Find the Roots/Zeros Using the Rational Roots Test 2x^4-5x^3-4x^2+15x-6

$2 x^{4} - 5 x^{3} - 4 x^{2} + 15 x - 6$

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.

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

$q = \pm 1, \pm 2$

Step 2

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

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

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.

$2 {(\frac{1}{2})}^{4} - 5 {(\frac{1}{2})}^{3} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{1}{2}$ is a root of the polynomial.

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

$2 \frac{1^{4}}{2^{4}} - 5 {(\frac{1}{2})}^{3} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

One to any power is one.

$2 \frac{1}{2^{4}} - 5 {(\frac{1}{2})}^{3} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Raise $2$ to the power of $4$ .

$2 (\frac{1}{16}) - 5 {(\frac{1}{2})}^{3} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Cancel the common factor of $2$ .

Factor $2$ out of $16$ .

$2 \frac{1}{2 (8)} - 5 {(\frac{1}{2})}^{3} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Cancel the common factor.

$2 \frac{1}{2 \cdot 8} - 5 {(\frac{1}{2})}^{3} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Rewrite the expression.

$\frac{1}{8} - 5 {(\frac{1}{2})}^{3} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Apply the product rule to $\frac{1}{2}$ .

$\frac{1}{8} - 5 \frac{1^{3}}{2^{3}} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

One to any power is one.

$\frac{1}{8} - 5 \frac{1}{2^{3}} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

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

$\frac{1}{8} - 5 (\frac{1}{8}) - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Combine $- 5$ and $\frac{1}{8}$ .

$\frac{1}{8} + \frac{- 5}{8} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Move the negative in front of the fraction.

$\frac{1}{8} - \frac{5}{8} - 4 {(\frac{1}{2})}^{2} + 15 (\frac{1}{2}) - 6$

Apply the product rule to $\frac{1}{2}$ .

$\frac{1}{8} - \frac{5}{8} - 4 \frac{1^{2}}{2^{2}} + 15 (\frac{1}{2}) - 6$

One to any power is one.

$\frac{1}{8} - \frac{5}{8} - 4 \frac{1}{2^{2}} + 15 (\frac{1}{2}) - 6$

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

$\frac{1}{8} - \frac{5}{8} - 4 (\frac{1}{4}) + 15 (\frac{1}{2}) - 6$

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

$\frac{1}{8} - \frac{5}{8} + 4 (- 1) \frac{1}{4} + 15 (\frac{1}{2}) - 6$

Cancel the common factor.

$\frac{1}{8} - \frac{5}{8} + 4 \cdot - 1 \frac{1}{4} + 15 (\frac{1}{2}) - 6$

Rewrite the expression.

$\frac{1}{8} - \frac{5}{8} - 1 + 15 (\frac{1}{2}) - 6$

Combine $15$ and $\frac{1}{2}$ .

$\frac{1}{8} - \frac{5}{8} - 1 + \frac{15}{2} - 6$

Combine fractions.

Combine the numerators over the common denominator.

$- 1 - 6 + \frac{1 - 5}{8} + \frac{15}{2}$

Subtract $5$ from $1$ .

$- 1 - 6 + \frac{- 4}{8} + \frac{15}{2}$

Find the common denominator.

Write $- 1$ as a fraction with denominator $1$ .

$\frac{- 1}{1} - 6 + \frac{- 4}{8} + \frac{15}{2}$

Multiply $\frac{- 1}{1}$ by $\frac{8}{8}$ .

$\frac{- 1}{1} \cdot \frac{8}{8} - 6 + \frac{- 4}{8} + \frac{15}{2}$

Multiply $\frac{- 1}{1}$ by $\frac{8}{8}$ .

$\frac{- 1 \cdot 8}{8} - 6 + \frac{- 4}{8} + \frac{15}{2}$

Write $- 6$ as a fraction with denominator $1$ .

$\frac{- 1 \cdot 8}{8} + \frac{- 6}{1} + \frac{- 4}{8} + \frac{15}{2}$

Multiply $\frac{- 6}{1}$ by $\frac{8}{8}$ .

$\frac{- 1 \cdot 8}{8} + \frac{- 6}{1} \cdot \frac{8}{8} + \frac{- 4}{8} + \frac{15}{2}$

Multiply $\frac{- 6}{1}$ by $\frac{8}{8}$ .

$\frac{- 1 \cdot 8}{8} + \frac{- 6 \cdot 8}{8} + \frac{- 4}{8} + \frac{15}{2}$

Multiply $\frac{15}{2}$ by $\frac{4}{4}$ .

$\frac{- 1 \cdot 8}{8} + \frac{- 6 \cdot 8}{8} + \frac{- 4}{8} + \frac{15}{2} \cdot \frac{4}{4}$

Multiply $\frac{15}{2}$ by $\frac{4}{4}$ .

$\frac{- 1 \cdot 8}{8} + \frac{- 6 \cdot 8}{8} + \frac{- 4}{8} + \frac{15 \cdot 4}{2 \cdot 4}$

Multiply $2$ by $4$ .

$\frac{- 1 \cdot 8}{8} + \frac{- 6 \cdot 8}{8} + \frac{- 4}{8} + \frac{15 \cdot 4}{8}$

Combine the numerators over the common denominator.

$\frac{- 1 \cdot 8 - 6 \cdot 8 - 4 + 15 \cdot 4}{8}$

Simplify each term.

Multiply $- 1$ by $8$ .

$\frac{- 8 - 6 \cdot 8 - 4 + 15 \cdot 4}{8}$

Multiply $- 6$ by $8$ .

$\frac{- 8 - 48 - 4 + 15 \cdot 4}{8}$

Multiply $15$ by $4$ .

$\frac{- 8 - 48 - 4 + 60}{8}$

Simplify the expression.

Subtract $48$ from $- 8$ .

$\frac{- 56 - 4 + 60}{8}$

Subtract $4$ from $- 56$ .

$\frac{- 60 + 60}{8}$

Add $- 60$ and $60$ .

$\frac{0}{8}$

Divide $0$ by $8$ .

$0$

Step 5

Since $\frac{1}{2}$ is a known root, divide the polynomial by $x - \frac{1}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{4} - 5 x^{3} - 4 x^{2} + 15 x - 6}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0				\xa0

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

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0				\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(\frac{1}{2})$ and place the result of $(1)$ under the next term in the dividend $(- 5)$ .

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0	$1$			\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.

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0	$1$			\xa0
	$2$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 2)$ under the next term in the dividend $(- 4)$ .

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0	$1$	$- 2$		\xa0
	$2$	$- 4$

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

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0	$1$	$- 2$		\xa0
	$2$	$- 4$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 3)$ under the next term in the dividend $(15)$ .

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0	$1$	$- 2$	$- 3$	\xa0
	$2$	$- 4$	$- 6$

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

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0	$1$	$- 2$	$- 3$	\xa0
	$2$	$- 4$	$- 6$	$12$

Multiply the newest entry in the result $(12)$ by the divisor $(\frac{1}{2})$ and place the result of $(6)$ under the next term in the dividend $(- 6)$ .

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0	$1$	$- 2$	$- 3$	$6$
	$2$	$- 4$	$- 6$	$12$

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

$\frac{1}{2}$	$2$	$- 5$	$- 4$	$15$	$- 6$
	\xa0	$1$	$- 2$	$- 3$	$6$
	$2$	$- 4$	$- 6$	$12$	$0$

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

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

Simplify the quotient polynomial.

$2 x^{3} - 4 x^{2} - 6 x + 12$

Step 7

Factor $2$ out of $2 x^{3} - 4 x^{2} - 6 x + 12$ .

Factor $2$ out of $2 x^{3}$ .

$(x - \frac{1}{2}) (2 (x^{3}) - 4 x^{2} - 6 x + 12)$

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

$(x - \frac{1}{2}) (2 (x^{3}) + 2 (- 2 x^{2}) - 6 x + 12)$

Factor $2$ out of $- 6 x$ .

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

Factor $2$ out of $12$ .

$(x - \frac{1}{2}) (2 x^{3} + 2 (- 2 x^{2}) + 2 (- 3 x) + 2 \cdot 6)$

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

$(x - \frac{1}{2}) (2 (x^{3} - 2 x^{2}) + 2 (- 3 x) + 2 \cdot 6)$

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

$(x - \frac{1}{2}) (2 (x^{3} - 2 x^{2} - 3 x) + 2 \cdot 6)$

Factor $2$ out of $2 (x^{3} - 2 x^{2} - 3 x) + 2 \cdot 6$ .

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

Step 8

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$(x - \frac{1}{2}) (2 (x^{2} (x - 2) - 3 (x - 2)))$

Step 9

Factor.

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

$(x - \frac{1}{2}) (2 ((x - 2) (x^{2} - 3)))$

Remove unnecessary parentheses.

$(x - \frac{1}{2}) (2 (x - 2) (x^{2} - 3))$

Step 10

Factor the left side of the equation.

Regroup terms.

$- 5 x^{3} + 15 x + 2 x^{4} - 4 x^{2} - 6 = 0$

Factor $- 5 x$ out of $- 5 x^{3} + 15 x$ .

Factor $- 5 x$ out of $- 5 x^{3}$ .

$- 5 x \cdot x^{2} + 15 x + 2 x^{4} - 4 x^{2} - 6 = 0$

Factor $- 5 x$ out of $15 x$ .

$- 5 x \cdot x^{2} - 5 x \cdot - 3 + 2 x^{4} - 4 x^{2} - 6 = 0$

Factor $- 5 x$ out of $- 5 x (x^{2}) - 5 x (- 3)$ .

$- 5 x (x^{2} - 3) + 2 x^{4} - 4 x^{2} - 6 = 0$

Factor $2$ out of $2 x^{4} - 4 x^{2} - 6$ .

Factor $2$ out of $2 x^{4}$ .

$- 5 x (x^{2} - 3) + 2 (x^{4}) - 4 x^{2} - 6 = 0$

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

$- 5 x (x^{2} - 3) + 2 (x^{4}) + 2 (- 2 x^{2}) - 6 = 0$

Factor $2$ out of $- 6$ .

$- 5 x (x^{2} - 3) + 2 x^{4} + 2 (- 2 x^{2}) + 2 \cdot - 3 = 0$

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

$- 5 x (x^{2} - 3) + 2 (x^{4} - 2 x^{2}) + 2 \cdot - 3 = 0$

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

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

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

$- 5 x (x^{2} - 3) + 2 ({(x^{2})}^{2} - 2 x^{2} - 3) = 0$

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

$- 5 x (x^{2} - 3) + 2 (u^{2} - 2 u - 3) = 0$

Factor $u^{2} - 2 u - 3$ 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 $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$- 5 x (x^{2} - 3) + 2 ((u - 3) (u + 1)) = 0$

Factor.

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

$- 5 x (x^{2} - 3) + 2 ((x^{2} - 3) (x^{2} + 1)) = 0$

Remove unnecessary parentheses.

$- 5 x (x^{2} - 3) + 2 (x^{2} - 3) (x^{2} + 1) = 0$

Factor $x^{2} - 3$ out of $- 5 x (x^{2} - 3) + 2 (x^{2} - 3) (x^{2} + 1)$ .

Factor $x^{2} - 3$ out of $- 5 x (x^{2} - 3)$ .

$(x^{2} - 3) (- 5 x) + 2 (x^{2} - 3) (x^{2} + 1) = 0$

Factor $x^{2} - 3$ out of $2 (x^{2} - 3) (x^{2} + 1)$ .

$(x^{2} - 3) (- 5 x) + (x^{2} - 3) (2 (x^{2} + 1)) = 0$

Factor $x^{2} - 3$ out of $(x^{2} - 3) (- 5 x) + (x^{2} - 3) (2 (x^{2} + 1))$ .

$(x^{2} - 3) (- 5 x + 2 (x^{2} + 1)) = 0$

Apply the distributive property.

$(x^{2} - 3) (- 5 x + 2 x^{2} + 2 \cdot 1) = 0$

Multiply $2$ by $1$ .

$(x^{2} - 3) (- 5 x + 2 x^{2} + 2) = 0$

Reorder terms.

$(x^{2} - 3) (2 x^{2} - 5 x + 2) = 0$

Factor.

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 $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$(x^{2} - 3) (2 x^{2} - 5 x + 2) = 0$

Rewrite $- 5$ as $- 1$ plus $- 4$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

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

Remove unnecessary parentheses.

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

Step 11

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

$x^{2} - 3 = 0$

$2 x - 1 = 0$

$x - 2 = 0$

Step 12

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Solve $x^{2} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

Step 13

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 14

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 15

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

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

Step 16

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 1.73205080 \dots, - 1.73205080 \dots, 0.5, 2$

Step 17

Find the Roots/Zeros Using the Rational Roots Test 2x^4-5x^3-4x^2+15x-6

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