Find the Roots/Zeros Using the Rational Roots Test f(x)=-4x^3+23x^2-14x-5

$f (x) = - 4 x^{3} + 23 x^{2} - 14 x - 5$

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.

$- 4 {(1)}^{3} + 23 {(1)}^{2} - 14 \cdot 1 - 5$

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.

$- 4 \cdot 1 + 23 {(1)}^{2} - 14 \cdot 1 - 5$

Multiply $- 4$ by $1$ .

$- 4 + 23 {(1)}^{2} - 14 \cdot 1 - 5$

One to any power is one.

$- 4 + 23 \cdot 1 - 14 \cdot 1 - 5$

Multiply $23$ by $1$ .

$- 4 + 23 - 14 \cdot 1 - 5$

Multiply $- 14$ by $1$ .

$- 4 + 23 - 14 - 5$

Simplify by adding and subtracting.

Add $- 4$ and $23$ .

$19 - 14 - 5$

Subtract $14$ from $19$ .

$5 - 5$

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.

$\frac{- 4 x^{3} + 23 x^{2} - 14 x - 5}{x - 1}$

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.

$1$	$- 4$	$23$	$- 14$	$- 5$
	\xa0			\xa0

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

$1$	$- 4$	$23$	$- 14$	$- 5$
	\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 $(23)$ .

$1$	$- 4$	$23$	$- 14$	$- 5$
	\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.

$1$	$- 4$	$23$	$- 14$	$- 5$
	\xa0	$- 4$		\xa0
	$- 4$	$19$

Multiply the newest entry in the result $(19)$ by the divisor $(1)$ and place the result of $(19)$ under the next term in the dividend $(- 14)$ .

$1$	$- 4$	$23$	$- 14$	$- 5$
	\xa0	$- 4$	$19$	\xa0
	$- 4$	$19$

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

$1$	$- 4$	$23$	$- 14$	$- 5$
	\xa0	$- 4$	$19$	\xa0
	$- 4$	$19$	$5$

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 $(- 5)$ .

$1$	$- 4$	$23$	$- 14$	$- 5$
	\xa0	$- 4$	$19$	$5$
	$- 4$	$19$	$5$

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

$1$	$- 4$	$23$	$- 14$	$- 5$
	\xa0	$- 4$	$19$	$5$
	$- 4$	$19$	$5$	$0$

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

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

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 $b = 19$ .

Factor $19$ out of $19 x$ .

$(x - 1) - 4 x^{2} + 19 (x) + 5$

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

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

Step 8

Factor the left side of the equation.

Factor $- 4 x^{3} + 23 x^{2} - 14 x - 5$ using the rational roots test.

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.

$p = \pm 1, \pm 5 q = \pm 1, \pm 4, \pm 2$

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 5, \pm 1.25, \pm 2.5$

Substitute $- 0.25$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.25$ is a root of the polynomial.

Substitute $- 0.25$ into the polynomial.

$- 4 {(- 0.25)}^{3} + 23 {(- 0.25)}^{2} - 14 \cdot - 0.25 - 5$

Raise $- 0.25$ to the power of $3$ .

$- 4 \cdot - 0.015625 + 23 {(- 0.25)}^{2} - 14 \cdot - 0.25 - 5$

Multiply $- 4$ by $- 0.015625$ .

$0.0625 + 23 {(- 0.25)}^{2} - 14 \cdot - 0.25 - 5$

Raise $- 0.25$ to the power of $2$ .

$0.0625 + 23 \cdot 0.0625 - 14 \cdot - 0.25 - 5$

Multiply $23$ by $0.0625$ .

$0.0625 + 1.4375 - 14 \cdot - 0.25 - 5$

Add $0.0625$ and $1.4375$ .

$1.5 - 14 \cdot - 0.25 - 5$

Multiply $- 14$ by $- 0.25$ .

$1.5 + 3.5 - 5$

Add $1.5$ and $3.5$ .

$5 - 5$

Subtract $5$ from $5$ .

$0$

Since $- 0.25$ is a known root, divide the polynomial by $4 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- 4 x^{3} + 23 x^{2} - 14 x - 5}{4 x + 1}$

Divide $- 4 x^{3} + 23 x^{2} - 14 x - 5$ by $4 x + 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$4 x$

$1$

$4 x^{3}$

$23 x^{2}$

$14 x$

$5$

Divide the highest order term in the dividend $- 4 x^{3}$ by the highest order term in divisor $4 x$ .

				-	$x^{2}$
	$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			-	$4 x^{3}$	-	$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{3} - x^{2}$

			-	$x^{2}$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$

Divide the highest order term in the dividend $24 x^{2}$ by the highest order term in divisor $4 x$ .

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					+	$24 x^{2}$	+	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $24 x^{2} + 6 x$

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$

Divide the highest order term in the dividend $- 20 x$ by the highest order term in divisor $4 x$ .

			-	$x^{2}$	+	$6 x$	-	$5$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$6 x$	-	$5$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$
							-	$20 x$	-	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $- 20 x - 5$

			-	$x^{2}$	+	$6 x$	-	$5$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$
							+	$20 x$	+	$5$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	+	$6 x$	-	$5$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$
							+	$20 x$	+	$5$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$- x^{2} + 6 x - 5$

Write $- 4 x^{3} + 23 x^{2} - 14 x - 5$ as a set of factors.

$(4 x + 1) (- x^{2} + 6 x - 5) = 0$

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 $a \cdot c = - 1 \cdot - 5 = 5$ and whose sum is $b = 6$ .

Factor $6$ out of $6 x$ .

$(4 x + 1) (- x^{2} + 6 (x) - 5) = 0$

Rewrite $6$ as $1$ plus $5$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

Step 9

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

$4 x + 1 = 0$

$- x + 1 = 0$

$x - 5 = 0$

Step 10

Set $4 x + 1$ equal to $0$ and solve for $x$ .

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Solve $4 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Divide each term in $4 x = - 1$ by $4$ and simplify.

Divide each term in $4 x = - 1$ by $4$ .

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 11

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

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

$- x + 1 = 0$

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

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Step 12

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 13

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

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

Step 14

Find the Roots/Zeros Using the Rational Roots Test f(x)=-4x^3+23x^2-14x-5

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