Find the Roots/Zeros Using the Rational Roots Test x^4+4x^3+7x^2+16x+12

$x^{4} + 4 x^{3} + 7 x^{2} + 16 x + 12$

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.

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

$q = \pm 1$

Step 2

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

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

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.

${(- 1)}^{4} + 4 {(- 1)}^{3} + 7 {(- 1)}^{2} + 16 (- 1) + 12$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

Simplify each term.

Raise $- 1$ to the power of $4$ .

$1 + 4 {(- 1)}^{3} + 7 {(- 1)}^{2} + 16 (- 1) + 12$

Raise $- 1$ to the power of $3$ .

$1 + 4 \cdot - 1 + 7 {(- 1)}^{2} + 16 (- 1) + 12$

Multiply $4$ by $- 1$ .

$1 - 4 + 7 {(- 1)}^{2} + 16 (- 1) + 12$

Raise $- 1$ to the power of $2$ .

$1 - 4 + 7 \cdot 1 + 16 (- 1) + 12$

Multiply $7$ by $1$ .

$1 - 4 + 7 + 16 (- 1) + 12$

Multiply $16$ by $- 1$ .

$1 - 4 + 7 - 16 + 12$

Simplify by adding and subtracting.

Subtract $4$ from $1$ .

$- 3 + 7 - 16 + 12$

Add $- 3$ and $7$ .

$4 - 16 + 12$

Subtract $16$ from $4$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

Step 5

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} + 4 x^{3} + 7 x^{2} + 16 x + 12}{x + 1}$

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.

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0				\xa0

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

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0				\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 1)$ and place the result of $(- 1)$ under the next term in the dividend $(4)$ .

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0	$- 1$			\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.

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0	$- 1$			\xa0
	$1$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(- 1)$ and place the result of $(- 3)$ under the next term in the dividend $(7)$ .

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0	$- 1$	$- 3$		\xa0
	$1$	$3$

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

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0	$- 1$	$- 3$		\xa0
	$1$	$3$	$4$

Multiply the newest entry in the result $(4)$ by the divisor $(- 1)$ and place the result of $(- 4)$ under the next term in the dividend $(16)$ .

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0	$- 1$	$- 3$	$- 4$	\xa0
	$1$	$3$	$4$

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

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0	$- 1$	$- 3$	$- 4$	\xa0
	$1$	$3$	$4$	$12$

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

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0	$- 1$	$- 3$	$- 4$	$- 12$
	$1$	$3$	$4$	$12$

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

$- 1$	$1$	$4$	$7$	$16$	$12$
	\xa0	$- 1$	$- 3$	$- 4$	$- 12$
	$1$	$3$	$4$	$12$	$0$

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

$1 x^{3} + 3 x^{2} + (4) x + 12$

Simplify the quotient polynomial.

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

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) (x^{3} + 3 x^{2}) + 4 x + 12$

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

$(x + 1) + x^{2} (x + 3) + 4 (x + 3)$

$(x + 1) x^{2} (x + 3) + 4 (x + 3)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

$(x + 1) (x + 3) (x^{2} + 4)$

Step 9

Factor the left side of the equation.

Regroup terms.

$4 x^{3} + 16 x + x^{4} + 7 x^{2} + 12 = 0$

Factor $4 x$ out of $4 x^{3} + 16 x$ .

Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) + 16 x + x^{4} + 7 x^{2} + 12 = 0$

Factor $4 x$ out of $16 x$ .

$4 x (x^{2}) + 4 x (4) + x^{4} + 7 x^{2} + 12 = 0$

Factor $4 x$ out of $4 x (x^{2}) + 4 x (4)$ .

$4 x (x^{2} + 4) + x^{4} + 7 x^{2} + 12 = 0$

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

$4 x (x^{2} + 4) + {(x^{2})}^{2} + 7 x^{2} + 12 = 0$

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

$4 x (x^{2} + 4) + u^{2} + 7 u + 12 = 0$

Factor $u^{2} + 7 u + 12$ 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 $7$ .

$3, 4$

Write the factored form using these integers.

$4 x (x^{2} + 4) + (u + 3) (u + 4) = 0$

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

$4 x (x^{2} + 4) + (x^{2} + 3) (x^{2} + 4) = 0$

Factor $x^{2} + 4$ out of $4 x (x^{2} + 4) + (x^{2} + 3) (x^{2} + 4)$ .

Factor $x^{2} + 4$ out of $4 x (x^{2} + 4)$ .

$(x^{2} + 4) (4 x) + (x^{2} + 3) (x^{2} + 4) = 0$

Factor $x^{2} + 4$ out of $(x^{2} + 3) (x^{2} + 4)$ .

$(x^{2} + 4) (4 x) + (x^{2} + 4) (x^{2} + 3) = 0$

Factor $x^{2} + 4$ out of $(x^{2} + 4) (4 x) + (x^{2} + 4) (x^{2} + 3)$ .

$(x^{2} + 4) (4 x + x^{2} + 3) = 0$

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

$(x^{2} + 4) (4 u + u^{2} + 3) = 0$

Factor $4 u + u^{2} + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

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

Factor.

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

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

Remove unnecessary parentheses.

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

Step 10

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

$x^{2} + 4 = 0$

$x + 1 = 0$

$x + 3 = 0$

Step 11

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4$ equal to $0$ .

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

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

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

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

$x = \pm i \cdot \sqrt{4}$

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

$x = \pm i \cdot \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

Step 12

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 13

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 14

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

$x = 2 i, - 2 i, - 1, - 3$

Step 15

Find the Roots/Zeros Using the Rational Roots Test x^4+4x^3+7x^2+16x+12

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