Find All Complex Solutions 2x^4-3x^3-24x^2+13x+12=0

$2 x^{4} - 3 x^{3} - 24 x^{2} + 13 x + 12 = 0$

Step 1

Factor the left side of the equation.

Factor $2 x^{4} - 3 x^{3} - 24 x^{2} + 13 x + 12$ 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.

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

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

$\pm 1, \pm 0.5, \pm 12, \pm 6, \pm 2, \pm 3, \pm 1.5, \pm 4$

Substitute $- 0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.5$ is a root of the polynomial.

Substitute $- 0.5$ into the polynomial.

$2 {(- 0.5)}^{4} - 3 {(- 0.5)}^{3} - 24 {(- 0.5)}^{2} + 13 \cdot - 0.5 + 12$

Raise $- 0.5$ to the power of $4$ .

$2 \cdot 0.0625 - 3 {(- 0.5)}^{3} - 24 {(- 0.5)}^{2} + 13 \cdot - 0.5 + 12$

Multiply $2$ by $0.0625$ .

$0.125 - 3 {(- 0.5)}^{3} - 24 {(- 0.5)}^{2} + 13 \cdot - 0.5 + 12$

Raise $- 0.5$ to the power of $3$ .

$0.125 - 3 \cdot - 0.125 - 24 {(- 0.5)}^{2} + 13 \cdot - 0.5 + 12$

Multiply $- 3$ by $- 0.125$ .

$0.125 + 0.375 - 24 {(- 0.5)}^{2} + 13 \cdot - 0.5 + 12$

Add $0.125$ and $0.375$ .

$0.5 - 24 {(- 0.5)}^{2} + 13 \cdot - 0.5 + 12$

Raise $- 0.5$ to the power of $2$ .

$0.5 - 24 \cdot 0.25 + 13 \cdot - 0.5 + 12$

Multiply $- 24$ by $0.25$ .

$0.5 - 6 + 13 \cdot - 0.5 + 12$

Subtract $6$ from $0.5$ .

$- 5.5 + 13 \cdot - 0.5 + 12$

Multiply $13$ by $- 0.5$ .

$- 5.5 - 6.5 + 12$

Subtract $6.5$ from $- 5.5$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

Since $- 0.5$ is a known root, divide the polynomial by $2 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{4} - 3 x^{3} - 24 x^{2} + 13 x + 12}{2 x + 1}$

Divide $2 x^{4} - 3 x^{3} - 24 x^{2} + 13 x + 12$ by $2 x + 1$ .

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

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

Divide the highest order term in the dividend $2 x^{4}$ by the highest order term in divisor $2 x$ .

$x^{3}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

Multiply the new quotient term by the divisor.

$x^{3}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

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

$x^{3}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

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

$x^{3}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

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

$x^{3}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

Divide the highest order term in the dividend $- 4 x^{3}$ by the highest order term in divisor $2 x$ .

$x^{3}$

$2 x^{2}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

Multiply the new quotient term by the divisor.

				$x^{3}$	-	$2 x^{2}$
$2 x$	+	$1$		$2 x^{4}$	-	$3 x^{3}$	-	$24 x^{2}$	+	$13 x$	+	$12$
			-	$2 x^{4}$	-	$x^{3}$
					-	$4 x^{3}$	-	$24 x^{2}$
					-	$4 x^{3}$	-	$2 x^{2}$

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

$x^{3}$

$2 x^{2}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

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

$x^{3}$

$2 x^{2}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

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

$x^{3}$

$2 x^{2}$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

Divide the highest order term in the dividend $- 22 x^{2}$ by the highest order term in divisor $2 x$ .

$x^{3}$

$2 x^{2}$

$11 x$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

Multiply the new quotient term by the divisor.

$x^{3}$

$2 x^{2}$

$11 x$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

$22 x^{2}$

$11 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 22 x^{2} - 11 x$

$x^{3}$

$2 x^{2}$

$11 x$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

$22 x^{2}$

$11 x$

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

$x^{3}$

$2 x^{2}$

$11 x$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

$22 x^{2}$

$11 x$

$24 x$

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

$x^{3}$

$2 x^{2}$

$11 x$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

$22 x^{2}$

$11 x$

$24 x$

$12$

Divide the highest order term in the dividend $24 x$ by the highest order term in divisor $2 x$ .

$x^{3}$

$2 x^{2}$

$11 x$

$12$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

$22 x^{2}$

$11 x$

$24 x$

$12$

Multiply the new quotient term by the divisor.

$x^{3}$

$2 x^{2}$

$11 x$

$12$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

$22 x^{2}$

$11 x$

$24 x$

$12$

$24 x$

$12$

The expression needs to be subtracted from the dividend, so change all the signs in $24 x + 12$

$x^{3}$

$2 x^{2}$

$11 x$

$12$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

$22 x^{2}$

$11 x$

$24 x$

$12$

$24 x$

$12$

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

$x^{3}$

$2 x^{2}$

$11 x$

$12$

$2 x$

$1$

$2 x^{4}$

$3 x^{3}$

$24 x^{2}$

$13 x$

$12$

$2 x^{4}$

$x^{3}$

$4 x^{3}$

$24 x^{2}$

$4 x^{3}$

$2 x^{2}$

$22 x^{2}$

$13 x$

$22 x^{2}$

$11 x$

$24 x$

$12$

$24 x$

$12$

$0$

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

$x^{3} - 2 x^{2} - 11 x + 12$

Write $2 x^{4} - 3 x^{3} - 24 x^{2} + 13 x + 12$ as a set of factors.

$(2 x + 1) (x^{3} - 2 x^{2} - 11 x + 12) = 0$

Factor $x^{3} - 2 x^{2} - 11 x + 12$ 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.

$1^{3} - 2 \cdot 1^{2} - 11 \cdot 1 + 12$

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

$1 - 2 \cdot 1^{2} - 11 \cdot 1 + 12$

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

$1 - 2 \cdot 1 - 11 \cdot 1 + 12$

Multiply $- 2$ by $1$ .

$1 - 2 - 11 \cdot 1 + 12$

Subtract $2$ from $1$ .

$- 1 - 11 \cdot 1 + 12$

Multiply $- 11$ by $1$ .

$- 1 - 11 + 12$

Subtract $11$ 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.

$\frac{x^{3} - 2 x^{2} - 11 x + 12}{x - 1}$

Divide $x^{3} - 2 x^{2} - 11 x + 12$ 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 $0$ .

$x$

$1$

$x^{3}$

$2 x^{2}$

$11 x$

$12$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$

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

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$
					-	$x^{2}$	+	$x$

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

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$
					+	$x^{2}$	-	$x$

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

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$
					+	$x^{2}$	-	$x$
							-	$12 x$

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

				$x^{2}$	-	$x$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$
					+	$x^{2}$	-	$x$
							-	$12 x$	+	$12$

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

				$x^{2}$	-	$x$	-	$12$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$
					+	$x^{2}$	-	$x$
							-	$12 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$12$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$
					+	$x^{2}$	-	$x$
							-	$12 x$	+	$12$
							-	$12 x$	+	$12$

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

				$x^{2}$	-	$x$	-	$12$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$
					+	$x^{2}$	-	$x$
							-	$12 x$	+	$12$
							+	$12 x$	-	$12$

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

				$x^{2}$	-	$x$	-	$12$
$x$	-	$1$		$x^{3}$	-	$2 x^{2}$	-	$11 x$	+	$12$
			-	$x^{3}$	+	$x^{2}$
					-	$x^{2}$	-	$11 x$
					+	$x^{2}$	-	$x$
							-	$12 x$	+	$12$
							+	$12 x$	-	$12$
										$0$

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

$x^{2} - x - 12$

Write $x^{3} - 2 x^{2} - 11 x + 12$ as a set of factors.

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

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 $- 1$ .

$- 4, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Step 2

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

$2 x + 1 = 0$

$x - 1 = 0$

$x - 4 = 0$

$x + 3 = 0$

Step 3

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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 $x - 4$ equal to $0$ and solve for $x$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 6

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.

$x = - 3$

Step 7

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

$x = - \frac{1}{2}, 1, 4, - 3$

Find All Complex Solutions 2x^4-3x^3-24x^2+13x+12=0

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