Find the Roots/Zeros Using the Rational Roots Test f(x)=4x^5-8x^4-5x^3+10x^2+x-2

$f (x) = 4 x^{5} - 8 x^{4} - 5 x^{3} + 10 x^{2} + x - 2$

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.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 2, \pm 4$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 2$

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.

$4 {(1)}^{5} - 8 {(1)}^{4} - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Step 4

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

Remove parentheses.

$4 {(1)}^{5} - 8 {(1)}^{4} - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Simplify each term.

One to any power is one.

$4 \cdot 1 - 8 {(1)}^{4} - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Multiply $4$ by $1$ .

$4 - 8 {(1)}^{4} - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

One to any power is one.

$4 - 8 \cdot 1 - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Multiply $- 8$ by $1$ .

$4 - 8 - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

One to any power is one.

$4 - 8 - 5 \cdot 1 + 10 {(1)}^{2} + 1 - 2$

Multiply $- 5$ by $1$ .

$4 - 8 - 5 + 10 {(1)}^{2} + 1 - 2$

One to any power is one.

$4 - 8 - 5 + 10 \cdot 1 + 1 - 2$

Multiply $10$ by $1$ .

$4 - 8 - 5 + 10 + 1 - 2$

$4 - 8 - 5 + 10 + 1 - 2$

Simplify by adding and subtracting.

Subtract $8$ from $4$ .

$- 4 - 5 + 10 + 1 - 2$

Subtract $5$ from $- 4$ .

$- 9 + 10 + 1 - 2$

Add $- 9$ and $10$ .

$1 + 1 - 2$

Add $1$ and $1$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

$0$

$0$

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.

$\frac{4 x^{5} - 8 x^{4} - 5 x^{3} + 10 x^{2} + x - 2}{x - 1}$

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.

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0					\xa0

The first number in the dividend $(4)$ is put into the first position of the result area (below the horizontal line).

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0					\xa0
	$4$

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 $(- 8)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$				\xa0
	$4$

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

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$				\xa0
	$4$	$- 4$

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 $(- 5)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$	$- 4$			\xa0
	$4$	$- 4$

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

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$	$- 4$			\xa0
	$4$	$- 4$	$- 9$

Multiply the newest entry in the result $(- 9)$ by the divisor $(1)$ and place the result of $(- 9)$ under the next term in the dividend $(10)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$	$- 4$	$- 9$		\xa0
	$4$	$- 4$	$- 9$

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

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$	$- 4$	$- 9$		\xa0
	$4$	$- 4$	$- 9$	$1$

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 $(1)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$	$- 4$	$- 9$	$1$	\xa0
	$4$	$- 4$	$- 9$	$1$

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

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$	$- 4$	$- 9$	$1$	\xa0
	$4$	$- 4$	$- 9$	$1$	$2$

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 $(- 2)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$	$- 4$	$- 9$	$1$	$2$
	$4$	$- 4$	$- 9$	$1$	$2$

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

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
	\xa0	$4$	$- 4$	$- 9$	$1$	$2$
	$4$	$- 4$	$- 9$	$1$	$2$	$0$

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

$4 x^{4} + - 4 x^{3} - 9 x^{2} + 1 x + 2$

Simplify the quotient polynomial.

$4 x^{4} - 4 x^{3} - 9 x^{2} + x + 2$

$4 x^{4} - 4 x^{3} - 9 x^{2} + x + 2$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Regroup terms.

$- 4 x^{3} + x + 4 x^{4} - 9 x^{2} + 2 = 0$

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

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

$- x (4 x^{2}) + x + 4 x^{4} - 9 x^{2} + 2 = 0$

Rewrite $x$ as $1 x$ .

$- x (4 x^{2}) + 1 x + 4 x^{4} - 9 x^{2} + 2 = 0$

Factor $- x$ out of $1 x$ .

$- x (4 x^{2}) - x \cdot - 1 + 4 x^{4} - 9 x^{2} + 2 = 0$

Factor $- x$ out of $- x (4 x^{2}) - x (- 1)$ .

$- x (4 x^{2} - 1) + 4 x^{4} - 9 x^{2} + 2 = 0$

$- x (4 x^{2} - 1) + 4 x^{4} - 9 x^{2} + 2 = 0$

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

$- x ({(2 x)}^{2} - 1) + 4 x^{4} - 9 x^{2} + 2 = 0$

Rewrite $1$ as $1^{2}$ .

$- x ({(2 x)}^{2} - 1^{2}) + 4 x^{4} - 9 x^{2} + 2 = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

$- x ((2 x + 1) (2 x - 1)) + 4 x^{4} - 9 x^{2} + 2 = 0$

Remove unnecessary parentheses.

$- x (2 x + 1) (2 x - 1) + 4 x^{4} - 9 x^{2} + 2 = 0$

$- x (2 x + 1) (2 x - 1) + 4 x^{4} - 9 x^{2} + 2 = 0$

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

$- x (2 x + 1) (2 x - 1) + 4 {(x^{2})}^{2} - 9 x^{2} + 2 = 0$

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

$- x (2 x + 1) (2 x - 1) + 4 u^{2} - 9 u + 2 = 0$

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 $a \cdot c = 4 \cdot 2 = 8$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 u$ .

$- x (2 x + 1) (2 x - 1) + 4 u^{2} - 9 u + 2 = 0$

Rewrite $- 9$ as $- 1$ plus $- 8$

$- x (2 x + 1) (2 x - 1) + 4 u^{2} + (- 1 - 8) u + 2 = 0$

Apply the distributive property.

$- x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u - 8 u + 2 = 0$

$- x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u - 8 u + 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u - 8 u + 2 = 0$

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

$- x (2 x + 1) (2 x - 1) + u (4 u - 1) - 2 (4 u - 1) = 0$

$- x (2 x + 1) (2 x - 1) + u (4 u - 1) - 2 (4 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$- x (2 x + 1) (2 x - 1) + (4 u - 1) (u - 2) = 0$

$- x (2 x + 1) (2 x - 1) + (4 u - 1) (u - 2) = 0$

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

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

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

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

Rewrite $1$ as $1^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

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

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

Factor $(2 x + 1) (2 x - 1)$ out of $- x (2 x + 1) (2 x - 1)$ .

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

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

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

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

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

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

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

$- 2, 1$

Write the factored form using these integers.

$(2 x + 1) (2 x - 1) ((u - 2) (u + 1)) = 0$

$(2 x + 1) (2 x - 1) ((u - 2) (u + 1)) = 0$

Factor.

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

$(2 x + 1) (2 x - 1) ((x - 2) (x + 1)) = 0$

Remove unnecessary parentheses.

$(2 x + 1) (2 x - 1) (x - 2) (x + 1) = 0$

$(2 x + 1) (2 x - 1) (x - 2) (x + 1) = 0$

$(2 x + 1) (2 x - 1) (x - 2) (x + 1) = 0$

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

$2 x + 1 = 0$

$2 x - 1 = 0$

$x - 2 = 0$

$x + 1 = 0$

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 $x$ .

Subtract $1$ from 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 $1$ .

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

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

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

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

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

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

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 $1$ .

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

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

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

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

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

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

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.

$x = 2$

$x = 2$

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.

$x = - 1$

$x = - 1$

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

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

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

Step 8

The polynomial can be written as a set of linear factors.

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

Step 9

These are the roots (zeros) of the polynomial $4 x^{5} - 8 x^{4} - 5 x^{3} + 10 x^{2} + x - 2$ .

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

Step 10

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