Find the Roots/Zeros Using the Rational Roots Test f(x)=3x^4+13x^3+x^2-13x-4

$f (x) = 3 x^{4} + 13 x^{3} + x^{2} - 13 x - 4$

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.

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

$q = \pm 1, \pm 3$

Step 2

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

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

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.

$3 {(1)}^{4} + 13 {(1)}^{3} + {(1)}^{2} - 13 \cdot 1 - 4$

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.

$3 \cdot 1 + 13 {(1)}^{3} + {(1)}^{2} - 13 \cdot 1 - 4$

Multiply $3$ by $1$ .

$3 + 13 {(1)}^{3} + {(1)}^{2} - 13 \cdot 1 - 4$

One to any power is one.

$3 + 13 \cdot 1 + {(1)}^{2} - 13 \cdot 1 - 4$

Multiply $13$ by $1$ .

$3 + 13 + {(1)}^{2} - 13 \cdot 1 - 4$

One to any power is one.

$3 + 13 + 1 - 13 \cdot 1 - 4$

Multiply $- 13$ by $1$ .

$3 + 13 + 1 - 13 - 4$

Simplify by adding and subtracting.

Add $3$ and $13$ .

$16 + 1 - 13 - 4$

Add $16$ and $1$ .

$17 - 13 - 4$

Subtract $13$ from $17$ .

$4 - 4$

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.

$\frac{3 x^{4} + 13 x^{3} + x^{2} - 13 x - 4}{x - 1}$

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.

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0				\xa0

The first number in the dividend $(3)$ is put into the first position of the result area (below the horizontal line).

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0				\xa0
	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(1)$ and place the result of $(3)$ under the next term in the dividend $(13)$ .

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0	$3$			\xa0
	$3$

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

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0	$3$			\xa0
	$3$	$16$

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 $(1)$ .

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0	$3$	$16$		\xa0
	$3$	$16$

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

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0	$3$	$16$		\xa0
	$3$	$16$	$17$

Multiply the newest entry in the result $(17)$ by the divisor $(1)$ and place the result of $(17)$ under the next term in the dividend $(- 13)$ .

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0	$3$	$16$	$17$	\xa0
	$3$	$16$	$17$

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

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0	$3$	$16$	$17$	\xa0
	$3$	$16$	$17$	$4$

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 $(- 4)$ .

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0	$3$	$16$	$17$	$4$
	$3$	$16$	$17$	$4$

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

$1$	$3$	$13$	$1$	$- 13$	$- 4$
	\xa0	$3$	$16$	$17$	$4$
	$3$	$16$	$17$	$4$	$0$

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

$3 x^{3} + 16 x^{2} + (17) x + 4$

Simplify the quotient polynomial.

$3 x^{3} + 16 x^{2} + 17 x + 4$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $3 x^{3} + 16 x^{2} + 17 x + 4$ 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.

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

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

$\pm 1, \pm 0.\overline{3}, \pm 4, \pm 1.\overline{3}, \pm 2, \pm 0.\overline{6}$

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.

$3 {(- 1)}^{3} + 16 {(- 1)}^{2} + 17 \cdot - 1 + 4$

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

$3 \cdot - 1 + 16 {(- 1)}^{2} + 17 \cdot - 1 + 4$

Multiply $3$ by $- 1$ .

$- 3 + 16 {(- 1)}^{2} + 17 \cdot - 1 + 4$

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

$- 3 + 16 \cdot 1 + 17 \cdot - 1 + 4$

Multiply $16$ by $1$ .

$- 3 + 16 + 17 \cdot - 1 + 4$

Add $- 3$ and $16$ .

$13 + 17 \cdot - 1 + 4$

Multiply $17$ by $- 1$ .

$13 - 17 + 4$

Subtract $17$ from $13$ .

$- 4 + 4$

Add $- 4$ and $4$ .

$0$

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.

$\frac{3 x^{3} + 16 x^{2} + 17 x + 4}{x + 1}$

Divide $3 x^{3} + 16 x^{2} + 17 x + 4$ 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 $0$ .

$x$

$1$

$3 x^{3}$

$16 x^{2}$

$17 x$

$4$

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

					$3 x^{2}$
	$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			+	$3 x^{3}$	+	$3 x^{2}$

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

				$3 x^{2}$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$

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

				$3 x^{2}$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$

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

				$3 x^{2}$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$

Divide the highest order term in the dividend $13 x^{2}$ by the highest order term in divisor $x$ .

				$3 x^{2}$	+	$13 x$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$13 x$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$
					+	$13 x^{2}$	+	$13 x$

The expression needs to be subtracted from the dividend, so change all the signs in $13 x^{2} + 13 x$

				$3 x^{2}$	+	$13 x$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$
					-	$13 x^{2}$	-	$13 x$

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

				$3 x^{2}$	+	$13 x$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$
					-	$13 x^{2}$	-	$13 x$
							+	$4 x$

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

				$3 x^{2}$	+	$13 x$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$
					-	$13 x^{2}$	-	$13 x$
							+	$4 x$	+	$4$

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

				$3 x^{2}$	+	$13 x$	+	$4$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$
					-	$13 x^{2}$	-	$13 x$
							+	$4 x$	+	$4$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$13 x$	+	$4$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$
					-	$13 x^{2}$	-	$13 x$
							+	$4 x$	+	$4$
							+	$4 x$	+	$4$

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

				$3 x^{2}$	+	$13 x$	+	$4$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$
					-	$13 x^{2}$	-	$13 x$
							+	$4 x$	+	$4$
							-	$4 x$	-	$4$

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

				$3 x^{2}$	+	$13 x$	+	$4$
$x$	+	$1$		$3 x^{3}$	+	$16 x^{2}$	+	$17 x$	+	$4$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$13 x^{2}$	+	$17 x$
					-	$13 x^{2}$	-	$13 x$
							+	$4 x$	+	$4$
							-	$4 x$	-	$4$
										$0$

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

$3 x^{2} + 13 x + 4$

Write $3 x^{3} + 16 x^{2} + 17 x + 4$ as a set of factors.

$(x + 1) (3 x^{2} + 13 x + 4) = 0$

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 $a \cdot c = 3 \cdot 4 = 12$ and whose sum is $b = 13$ .

Factor $13$ out of $13 x$ .

$(x + 1) (3 x^{2} + 13 (x) + 4) = 0$

Rewrite $13$ as $1$ plus $12$

$(x + 1) (3 x^{2} + (1 + 12) x + 4) = 0$

Apply the distributive property.

$(x + 1) (3 x^{2} + 1 x + 12 x + 4) = 0$

Multiply $x$ by $1$ .

$(x + 1) (3 x^{2} + x + 12 x + 4) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) ((3 x^{2} + x) + 12 x + 4) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

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

Remove unnecessary parentheses.

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

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

$x + 1 = 0$

$3 x + 1 = 0$

$x + 4 = 0$

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.

$x = - 1$

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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

Step 8

The polynomial can be written as a set of linear factors.

$(x - 1) (x + 1) (x + \frac{1}{3}) (x + 4)$

Step 9

These are the roots (zeros) of the polynomial $3 x^{4} + 13 x^{3} + x^{2} - 13 x - 4$ .

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

Step 10

Find the Roots/Zeros Using the Rational Roots Test f(x)=3x^4+13x^3+x^2-13x-4

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