Find the Roots (Zeros) f(x)=2x^3-5x^2-4x+12

$f (x) = 2 x^{3} - 5 x^{2} - 4 x + 12$

Step 1

Set $2 x^{3} - 5 x^{2} - 4 x + 12$ equal to $0$ .

$2 x^{3} - 5 x^{2} - 4 x + 12 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2 x^{3} - 5 x^{2} - 4 x + 12$ 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 $2$ 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.

$2 \cdot 2^{3} - 5 \cdot 2^{2} - 4 \cdot 2 + 12$

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

$2 \cdot 8 - 5 \cdot 2^{2} - 4 \cdot 2 + 12$

Multiply $2$ by $8$ .

$16 - 5 \cdot 2^{2} - 4 \cdot 2 + 12$

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

$16 - 5 \cdot 4 - 4 \cdot 2 + 12$

Multiply $- 5$ by $4$ .

$16 - 20 - 4 \cdot 2 + 12$

Subtract $20$ from $16$ .

$- 4 - 4 \cdot 2 + 12$

Multiply $- 4$ by $2$ .

$- 4 - 8 + 12$

Subtract $8$ from $- 4$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

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.

$\frac{2 x^{3} - 5 x^{2} - 4 x + 12}{x - 2}$

Divide $2 x^{3} - 5 x^{2} - 4 x + 12$ 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 $0$ .

$x$

$2$

$2 x^{3}$

$5 x^{2}$

$4 x$

$12$

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

					$2 x^{2}$
	$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			+	$2 x^{3}$	-	$4 x^{2}$

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

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$

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

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$

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

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$

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

				$2 x^{2}$	-	$x$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$x$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$
					-	$x^{2}$	+	$2 x$

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

				$2 x^{2}$	-	$x$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$
					+	$x^{2}$	-	$2 x$

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

				$2 x^{2}$	-	$x$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$
					+	$x^{2}$	-	$2 x$
							-	$6 x$

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

				$2 x^{2}$	-	$x$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$
					+	$x^{2}$	-	$2 x$
							-	$6 x$	+	$12$

Divide the highest order term in the dividend $- 6 x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$x$	-	$6$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$
					+	$x^{2}$	-	$2 x$
							-	$6 x$	+	$12$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$x$	-	$6$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$
					+	$x^{2}$	-	$2 x$
							-	$6 x$	+	$12$
							-	$6 x$	+	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x + 12$

				$2 x^{2}$	-	$x$	-	$6$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$
					+	$x^{2}$	-	$2 x$
							-	$6 x$	+	$12$
							+	$6 x$	-	$12$

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

				$2 x^{2}$	-	$x$	-	$6$
$x$	-	$2$		$2 x^{3}$	-	$5 x^{2}$	-	$4 x$	+	$12$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$x^{2}$	-	$4 x$
					+	$x^{2}$	-	$2 x$
							-	$6 x$	+	$12$
							+	$6 x$	-	$12$
										$0$

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

$2 x^{2} - x - 6$

Write $2 x^{3} - 5 x^{2} - 4 x + 12$ as a set of factors.

$(x - 2) (2 x^{2} - x - 6) = 0$

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 $a \cdot c = 2 \cdot - 6 = - 12$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$(x - 2) (2 x^{2} - (x) - 6) = 0$

Rewrite $- 1$ as $3$ plus $- 4$

$(x - 2) (2 x^{2} + (3 - 4) x - 6) = 0$

Apply the distributive property.

$(x - 2) (2 x^{2} + 3 x - 4 x - 6) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 2) ((2 x^{2} + 3 x) - 4 x - 6) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

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

Remove unnecessary parentheses.

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

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

$x - 2 = 0$

$2 x + 3 = 0$

$x - 2 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

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

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

Step 3

Find the Roots (Zeros) f(x)=2x^3-5x^2-4x+12

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