Find the Roots/Zeros Using the Rational Roots Test w^4-w^3-29w^2-w-30

$w^{4} - w^{3} - 29 w^{2} - w - 30$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$

$q = \pm 1$

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

$\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$

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.

${(- 5)}^{4} - {(- 5)}^{3} - 29 {(- 5)}^{2} - (- 5) - 30$

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

Simplify each term.

Raise $- 5$ to the power of $4$ .

$625 - {(- 5)}^{3} - 29 {(- 5)}^{2} - (- 5) - 30$

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

$625 - - 125 - 29 {(- 5)}^{2} - (- 5) - 30$

Multiply $- 1$ by $- 125$ .

$625 + 125 - 29 {(- 5)}^{2} - (- 5) - 30$

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

$625 + 125 - 29 \cdot 25 - (- 5) - 30$

Multiply $- 29$ by $25$ .

$625 + 125 - 725 - (- 5) - 30$

Multiply $- 1$ by $- 5$ .

$625 + 125 - 725 + 5 - 30$

Simplify by adding and subtracting.

Add $625$ and $125$ .

$750 - 725 + 5 - 30$

Subtract $725$ from $750$ .

$25 + 5 - 30$

Add $25$ and $5$ .

$30 - 30$

Subtract $30$ from $30$ .

$0$

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

$\frac{w^{4} - w^{3} - 29 w^{2} - w - 30}{w + 5}$

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.

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\xa0				\xa0

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

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\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 $(- 1)$ .

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\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.

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\xa0	$- 5$			\xa0
	$1$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(- 5)$ and place the result of $(30)$ under the next term in the dividend $(- 29)$ .

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\xa0	$- 5$	$30$		\xa0
	$1$	$- 6$

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

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\xa0	$- 5$	$30$		\xa0
	$1$	$- 6$	$1$

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 $(- 1)$ .

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\xa0	$- 5$	$30$	$- 5$	\xa0
	$1$	$- 6$	$1$

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

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\xa0	$- 5$	$30$	$- 5$	\xa0
	$1$	$- 6$	$1$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(- 5)$ and place the result of $(30)$ under the next term in the dividend $(- 30)$ .

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\xa0	$- 5$	$30$	$- 5$	$30$
	$1$	$- 6$	$1$	$- 6$

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

$- 5$	$1$	$- 1$	$- 29$	$- 1$	$- 30$
	\xa0	$- 5$	$30$	$- 5$	$30$
	$1$	$- 6$	$1$	$- 6$	$0$

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

$1 w^{3} + - 6 w^{2} + (1) w - 6$

Simplify the quotient polynomial.

$w^{3} - 6 w^{2} + w - 6$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(w + 5) (w^{3} - 6 w^{2}) + w - 6$

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

$(w + 5) + w^{2} (w - 6) + 1 (w - 6)$

$(w + 5) w^{2} (w - 6) + 1 (w - 6)$

Factor the polynomial by factoring out the greatest common factor, $w - 6$ .

$(w + 5) (w - 6) (w^{2} + 1)$

Factor the left side of the equation.

Regroup terms.

$- w^{3} - w + w^{4} - 29 w^{2} - 30 = 0$

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

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

$- w \cdot w^{2} - w + w^{4} - 29 w^{2} - 30 = 0$

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

$- w \cdot w^{2} - w \cdot 1 + w^{4} - 29 w^{2} - 30 = 0$

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

$- w (w^{2} + 1) + w^{4} - 29 w^{2} - 30 = 0$

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

$- w (w^{2} + 1) + {(w^{2})}^{2} - 29 w^{2} - 30 = 0$

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

$- w (w^{2} + 1) + u^{2} - 29 u - 30 = 0$

Factor $u^{2} - 29 u - 30$ 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 $- 30$ and whose sum is $- 29$ .

$- 30, 1$

Write the factored form using these integers.

$- w (w^{2} + 1) + (u - 30) (u + 1) = 0$

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

$- w (w^{2} + 1) + (w^{2} - 30) (w^{2} + 1) = 0$

Factor $w^{2} + 1$ out of $- w (w^{2} + 1) + (w^{2} - 30) (w^{2} + 1)$ .

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

$(w^{2} + 1) (- w) + (w^{2} - 30) (w^{2} + 1) = 0$

Factor $w^{2} + 1$ out of $(w^{2} - 30) (w^{2} + 1)$ .

$(w^{2} + 1) (- w) + (w^{2} + 1) (w^{2} - 30) = 0$

Factor $w^{2} + 1$ out of $(w^{2} + 1) (- w) + (w^{2} + 1) (w^{2} - 30)$ .

$(w^{2} + 1) (- w + w^{2} - 30) = 0$

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

$(w^{2} + 1) (- u + u^{2} - 30) = 0$

Factor $- u + u^{2} - 30$ 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 $- 30$ and whose sum is $- 1$ .

$- 6, 5$

Write the factored form using these integers.

$(w^{2} + 1) ((u - 6) (u + 5)) = 0$

Factor.

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

$(w^{2} + 1) ((w - 6) (w + 5)) = 0$

Remove unnecessary parentheses.

$(w^{2} + 1) (w - 6) (w + 5) = 0$

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

$w^{2} + 1 = 0$

$w - 6 = 0$

$w + 5 = 0$

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 $w - 6$ equal to $0$ and solve for $w$ .

Set $w - 6$ equal to $0$ .

$w - 6 = 0$

Add $6$ to both sides of the equation.

$w = 6$

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

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

$w + 5 = 0$

Subtract $5$ from both sides of the equation.

$w = - 5$

The final solution is all the values that make $(w^{2} + 1) (w - 6) (w + 5) = 0$ true.

$w = i, - i, 6, - 5$

Find the Roots/Zeros Using the Rational Roots Test w^4-w^3-29w^2-w-30

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