Find the Roots/Zeros Using the Rational Roots Test x^4+13x^3+53x^2+83x+42

$x^{4} + 13 x^{3} + 53 x^{2} + 83 x + 42$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42$

$q = \pm 1$

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

$\pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42$

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.

${(- 1)}^{4} + 13 {(- 1)}^{3} + 53 {(- 1)}^{2} + 83 (- 1) + 42$

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 $4$ .

$1 + 13 {(- 1)}^{3} + 53 {(- 1)}^{2} + 83 (- 1) + 42$

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

$1 + 13 \cdot - 1 + 53 {(- 1)}^{2} + 83 (- 1) + 42$

Multiply $13$ by $- 1$ .

$1 - 13 + 53 {(- 1)}^{2} + 83 (- 1) + 42$

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

$1 - 13 + 53 \cdot 1 + 83 (- 1) + 42$

Multiply $53$ by $1$ .

$1 - 13 + 53 + 83 (- 1) + 42$

Multiply $83$ by $- 1$ .

$1 - 13 + 53 - 83 + 42$

Simplify by adding and subtracting.

Subtract $13$ from $1$ .

$- 12 + 53 - 83 + 42$

Add $- 12$ and $53$ .

$41 - 83 + 42$

Subtract $83$ from $41$ .

$- 42 + 42$

Add $- 42$ and $42$ .

$0$

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.

$\frac{x^{4} + 13 x^{3} + 53 x^{2} + 83 x + 42}{x + 1}$

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.

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0				\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0				\xa0
	$1$

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

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0	$- 1$			\xa0
	$1$

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

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0	$- 1$			\xa0
	$1$	$12$

Multiply the newest entry in the result $(12)$ by the divisor $(- 1)$ and place the result of $(- 12)$ under the next term in the dividend $(53)$ .

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0	$- 1$	$- 12$		\xa0
	$1$	$12$

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

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0	$- 1$	$- 12$		\xa0
	$1$	$12$	$41$

Multiply the newest entry in the result $(41)$ by the divisor $(- 1)$ and place the result of $(- 41)$ under the next term in the dividend $(83)$ .

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0	$- 1$	$- 12$	$- 41$	\xa0
	$1$	$12$	$41$

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

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0	$- 1$	$- 12$	$- 41$	\xa0
	$1$	$12$	$41$	$42$

Multiply the newest entry in the result $(42)$ by the divisor $(- 1)$ and place the result of $(- 42)$ under the next term in the dividend $(42)$ .

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0	$- 1$	$- 12$	$- 41$	$- 42$
	$1$	$12$	$41$	$42$

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

$- 1$	$1$	$13$	$53$	$83$	$42$
	\xa0	$- 1$	$- 12$	$- 41$	$- 42$
	$1$	$12$	$41$	$42$	$0$

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

$1 x^{3} + 12 x^{2} + (41) x + 42$

Simplify the quotient polynomial.

$x^{3} + 12 x^{2} + 41 x + 42$

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $x^{3} + 12 x^{2} + 41 x + 42$ 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.

$p = \pm 1, \pm 42, \pm 2, \pm 21, \pm 3, \pm 14, \pm 6, \pm 7 q = \pm 1$

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

$\pm 1, \pm 42, \pm 2, \pm 21, \pm 3, \pm 14, \pm 6, \pm 7$

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

Substitute $- 2$ into the polynomial.

${(- 2)}^{3} + 12 {(- 2)}^{2} + 41 \cdot - 2 + 42$

Raise $- 2$ to the power of $3$ .

$- 8 + 12 {(- 2)}^{2} + 41 \cdot - 2 + 42$

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

$- 8 + 12 \cdot 4 + 41 \cdot - 2 + 42$

Multiply $12$ by $4$ .

$- 8 + 48 + 41 \cdot - 2 + 42$

Add $- 8$ and $48$ .

$40 + 41 \cdot - 2 + 42$

Multiply $41$ by $- 2$ .

$40 - 82 + 42$

Subtract $82$ from $40$ .

$- 42 + 42$

Add $- 42$ and $42$ .

$0$

Since $- 2$ is a known root, divide the polynomial by $x + 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} + 12 x^{2} + 41 x + 42}{x + 2}$

Divide $x^{3} + 12 x^{2} + 41 x + 42$ by $x + 2$ .

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

$x$

$2$

$x^{3}$

$12 x^{2}$

$41 x$

$42$

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

					$x^{2}$
	$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			+	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$

Divide the highest order term in the dividend $10 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$10 x$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$10 x$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$
					+	$10 x^{2}$	+	$20 x$

The expression needs to be subtracted from the dividend, so change all the signs in $10 x^{2} + 20 x$

				$x^{2}$	+	$10 x$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$
					-	$10 x^{2}$	-	$20 x$

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

				$x^{2}$	+	$10 x$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$
					-	$10 x^{2}$	-	$20 x$
							+	$21 x$

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

				$x^{2}$	+	$10 x$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$
					-	$10 x^{2}$	-	$20 x$
							+	$21 x$	+	$42$

Divide the highest order term in the dividend $21 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$10 x$	+	$21$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$
					-	$10 x^{2}$	-	$20 x$
							+	$21 x$	+	$42$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$10 x$	+	$21$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$
					-	$10 x^{2}$	-	$20 x$
							+	$21 x$	+	$42$
							+	$21 x$	+	$42$

The expression needs to be subtracted from the dividend, so change all the signs in $21 x + 42$

				$x^{2}$	+	$10 x$	+	$21$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$
					-	$10 x^{2}$	-	$20 x$
							+	$21 x$	+	$42$
							-	$21 x$	-	$42$

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

				$x^{2}$	+	$10 x$	+	$21$
$x$	+	$2$		$x^{3}$	+	$12 x^{2}$	+	$41 x$	+	$42$
			-	$x^{3}$	-	$2 x^{2}$
					+	$10 x^{2}$	+	$41 x$
					-	$10 x^{2}$	-	$20 x$
							+	$21 x$	+	$42$
							-	$21 x$	-	$42$
										$0$

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

$x^{2} + 10 x + 21$

Write $x^{3} + 12 x^{2} + 41 x + 42$ as a set of factors.

$(x + 2) (x^{2} + 10 x + 21) = 0$

Factor $x^{2} + 10 x + 21$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $21$ and whose sum is $10$ .

$3, 7$

Write the factored form using these integers.

$(x + 2) ((x + 3) (x + 7)) = 0$

Remove unnecessary parentheses.

$(x + 2) (x + 3) (x + 7) = 0$

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

$x + 2 = 0$

$x + 3 = 0$

$x + 7 = 0$

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.

$x = - 2$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x + 2) (x + 3) (x + 7) = 0$ true.

$x = - 2, - 3, - 7$

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

$(x + 1) (x + 2) (x + 3) (x + 7)$

These are the roots (zeros) of the polynomial $x^{4} + 13 x^{3} + 53 x^{2} + 83 x + 42$ .

$x = - 1, - 2, - 3, - 7$

Find the Roots/Zeros Using the Rational Roots Test x^4+13x^3+53x^2+83x+42

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