Find the Roots/Zeros Using the Rational Roots Test 5x^3+x^2-20x-4

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

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.

$5 {(2)}^{3} + {(2)}^{2} - 20 \cdot 2 - 4$

Step 4

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

Simplify each term.

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

$5 \cdot 8 + {(2)}^{2} - 20 \cdot 2 - 4$

Multiply $5$ by $8$ .

$40 + {(2)}^{2} - 20 \cdot 2 - 4$

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

$40 + 4 - 20 \cdot 2 - 4$

Multiply $- 20$ by $2$ .

$40 + 4 - 40 - 4$

Simplify by adding and subtracting.

Add $40$ and $4$ .

$44 - 40 - 4$

Subtract $40$ from $44$ .

$4 - 4$

Subtract $4$ from $4$ .

$0$

Step 5

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

$\frac{5 x^{3} + x^{2} - 20 x - 4}{x - 2}$

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.

$2$	$5$	$1$	$- 20$	$- 4$
	\xa0			\xa0

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

$2$	$5$	$1$	$- 20$	$- 4$
	\xa0			\xa0
	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(2)$ and place the result of $(10)$ under the next term in the dividend $(1)$ .

$2$	$5$	$1$	$- 20$	$- 4$
	\xa0	$10$		\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.

$2$	$5$	$1$	$- 20$	$- 4$
	\xa0	$10$		\xa0
	$5$	$11$

Multiply the newest entry in the result $(11)$ by the divisor $(2)$ and place the result of $(22)$ under the next term in the dividend $(- 20)$ .

$2$	$5$	$1$	$- 20$	$- 4$
	\xa0	$10$	$22$	\xa0
	$5$	$11$

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

$2$	$5$	$1$	$- 20$	$- 4$
	\xa0	$10$	$22$	\xa0
	$5$	$11$	$2$

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

$2$	$5$	$1$	$- 20$	$- 4$
	\xa0	$10$	$22$	$4$
	$5$	$11$	$2$

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

$2$	$5$	$1$	$- 20$	$- 4$
	\xa0	$10$	$22$	$4$
	$5$	$11$	$2$	$0$

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

$5 x^{2} + 11 x + 2$

Step 7

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 $a \cdot c = 5 \cdot 2 = 10$ and whose sum is $b = 11$ .

Factor $11$ out of $11 x$ .

$(x - 2) + 5 x^{2} + 11 (x) + 2$

Rewrite $11$ as $1$ plus $10$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

$(x - 2) + 5 x^{2} + x + 10 x + 2$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 2) (5 x^{2} + x) + 10 x + 2$

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

$(x - 2) + x (5 x + 1) + 2 (5 x + 1)$

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

$(x - 2) + (5 x + 1) (x + 2)$

$(x - 2) (5 x + 1) (x + 2)$

Step 8

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 x^{3} + x^{2}) - 20 x - 4 = 0$

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

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

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

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

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

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

Factor.

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

$(5 x + 1) ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$(5 x + 1) (x + 2) (x - 2) = 0$

Step 9

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

$5 x + 1 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 10

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 11

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 12

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 13

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

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

Step 14

Find the Roots/Zeros Using the Rational Roots Test 5x^3+x^2-20x-4

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