Find the Roots/Zeros Using the Rational Roots Test 2x^4-x^3+5x^2-4x-12

$2 x^{4} - x^{3} + 5 x^{2} - 4 x - 12$

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.

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

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 4, \pm 6, \pm 12$

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.

$2 {(- 1)}^{4} - {(- 1)}^{3} + 5 {(- 1)}^{2} - 4 \cdot - 1 - 12$

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 $4$ .

$2 \cdot 1 - {(- 1)}^{3} + 5 {(- 1)}^{2} - 4 \cdot - 1 - 12$

Multiply $2$ by $1$ .

$2 - {(- 1)}^{3} + 5 {(- 1)}^{2} - 4 \cdot - 1 - 12$

Multiply $- 1$ by ${(- 1)}^{3}$ by adding the exponents.

Multiply $- 1$ by ${(- 1)}^{3}$ .

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

$2 + {(- 1)}^{1} {(- 1)}^{3} + 5 {(- 1)}^{2} - 4 \cdot - 1 - 12$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 + {(- 1)}^{1 + 3} + 5 {(- 1)}^{2} - 4 \cdot - 1 - 12$

Add $1$ and $3$ .

$2 + {(- 1)}^{4} + 5 {(- 1)}^{2} - 4 \cdot - 1 - 12$

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

$2 + 1 + 5 {(- 1)}^{2} - 4 \cdot - 1 - 12$

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

$2 + 1 + 5 \cdot 1 - 4 \cdot - 1 - 12$

Multiply $5$ by $1$ .

$2 + 1 + 5 - 4 \cdot - 1 - 12$

Multiply $- 4$ by $- 1$ .

$2 + 1 + 5 + 4 - 12$

Simplify by adding and subtracting.

Add $2$ and $1$ .

$3 + 5 + 4 - 12$

Add $3$ and $5$ .

$8 + 4 - 12$

Add $8$ and $4$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

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.

$\frac{2 x^{4} - x^{3} + 5 x^{2} - 4 x - 12}{x + 1}$

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.

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0				\xa0

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0				\xa0
	$2$

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

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0	$- 2$			\xa0
	$2$

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

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0	$- 2$			\xa0
	$2$	$- 3$

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 $(5)$ .

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0	$- 2$	$3$		\xa0
	$2$	$- 3$

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

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0	$- 2$	$3$		\xa0
	$2$	$- 3$	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(- 1)$ and place the result of $(- 8)$ under the next term in the dividend $(- 4)$ .

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0	$- 2$	$3$	$- 8$	\xa0
	$2$	$- 3$	$8$

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

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0	$- 2$	$3$	$- 8$	\xa0
	$2$	$- 3$	$8$	$- 12$

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 $(- 12)$ .

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0	$- 2$	$3$	$- 8$	$12$
	$2$	$- 3$	$8$	$- 12$

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

$- 1$	$2$	$- 1$	$5$	$- 4$	$- 12$
	\xa0	$- 2$	$3$	$- 8$	$12$
	$2$	$- 3$	$8$	$- 12$	$0$

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

$2 x^{3} + - 3 x^{2} + (8) x - 12$

Simplify the quotient polynomial.

$2 x^{3} - 3 x^{2} + 8 x - 12$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) (2 x^{3} - 3 x^{2}) + 8 x - 12$

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

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

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

Step 8

Factor the polynomial by factoring out the greatest common factor, $2 x - 3$ .

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

Step 9

Factor the left side of the equation.

Regroup terms.

$- x^{3} - 4 x + 2 x^{4} + 5 x^{2} - 12 = 0$

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

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

$- x \cdot x^{2} - 4 x + 2 x^{4} + 5 x^{2} - 12 = 0$

Factor $- x$ out of $- 4 x$ .

$- x \cdot x^{2} - x \cdot 4 + 2 x^{4} + 5 x^{2} - 12 = 0$

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

$- x (x^{2} + 4) + 2 x^{4} + 5 x^{2} - 12 = 0$

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

$- x (x^{2} + 4) + 2 {(x^{2})}^{2} + 5 x^{2} - 12 = 0$

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

$- x (x^{2} + 4) + 2 u^{2} + 5 u - 12 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 12 = - 24$ and whose sum is $b = 5$ .

Factor $5$ out of $5 u$ .

$- x (x^{2} + 4) + 2 u^{2} + 5 (u) - 12 = 0$

Rewrite $5$ as $- 3$ plus $8$

$- x (x^{2} + 4) + 2 u^{2} + (- 3 + 8) u - 12 = 0$

Apply the distributive property.

$- x (x^{2} + 4) + 2 u^{2} - 3 u + 8 u - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- x (x^{2} + 4) + 2 u^{2} - 3 u + 8 u - 12 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 u - 3$ .

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

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

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

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

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

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

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

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

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

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

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

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

Factor by grouping.

Reorder terms.

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

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- u$ .

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

Rewrite $- 1$ as $2$ plus $- 3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $u + 1$ .

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

Factor.

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

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

Remove unnecessary parentheses.

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

Step 10

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

$x^{2} + 4 = 0$

$x + 1 = 0$

$2 x - 3 = 0$

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 $2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 14

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

$x = 2 i, - 2 i, - 1, \frac{3}{2}$

Step 15

Find the Roots/Zeros Using the Rational Roots Test 2x^4-x^3+5x^2-4x-12

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