Find the Roots/Zeros Using the Rational Roots Test f(x)=7x^3+9x^2-12x-4

$f (x) = 7 x^{3} + 9 x^{2} - 12 x - 4$

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.

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

$q = \pm 1, \pm 7$

Step 2

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

$\pm 1, \pm \frac{1}{7}, \pm 2, \pm \frac{2}{7}, \pm 4, \pm \frac{4}{7}$

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.

$7 {(1)}^{3} + 9 {(1)}^{2} - 12 \cdot 1 - 4$

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.

$7 \cdot 1 + 9 {(1)}^{2} - 12 \cdot 1 - 4$

Multiply $7$ by $1$ .

$7 + 9 {(1)}^{2} - 12 \cdot 1 - 4$

One to any power is one.

$7 + 9 \cdot 1 - 12 \cdot 1 - 4$

Multiply $9$ by $1$ .

$7 + 9 - 12 \cdot 1 - 4$

Multiply $- 12$ by $1$ .

$7 + 9 - 12 - 4$

Simplify by adding and subtracting.

Add $7$ and $9$ .

$16 - 12 - 4$

Subtract $12$ from $16$ .

$4 - 4$

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.

$\frac{7 x^{3} + 9 x^{2} - 12 x - 4}{x - 1}$

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.

$1$	$7$	$9$	$- 12$	$- 4$
	\xa0			\xa0

The first number in the dividend $(7)$ is put into the first position of the result area (below the horizontal line).

$1$	$7$	$9$	$- 12$	$- 4$
	\xa0			\xa0
	$7$

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 $(9)$ .

$1$	$7$	$9$	$- 12$	$- 4$
	\xa0	$7$		\xa0
	$7$

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

$1$	$7$	$9$	$- 12$	$- 4$
	\xa0	$7$		\xa0
	$7$	$16$

Multiply the newest entry in the result $(16)$ by the divisor $(1)$ and place the result of $(16)$ under the next term in the dividend $(- 12)$ .

$1$	$7$	$9$	$- 12$	$- 4$
	\xa0	$7$	$16$	\xa0
	$7$	$16$

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

$1$	$7$	$9$	$- 12$	$- 4$
	\xa0	$7$	$16$	\xa0
	$7$	$16$	$4$

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 $(- 4)$ .

$1$	$7$	$9$	$- 12$	$- 4$
	\xa0	$7$	$16$	$4$
	$7$	$16$	$4$

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

$1$	$7$	$9$	$- 12$	$- 4$
	\xa0	$7$	$16$	$4$
	$7$	$16$	$4$	$0$

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

$7 x^{2} + 16 x + 4$

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 $a \cdot c = 7 \cdot 4 = 28$ and whose sum is $b = 16$ .

Factor $16$ out of $16 x$ .

$(x - 1) + 7 x^{2} + 16 (x) + 4$

Rewrite $16$ as $2$ plus $14$

$(x - 1) + 7 x^{2} + (2 + 14) x + 4$

Apply the distributive property.

$(x - 1) + 7 x^{2} + 2 x + 14 x + 4$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (7 x^{2} + 2 x) + 14 x + 4$

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

$(x - 1) + x (7 x + 2) + 2 (7 x + 2)$

Factor the polynomial by factoring out the greatest common factor, $7 x + 2$ .

$(x - 1) + (7 x + 2) (x + 2)$

$(x - 1) (7 x + 2) (x + 2)$

Step 8

Factor the left side of the equation.

Factor $7 x^{3} + 9 x^{2} - 12 x - 4$ 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.

$p = \pm 1, \pm 4, \pm 2 q = \pm 1, \pm 7$

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

$\pm 1, \pm 0.\overline{142857}, \pm 4, \pm 0.\overline{571428}, \pm 2, \pm 0.\overline{285714}$

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.

$7 \cdot 1^{3} + 9 \cdot 1^{2} - 12 \cdot 1 - 4$

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

$7 \cdot 1 + 9 \cdot 1^{2} - 12 \cdot 1 - 4$

Multiply $7$ by $1$ .

$7 + 9 \cdot 1^{2} - 12 \cdot 1 - 4$

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

$7 + 9 \cdot 1 - 12 \cdot 1 - 4$

Multiply $9$ by $1$ .

$7 + 9 - 12 \cdot 1 - 4$

Add $7$ and $9$ .

$16 - 12 \cdot 1 - 4$

Multiply $- 12$ by $1$ .

$16 - 12 - 4$

Subtract $12$ from $16$ .

$4 - 4$

Subtract $4$ from $4$ .

$0$

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.

$\frac{7 x^{3} + 9 x^{2} - 12 x - 4}{x - 1}$

Divide $7 x^{3} + 9 x^{2} - 12 x - 4$ 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 $0$ .

$x$

$1$

$7 x^{3}$

$9 x^{2}$

$12 x$

$4$

Divide the highest order term in the dividend $7 x^{3}$ by the highest order term in divisor $x$ .

					$7 x^{2}$
	$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$

Multiply the new quotient term by the divisor.

				$7 x^{2}$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			+	$7 x^{3}$	-	$7 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{3} - 7 x^{2}$

				$7 x^{2}$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$

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

				$7 x^{2}$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$

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

				$7 x^{2}$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$

Divide the highest order term in the dividend $16 x^{2}$ by the highest order term in divisor $x$ .

				$7 x^{2}$	+	$16 x$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$

Multiply the new quotient term by the divisor.

				$7 x^{2}$	+	$16 x$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$
					+	$16 x^{2}$	-	$16 x$

The expression needs to be subtracted from the dividend, so change all the signs in $16 x^{2} - 16 x$

				$7 x^{2}$	+	$16 x$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$
					-	$16 x^{2}$	+	$16 x$

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

				$7 x^{2}$	+	$16 x$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$
					-	$16 x^{2}$	+	$16 x$
							+	$4 x$

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

				$7 x^{2}$	+	$16 x$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$
					-	$16 x^{2}$	+	$16 x$
							+	$4 x$	-	$4$

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

				$7 x^{2}$	+	$16 x$	+	$4$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$
					-	$16 x^{2}$	+	$16 x$
							+	$4 x$	-	$4$

Multiply the new quotient term by the divisor.

				$7 x^{2}$	+	$16 x$	+	$4$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$
					-	$16 x^{2}$	+	$16 x$
							+	$4 x$	-	$4$
							+	$4 x$	-	$4$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x - 4$

				$7 x^{2}$	+	$16 x$	+	$4$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$
					-	$16 x^{2}$	+	$16 x$
							+	$4 x$	-	$4$
							-	$4 x$	+	$4$

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

				$7 x^{2}$	+	$16 x$	+	$4$
$x$	-	$1$		$7 x^{3}$	+	$9 x^{2}$	-	$12 x$	-	$4$
			-	$7 x^{3}$	+	$7 x^{2}$
					+	$16 x^{2}$	-	$12 x$
					-	$16 x^{2}$	+	$16 x$
							+	$4 x$	-	$4$
							-	$4 x$	+	$4$
										$0$

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

$7 x^{2} + 16 x + 4$

Write $7 x^{3} + 9 x^{2} - 12 x - 4$ as a set of factors.

$(x - 1) (7 x^{2} + 16 x + 4) = 0$

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 $a \cdot c = 7 \cdot 4 = 28$ and whose sum is $b = 16$ .

Factor $16$ out of $16 x$ .

$(x - 1) (7 x^{2} + 16 (x) + 4) = 0$

Rewrite $16$ as $2$ plus $14$

$(x - 1) (7 x^{2} + (2 + 14) x + 4) = 0$

Apply the distributive property.

$(x - 1) (7 x^{2} + 2 x + 14 x + 4) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) ((7 x^{2} + 2 x) + 14 x + 4) = 0$

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

$(x - 1) (x (7 x + 2) + 2 (7 x + 2)) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 x + 2$ .

$(x - 1) ((7 x + 2) (x + 2)) = 0$

Remove unnecessary parentheses.

$(x - 1) (7 x + 2) (x + 2) = 0$

Step 9

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

$x - 1 = 0$

$7 x + 2 = 0$

$x + 2 = 0$

Step 10

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 11

Set $7 x + 2$ equal to $0$ and solve for $x$ .

Set $7 x + 2$ equal to $0$ .

$7 x + 2 = 0$

Solve $7 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$7 x = - 2$

Divide each term in $7 x = - 2$ by $7$ and simplify.

Divide each term in $7 x = - 2$ by $7$ .

$\frac{7 x}{7} = \frac{- 2}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 2}{7}$

Divide $x$ by $1$ .

$x = \frac{- 2}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{7}$

Step 12

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 13

The final solution is all the values that make $(x - 1) (7 x + 2) (x + 2) = 0$ true.

$x = 1, - \frac{2}{7}, - 2$

Step 14

Find the Roots/Zeros Using the Rational Roots Test f(x)=7x^3+9x^2-12x-4

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