Find the Roots/Zeros Using the Rational Roots Test 5x^4-4x^3+19x^2-16x-4

$5 x^{4} - 4 x^{3} + 19 x^{2} - 16 x - 4$

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.

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

$q = \pm 1, \pm 5$

Step 2

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

$\pm 1, \pm \frac{1}{5}, \pm 2, \pm \frac{2}{5}, \pm 4, \pm \frac{4}{5}$

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.

$5 {(1)}^{4} - 4 {(1)}^{3} + 19 {(1)}^{2} - 16 \cdot 1 - 4$

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.

One to any power is one.

$5 \cdot 1 - 4 {(1)}^{3} + 19 {(1)}^{2} - 16 \cdot 1 - 4$

Multiply $5$ by $1$ .

$5 - 4 {(1)}^{3} + 19 {(1)}^{2} - 16 \cdot 1 - 4$

One to any power is one.

$5 - 4 \cdot 1 + 19 {(1)}^{2} - 16 \cdot 1 - 4$

Multiply $- 4$ by $1$ .

$5 - 4 + 19 {(1)}^{2} - 16 \cdot 1 - 4$

One to any power is one.

$5 - 4 + 19 \cdot 1 - 16 \cdot 1 - 4$

Multiply $19$ by $1$ .

$5 - 4 + 19 - 16 \cdot 1 - 4$

Multiply $- 16$ by $1$ .

$5 - 4 + 19 - 16 - 4$

Simplify by adding and subtracting.

Subtract $4$ from $5$ .

$1 + 19 - 16 - 4$

Add $1$ and $19$ .

$20 - 16 - 4$

Subtract $16$ from $20$ .

$4 - 4$

Subtract $4$ from $4$ .

$0$

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.

$\frac{5 x^{4} - 4 x^{3} + 19 x^{2} - 16 x - 4}{x - 1}$

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.

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0				\xa0

The first number in the dividend $(5)$ is put into the first position of the result area (below the horizontal line).

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0				\xa0
	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(1)$ and place the result of $(5)$ under the next term in the dividend $(- 4)$ .

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0	$5$			\xa0
	$5$

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

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0	$5$			\xa0
	$5$	$1$

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 $(19)$ .

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0	$5$	$1$		\xa0
	$5$	$1$

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

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0	$5$	$1$		\xa0
	$5$	$1$	$20$

Multiply the newest entry in the result $(20)$ by the divisor $(1)$ and place the result of $(20)$ under the next term in the dividend $(- 16)$ .

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0	$5$	$1$	$20$	\xa0
	$5$	$1$	$20$

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

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0	$5$	$1$	$20$	\xa0
	$5$	$1$	$20$	$4$

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 $(- 4)$ .

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0	$5$	$1$	$20$	$4$
	$5$	$1$	$20$	$4$

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

$1$	$5$	$- 4$	$19$	$- 16$	$- 4$
	\xa0	$5$	$1$	$20$	$4$
	$5$	$1$	$20$	$4$	$0$

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

$5 x^{3} + 1 x^{2} + (20) x + 4$

Simplify the quotient polynomial.

$5 x^{3} + x^{2} + 20 x + 4$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (5 x^{3} + x^{2}) + 20 x + 4$

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

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

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

Step 8

Factor the polynomial by factoring out the greatest common factor, $5 x + 1$ .

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

Step 9

Factor the left side of the equation.

Regroup terms.

$- 4 x^{3} - 16 x + 5 x^{4} + 19 x^{2} - 4 = 0$

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

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

$- 4 x \cdot x^{2} - 16 x + 5 x^{4} + 19 x^{2} - 4 = 0$

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

$- 4 x \cdot x^{2} - 4 x \cdot 4 + 5 x^{4} + 19 x^{2} - 4 = 0$

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

$- 4 x (x^{2} + 4) + 5 x^{4} + 19 x^{2} - 4 = 0$

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

$- 4 x (x^{2} + 4) + 5 {(x^{2})}^{2} + 19 x^{2} - 4 = 0$

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

$- 4 x (x^{2} + 4) + 5 u^{2} + 19 u - 4 = 0$

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 $a \cdot c = 5 \cdot - 4 = - 20$ and whose sum is $b = 19$ .

Factor $19$ out of $19 u$ .

$- 4 x (x^{2} + 4) + 5 u^{2} + 19 (u) - 4 = 0$

Rewrite $19$ as $- 1$ plus $20$

$- 4 x (x^{2} + 4) + 5 u^{2} + (- 1 + 20) u - 4 = 0$

Apply the distributive property.

$- 4 x (x^{2} + 4) + 5 u^{2} - 1 u + 20 u - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- 4 x (x^{2} + 4) + 5 u^{2} - 1 u + 20 u - 4 = 0$

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

$- 4 x (x^{2} + 4) + u (5 u - 1) + 4 (5 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 u - 1$ .

$- 4 x (x^{2} + 4) + (5 u - 1) (u + 4) = 0$

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

$- 4 x (x^{2} + 4) + (5 x^{2} - 1) (x^{2} + 4) = 0$

Factor $x^{2} + 4$ out of $- 4 x (x^{2} + 4) + (5 x^{2} - 1) (x^{2} + 4)$ .

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

$(x^{2} + 4) (- 4 x) + (5 x^{2} - 1) (x^{2} + 4) = 0$

Factor $x^{2} + 4$ out of $(5 x^{2} - 1) (x^{2} + 4)$ .

$(x^{2} + 4) (- 4 x) + (x^{2} + 4) (5 x^{2} - 1) = 0$

Factor $x^{2} + 4$ out of $(x^{2} + 4) (- 4 x) + (x^{2} + 4) (5 x^{2} - 1)$ .

$(x^{2} + 4) (- 4 x + 5 x^{2} - 1) = 0$

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

$(x^{2} + 4) (- 4 u + 5 u^{2} - 1) = 0$

Factor by grouping.

Reorder terms.

$(x^{2} + 4) (5 u^{2} - 4 u - 1) = 0$

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 $a \cdot c = 5 \cdot - 1 = - 5$ and whose sum is $b = - 4$ .

Factor $- 4$ out of $- 4 u$ .

$(x^{2} + 4) (5 u^{2} - 4 u - 1) = 0$

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

$(x^{2} + 4) (5 u^{2} + (1 - 5) u - 1) = 0$

Apply the distributive property.

$(x^{2} + 4) (5 u^{2} + 1 u - 5 u - 1) = 0$

Multiply $u$ by $1$ .

$(x^{2} + 4) (5 u^{2} + u - 5 u - 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{2} + 4) ((5 u^{2} + u) - 5 u - 1) = 0$

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

$(x^{2} + 4) (u (5 u + 1) - (5 u + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 u + 1$ .

$(x^{2} + 4) ((5 u + 1) (u - 1)) = 0$

Factor.

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

$(x^{2} + 4) ((5 x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 4) (5 x + 1) (x - 1) = 0$

Step 10

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

$x^{2} + 4 = 0$

$5 x + 1 = 0$

$x - 1 = 0$

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 $5 x + 1$ equal to $0$ and solve for $x$ .

Set $5 x + 1$ equal to $0$ .

$5 x + 1 = 0$

Solve $5 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

Divide each term in $5 x = - 1$ by $5$ .

$\frac{5 x}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 1}{5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

Step 13

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 14

The final solution is all the values that make $(x^{2} + 4) (5 x + 1) (x - 1) = 0$ true.

$x = 2 i, - 2 i, - \frac{1}{5}, 1$

Step 15

Find the Roots/Zeros Using the Rational Roots Test 5x^4-4x^3+19x^2-16x-4

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