Find the Roots/Zeros Using the Rational Roots Test 4x^3-3x^2-22x-15

$4 x^{3} - 3 x^{2} - 22 x - 15$

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.

$p = \pm 1, \pm 3, \pm 5, \pm 15$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm 15, \pm \frac{15}{2}, \pm \frac{15}{4}$

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.

$4 {(- 1)}^{3} - 3 {(- 1)}^{2} - 22 \cdot - 1 - 15$

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.

Raise $- 1$ to the power of $3$ .

$4 \cdot - 1 - 3 {(- 1)}^{2} - 22 \cdot - 1 - 15$

Multiply $4$ by $- 1$ .

$- 4 - 3 {(- 1)}^{2} - 22 \cdot - 1 - 15$

Raise $- 1$ to the power of $2$ .

$- 4 - 3 \cdot 1 - 22 \cdot - 1 - 15$

Multiply $- 3$ by $1$ .

$- 4 - 3 - 22 \cdot - 1 - 15$

Multiply $- 22$ by $- 1$ .

$- 4 - 3 + 22 - 15$

Simplify by adding and subtracting.

Subtract $3$ from $- 4$ .

$- 7 + 22 - 15$

Add $- 7$ and $22$ .

$15 - 15$

Subtract $15$ from $15$ .

$0$

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.

$\frac{4 x^{3} - 3 x^{2} - 22 x - 15}{x + 1}$

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.

$- 1$	$4$	$- 3$	$- 22$	$- 15$
	\xa0			\xa0

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

$- 1$	$4$	$- 3$	$- 22$	$- 15$
	\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 $(- 3)$ .

$- 1$	$4$	$- 3$	$- 22$	$- 15$
	\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.

$- 1$	$4$	$- 3$	$- 22$	$- 15$
	\xa0	$- 4$		\xa0
	$4$	$- 7$

Multiply the newest entry in the result $(- 7)$ by the divisor $(- 1)$ and place the result of $(7)$ under the next term in the dividend $(- 22)$ .

$- 1$	$4$	$- 3$	$- 22$	$- 15$
	\xa0	$- 4$	$7$	\xa0
	$4$	$- 7$

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

$- 1$	$4$	$- 3$	$- 22$	$- 15$
	\xa0	$- 4$	$7$	\xa0
	$4$	$- 7$	$- 15$

Multiply the newest entry in the result $(- 15)$ by the divisor $(- 1)$ and place the result of $(15)$ under the next term in the dividend $(- 15)$ .

$- 1$	$4$	$- 3$	$- 22$	$- 15$
	\xa0	$- 4$	$7$	$15$
	$4$	$- 7$	$- 15$

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

$- 1$	$4$	$- 3$	$- 22$	$- 15$
	\xa0	$- 4$	$7$	$15$
	$4$	$- 7$	$- 15$	$0$

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

$4 x^{2} + - 7 x - 15$

Simplify the quotient polynomial.

$4 x^{2} - 7 x - 15$

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 $a \cdot c = 4 \cdot - 15 = - 60$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$(x + 1) + 4 x^{2} - 7 x - 15$

Rewrite $- 7$ as $5$ plus $- 12$

$(x + 1) + 4 x^{2} + (5 - 12) x - 15$

Apply the distributive property.

$(x + 1) + 4 x^{2} + 5 x - 12 x - 15$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) (4 x^{2} + 5 x) - 12 x - 15$

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

$(x + 1) + x (4 x + 5) - 3 (4 x + 5)$

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

$(x + 1) + (4 x + 5) (x - 3)$

$(x + 1) (4 x + 5) (x - 3)$

Step 8

Factor the left side of the equation.

Factor $4 x^{3} - 3 x^{2} - 22 x - 15$ 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.

$p = \pm 1, \pm 15, \pm 3, \pm 5 q = \pm 1, \pm 4, \pm 2$

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 15, \pm 3.75, \pm 7.5, \pm 3, \pm 0.75, \pm 1.5, \pm 5, \pm 1.25, \pm 2.5$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

$4 {(- 1)}^{3} - 3 {(- 1)}^{2} - 22 \cdot - 1 - 15$

Raise $- 1$ to the power of $3$ .

$4 \cdot - 1 - 3 {(- 1)}^{2} - 22 \cdot - 1 - 15$

Multiply $4$ by $- 1$ .

$- 4 - 3 {(- 1)}^{2} - 22 \cdot - 1 - 15$

Raise $- 1$ to the power of $2$ .

$- 4 - 3 \cdot 1 - 22 \cdot - 1 - 15$

Multiply $- 3$ by $1$ .

$- 4 - 3 - 22 \cdot - 1 - 15$

Subtract $3$ from $- 4$ .

$- 7 - 22 \cdot - 1 - 15$

Multiply $- 22$ by $- 1$ .

$- 7 + 22 - 15$

Add $- 7$ and $22$ .

$15 - 15$

Subtract $15$ from $15$ .

$0$

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.

$\frac{4 x^{3} - 3 x^{2} - 22 x - 15}{x + 1}$

Divide $4 x^{3} - 3 x^{2} - 22 x - 15$ by $x + 1$ .

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

$x$

$1$

$4 x^{3}$

$3 x^{2}$

$22 x$

$15$

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

					$4 x^{2}$
	$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			+	$4 x^{3}$	+	$4 x^{2}$

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

				$4 x^{2}$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$

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

				$4 x^{2}$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$

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

				$4 x^{2}$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$

Divide the highest order term in the dividend $- 7 x^{2}$ by the highest order term in divisor $x$ .

				$4 x^{2}$	-	$7 x$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$7 x$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$
					-	$7 x^{2}$	-	$7 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 x^{2} - 7 x$

				$4 x^{2}$	-	$7 x$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$
					+	$7 x^{2}$	+	$7 x$

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

				$4 x^{2}$	-	$7 x$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$
					+	$7 x^{2}$	+	$7 x$
							-	$15 x$

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

				$4 x^{2}$	-	$7 x$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$
					+	$7 x^{2}$	+	$7 x$
							-	$15 x$	-	$15$

Divide the highest order term in the dividend $- 15 x$ by the highest order term in divisor $x$ .

				$4 x^{2}$	-	$7 x$	-	$15$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$
					+	$7 x^{2}$	+	$7 x$
							-	$15 x$	-	$15$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$7 x$	-	$15$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$
					+	$7 x^{2}$	+	$7 x$
							-	$15 x$	-	$15$
							-	$15 x$	-	$15$

The expression needs to be subtracted from the dividend, so change all the signs in $- 15 x - 15$

				$4 x^{2}$	-	$7 x$	-	$15$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$
					+	$7 x^{2}$	+	$7 x$
							-	$15 x$	-	$15$
							+	$15 x$	+	$15$

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

				$4 x^{2}$	-	$7 x$	-	$15$
$x$	+	$1$		$4 x^{3}$	-	$3 x^{2}$	-	$22 x$	-	$15$
			-	$4 x^{3}$	-	$4 x^{2}$
					-	$7 x^{2}$	-	$22 x$
					+	$7 x^{2}$	+	$7 x$
							-	$15 x$	-	$15$
							+	$15 x$	+	$15$
										$0$

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

$4 x^{2} - 7 x - 15$

Write $4 x^{3} - 3 x^{2} - 22 x - 15$ as a set of factors.

$(x + 1) (4 x^{2} - 7 x - 15) = 0$

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 $a \cdot c = 4 \cdot - 15 = - 60$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$(x + 1) (4 x^{2} - 7 x - 15) = 0$

Rewrite $- 7$ as $5$ plus $- 12$

$(x + 1) (4 x^{2} + (5 - 12) x - 15) = 0$

Apply the distributive property.

$(x + 1) (4 x^{2} + 5 x - 12 x - 15) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) ((4 x^{2} + 5 x) - 12 x - 15) = 0$

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

$(x + 1) (x (4 x + 5) - 3 (4 x + 5)) = 0$

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

$(x + 1) ((4 x + 5) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x + 1) (4 x + 5) (x - 3) = 0$

Step 9

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

$x + 1 = 0$

$4 x + 5 = 0$

$x - 3 = 0$

Step 10

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 11

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 12

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 13

The final solution is all the values that make $(x + 1) (4 x + 5) (x - 3) = 0$ true.

$x = - 1, - \frac{5}{4}, 3$

Step 14

Find the Roots/Zeros Using the Rational Roots Test 4x^3-3x^2-22x-15

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