{2} + 1) (- w) + (w^{2} - 12) (w^{2} + 1) = 0$

Factor $w^{2} + 1$ out of $(w^{2} - 12) (w^{2} + 1)$ .

$(w^{2} + 1) (- w) + (w^{2} + 1) (w^{2} - 12) = 0$

Factor $w^{2} + 1$ out of $(w^{2} + 1) (- w) + (w^{2} + 1) (w^{2} - 12)$ .

$(w^{2} + 1) (- w + w^{2} - 12) = 0$

Let $u = w$ . Substitute $u$ for all occurrences of $w$ .

$(w^{2} + 1) (- u + u^{2} - 12) = 0$

Factor $- u + u^{2} - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$(w^{2} + 1) ((u - 4) (u + 3)) = 0$

Factor.

Replace all occurrences of $u$ with $w$ .

$(w^{2} + 1) ((w - 4) (w + 3)) = 0$

Remove unnecessary parentheses.

$(w^{2} + 1) (w - 4) (w + 3) = 0$

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

$w^{2} + 1 = 0$

$w - 4 = 0$

$w + 3 = 0$

Set $w^{2} + 1$ equal to $0$ and solve for $w$ .

Set $w^{2} + 1$ equal to $0$ .

$w^{2} + 1 = 0$

Solve $w^{2} + 1 = 0$ for $w$ .

Subtract $1$ from both sides of the equation.

$w^{2} = - 1$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$w = \pm \sqrt{- 1}$

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

$w = \pm i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$w = i$

Next, use the negative value of the $\pm$ to find the second solution.

$w = - i$

The complete solution is the result of both the positive and negative portions of the solution.

$w = i, - i$

Set $w - 4$ equal to $0$ and solve for $w$ .

Set $w - 4$ equal to $0$ .

$w - 4 = 0$

Add $4$ to both sides of the equation.

$w = 4$

Set $w + 3$ equal to $0$ and solve for $w$ .

Set $w + 3$ equal to $0$ .

$w + 3 = 0$

Subtract $3$ from both sides of the equation.

$w = - 3$

The final solution is all the values that make $(w^{2} + 1) (w - 4) (w + 3) = 0$ true.

$w = i, - i, 4, - 3$

Find the Roots/Zeros Using the Rational Roots Test w^4-w^3-11w^2-w-12

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