Find the Roots (Zeros) x^4+2x^3-13x^2-14x+24=0

$x^{4} + 2 x^{3} - 13 x^{2} - 14 x + 24 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$x^{4} + 2 x^{3} - 13 x^{2} - 14 x + 24 = 0$

Factor $x^{3}$ out of $x^{4} + 2 x^{3}$ .

Factor $x^{3}$ out of $x^{4}$ .

$x^{3} x + 2 x^{3} - 13 x^{2} - 14 x + 24 = 0$

Factor $x^{3}$ out of $2 x^{3}$ .

$x^{3} x + x^{3} \cdot 2 - 13 x^{2} - 14 x + 24 = 0$

Factor $x^{3}$ out of $x^{3} x + x^{3} \cdot 2$ .

$x^{3} (x + 2) - 13 x^{2} - 14 x + 24 = 0$

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 $a \cdot c = - 13 \cdot 24 = - 312$ and whose sum is $b = - 14$ .

Factor $- 14$ out of $- 14 x$ .

$x^{3} (x + 2) - 13 x^{2} - 14 x + 24 = 0$

Rewrite $- 14$ as $12$ plus $- 26$

$x^{3} (x + 2) - 13 x^{2} + (12 - 26) x + 24 = 0$

Apply the distributive property.

$x^{3} (x + 2) - 13 x^{2} + 12 x - 26 x + 24 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x^{3} (x + 2) - 13 x^{2} + 12 x - 26 x + 24 = 0$

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

$x^{3} (x + 2) + x (- 13 x + 12) + 2 (- 13 x + 12) = 0$

Factor the polynomial by factoring out the greatest common factor, $- 13 x + 12$ .

$x^{3} (x + 2) + (- 13 x + 12) (x + 2) = 0$

Factor $x + 2$ out of $x^{3} (x + 2) + (- 13 x + 12) (x + 2)$ .

Factor $x + 2$ out of $x^{3} (x + 2)$ .

$(x + 2) x^{3} + (- 13 x + 12) (x + 2) = 0$

Factor $x + 2$ out of $(- 13 x + 12) (x + 2)$ .

$(x + 2) x^{3} + (x + 2) (- 13 x + 12) = 0$

Factor $x + 2$ out of $(x + 2) x^{3} + (x + 2) (- 13 x + 12)$ .

$(x + 2) (x^{3} - 13 x + 12) = 0$

Factor.

Rewrite $x^{3} - 13 x + 12$ in a factored form.

Factor $x^{3} - 13 x + 12$ 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.

$p = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4 q = \pm 1$

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

$\pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

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.

$1^{3} - 13 \cdot 1 + 12$

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

$1 - 13 \cdot 1 + 12$

Multiply $- 13$ by $1$ .

$1 - 13 + 12$

Subtract $13$ from $1$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

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.

$\frac{x^{3} - 13 x + 12}{x - 1}$

Divide $x^{3} - 13 x + 12$ 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 $0$ .

$x$

$1$

$x^{3}$

$0 x^{2}$

$13 x$

$12$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$
					+	$x^{2}$	-	$x$

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$
					-	$x^{2}$	+	$x$

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$
					-	$x^{2}$	+	$x$
							-	$12 x$

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$
					-	$x^{2}$	+	$x$
							-	$12 x$	+	$12$

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$x$	-	$12$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$
					-	$x^{2}$	+	$x$
							-	$12 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	-	$12$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$
					-	$x^{2}$	+	$x$
							-	$12 x$	+	$12$
							-	$12 x$	+	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 x + 12$

				$x^{2}$	+	$x$	-	$12$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$
					-	$x^{2}$	+	$x$
							-	$12 x$	+	$12$
							+	$12 x$	-	$12$

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

				$x^{2}$	+	$x$	-	$12$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$13 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$13 x$
					-	$x^{2}$	+	$x$
							-	$12 x$	+	$12$
							+	$12 x$	-	$12$
										$0$

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

$x^{2} + x - 12$

Write $x^{3} - 13 x + 12$ as a set of factors.

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

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$(x + 2) ((x - 1) ((x - 3) (x + 4))) = 0$

Remove unnecessary parentheses.

$(x + 2) ((x - 1) (x - 3) (x + 4)) = 0$

Remove unnecessary parentheses.

$(x + 2) (x - 1) (x - 3) (x + 4) = 0$

Step 2

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

$x + 2 = 0$

$x - 1 = 0$

$x - 3 = 0$

$x + 4 = 0$

Step 3

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 4

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 5

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 6

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.

$x = - 4$

Step 7

The final solution is all the values that make $(x + 2) (x - 1) (x - 3) (x + 4) = 0$ true.

$x = - 2, 1, 3, - 4$

Step 8

Find the Roots (Zeros) x^4+2x^3-13x^2-14x+24=0

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