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

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

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.

$p = \pm 1, \pm 5$

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

Step 2

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

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

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.

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

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.

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

Multiply $4$ by $1$ .

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

One to any power is one.

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

Multiply $5$ by $1$ .

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

Multiply $- 4$ by $1$ .

$4 + 5 - 4 - 5$

Simplify by adding and subtracting.

Add $4$ and $5$ .

$9 - 4 - 5$

Subtract $4$ from $9$ .

$5 - 5$

Subtract $5$ from $5$ .

$0$

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.

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

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.

$1$	$4$	$5$	$- 4$	$- 5$
	\xa0			\xa0

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

$1$	$4$	$5$	$- 4$	$- 5$
	\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 $(5)$ .

$1$	$4$	$5$	$- 4$	$- 5$
	\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.

$1$	$4$	$5$	$- 4$	$- 5$
	\xa0	$4$		\xa0
	$4$	$9$

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 $(- 4)$ .

$1$	$4$	$5$	$- 4$	$- 5$
	\xa0	$4$	$9$	\xa0
	$4$	$9$

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

$1$	$4$	$5$	$- 4$	$- 5$
	\xa0	$4$	$9$	\xa0
	$4$	$9$	$5$

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 $(- 5)$ .

$1$	$4$	$5$	$- 4$	$- 5$
	\xa0	$4$	$9$	$5$
	$4$	$9$	$5$

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

$1$	$4$	$5$	$- 4$	$- 5$
	\xa0	$4$	$9$	$5$
	$4$	$9$	$5$	$0$

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

$4 x^{2} + 9 x + 5$

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 $a \cdot c = 4 \cdot 5 = 20$ and whose sum is $b = 9$ .

Factor $9$ out of $9 x$ .

$(x - 1) + 4 x^{2} + 9 (x) + 5$

Rewrite $9$ as $4$ plus $5$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$(x - 1) + 4 x (x + 1) + 5 (x + 1)$

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

$(x - 1) + (x + 1) (4 x + 5)$

$(x - 1) (x + 1) (4 x + 5)$

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.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $4 x + 5$ .

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

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

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

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 $b = 1$ .

$(4 x + 5) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(4 x + 5) (x + 1) (x - 1) = 0$

Step 9

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

$4 x + 5 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 10

Set $4 x + 5$ equal to $0$ and solve for $x$ .

Set $4 x + 5$ equal to $0$ .

$4 x + 5 = 0$

Solve $4 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$4 x = - 5$

Divide each term in $4 x = - 5$ by $4$ and simplify.

Divide each term in $4 x = - 5$ by $4$ .

$\frac{4 x}{4} = \frac{- 5}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 5}{4}$

Divide $x$ by $1$ .

$x = \frac{- 5}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{4}$

Step 11

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 12

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 13

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

$x = - \frac{5}{4}, - 1, 1$

Step 14

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

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