Find the Roots/Zeros Using the Rational Roots Test x^4+12x^3+37x^2+12x+36

$x^{4} + 12 x^{3} + 37 x^{2} + 12 x + 36$

Step 1

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$

Step 3

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.

${(- 6)}^{4} + 12 {(- 6)}^{3} + 37 {(- 6)}^{2} + 12 (- 6) + 36$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 6$ is a root of the polynomial.

Simplify each term.

Raise $- 6$ to the power of $4$ .

$1296 + 12 {(- 6)}^{3} + 37 {(- 6)}^{2} + 12 (- 6) + 36$

Raise $- 6$ to the power of $3$ .

$1296 + 12 \cdot - 216 + 37 {(- 6)}^{2} + 12 (- 6) + 36$

Multiply $12$ by $- 216$ .

$1296 - 2592 + 37 {(- 6)}^{2} + 12 (- 6) + 36$

Raise $- 6$ to the power of $2$ .

$1296 - 2592 + 37 \cdot 36 + 12 (- 6) + 36$

Multiply $37$ by $36$ .

$1296 - 2592 + 1332 + 12 (- 6) + 36$

Multiply $12$ by $- 6$ .

$1296 - 2592 + 1332 - 72 + 36$

Simplify by adding and subtracting.

Subtract $2592$ from $1296$ .

$- 1296 + 1332 - 72 + 36$

Add $- 1296$ and $1332$ .

$36 - 72 + 36$

Subtract $72$ from $36$ .

$- 36 + 36$

Add $- 36$ and $36$ .

$0$

Step 5

Since $- 6$ is a known root, divide the polynomial by $x + 6$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} + 12 x^{3} + 37 x^{2} + 12 x + 36}{x + 6}$

Step 6

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.

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0				\xa0

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

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0				\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 6)$ and place the result of $(- 6)$ under the next term in the dividend $(12)$ .

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0	$- 6$			\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.

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0	$- 6$			\xa0
	$1$	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(- 6)$ and place the result of $(- 36)$ under the next term in the dividend $(37)$ .

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0	$- 6$	$- 36$		\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.

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0	$- 6$	$- 36$		\xa0
	$1$	$6$	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 6)$ and place the result of $(- 6)$ under the next term in the dividend $(12)$ .

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0	$- 6$	$- 36$	$- 6$	\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.

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0	$- 6$	$- 36$	$- 6$	\xa0
	$1$	$6$	$1$	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(- 6)$ and place the result of $(- 36)$ under the next term in the dividend $(36)$ .

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0	$- 6$	$- 36$	$- 6$	$- 36$
	$1$	$6$	$1$	$6$

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

$- 6$	$1$	$12$	$37$	$12$	$36$
	\xa0	$- 6$	$- 36$	$- 6$	$- 36$
	$1$	$6$	$1$	$6$	$0$

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

$1 x^{3} + 6 x^{2} + (1) x + 6$

Simplify the quotient polynomial.

$x^{3} + 6 x^{2} + x + 6$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 6) (x^{3} + 6 x^{2}) + x + 6$

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

$(x + 6) + x^{2} (x + 6) + 1 (x + 6)$

$(x + 6) x^{2} (x + 6) + 1 (x + 6)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 6$ .

$(x + 6) (x + 6) (x^{2} + 1)$

Step 9

Factor the left side of the equation.

Regroup terms.

$12 x^{3} + 12 x + x^{4} + 37 x^{2} + 36 = 0$

Factor $12 x$ out of $12 x^{3} + 12 x$ .

Factor $12 x$ out of $12 x^{3}$ .

$12 x (x^{2}) + 12 x + x^{4} + 37 x^{2} + 36 = 0$

Factor $12 x$ out of $12 x$ .

$12 x (x^{2}) + 12 x (1) + x^{4} + 37 x^{2} + 36 = 0$

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

$12 x (x^{2} + 1) + x^{4} + 37 x^{2} + 36 = 0$

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

$12 x (x^{2} + 1) + {(x^{2})}^{2} + 37 x^{2} + 36 = 0$

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

$12 x (x^{2} + 1) + u^{2} + 37 u + 36 = 0$

Factor $u^{2} + 37 u + 36$ 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 $36$ and whose sum is $37$ .

$1, 36$

Write the factored form using these integers.

$12 x (x^{2} + 1) + (u + 1) (u + 36) = 0$

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

$12 x (x^{2} + 1) + (x^{2} + 1) (x^{2} + 36) = 0$

Factor $x^{2} + 1$ out of $12 x (x^{2} + 1) + (x^{2} + 1) (x^{2} + 36)$ .

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

$(x^{2} + 1) (12 x) + (x^{2} + 1) (x^{2} + 36) = 0$

Factor $x^{2} + 1$ out of $(x^{2} + 1) (12 x) + (x^{2} + 1) (x^{2} + 36)$ .

$(x^{2} + 1) (12 x + x^{2} + 36) = 0$

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

$(x^{2} + 1) (12 u + u^{2} + 36) = 0$

Factor using the perfect square rule.

Rearrange terms.

$(x^{2} + 1) (u^{2} + 12 u + 36) = 0$

Rewrite $36$ as $6^{2}$ .

$(x^{2} + 1) (u^{2} + 12 u + 6^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 u = 2 \cdot u \cdot 6$

Rewrite the polynomial.

$(x^{2} + 1) (u^{2} + 2 \cdot u \cdot 6 + 6^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 6$ .

$(x^{2} + 1) {(u + 6)}^{2} = 0$

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

$(x^{2} + 1) {(x + 6)}^{2} = 0$

Step 10

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

$x^{2} + 1 = 0$

${(x + 6)}^{2} = 0$

Step 11

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

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

$x^{2} + 1 = 0$

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

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

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

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

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

Step 12

Set ${(x + 6)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 6)}^{2}$ equal to $0$ .

${(x + 6)}^{2} = 0$

Solve ${(x + 6)}^{2} = 0$ for $x$ .

Set the $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 13

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

$x = i, - i, - 6$

Step 14

Find the Roots/Zeros Using the Rational Roots Test x^4+12x^3+37x^2+12x+36

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