Identify the Zeros and Their Multiplicities f(x)=2x^3-3x^2-11x+6

$f (x) = 2 x^{3} - 3 x^{2} - 11 x + 6$

Step 1

Set $2 x^{3} - 3 x^{2} - 11 x + 6$ equal to $0$ .

$2 x^{3} - 3 x^{2} - 11 x + 6 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2 x^{3} - 3 x^{2} - 11 x + 6$ 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.

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

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

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

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.

$2 \cdot {0.5}^{3} - 3 \cdot {0.5}^{2} - 11 \cdot 0.5 + 6$

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

$2 \cdot 0.125 - 3 \cdot {0.5}^{2} - 11 \cdot 0.5 + 6$

Multiply $2$ by $0.125$ .

$0.25 - 3 \cdot {0.5}^{2} - 11 \cdot 0.5 + 6$

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

$0.25 - 3 \cdot 0.25 - 11 \cdot 0.5 + 6$

Multiply $- 3$ by $0.25$ .

$0.25 - 0.75 - 11 \cdot 0.5 + 6$

Subtract $0.75$ from $0.25$ .

$- 0.5 - 11 \cdot 0.5 + 6$

Multiply $- 11$ by $0.5$ .

$- 0.5 - 5.5 + 6$

Subtract $5.5$ from $- 0.5$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

$\frac{2 x^{3} - 3 x^{2} - 11 x + 6}{2 x - 1}$

Divide $2 x^{3} - 3 x^{2} - 11 x + 6$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$3 x^{2}$

$11 x$

$6$

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

					$x^{2}$
	$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			+	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$

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

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					-	$2 x^{2}$	+	$x$

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

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$

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

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$

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

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	-	$x$	-	$6$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$6$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$
							-	$12 x$	+	$6$

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

				$x^{2}$	-	$x$	-	$6$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$
							+	$12 x$	-	$6$

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

				$x^{2}$	-	$x$	-	$6$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$
							+	$12 x$	-	$6$
										$0$

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

$x^{2} - x - 6$

Write $2 x^{3} - 3 x^{2} - 11 x + 6$ as a set of factors.

$(2 x - 1) (x^{2} - x - 6) = 0$

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$2 x - 1 = 0$

$x - 3 = 0$

$x + 2 = 0$

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 $x$ .

Add $1$ to 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 $1$ .

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

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.

$x = 3$

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.

$x = - 2$

The final solution is all the values that make $(2 x - 1) (x - 3) (x + 2) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = \frac{1}{2}$ (Multiplicity of $1$ )

$x = 3$ (Multiplicity of $1$ )

$x = - 2$ (Multiplicity of $1$ )

$x = \frac{1}{2}$ (Multiplicity of $1$ )

$x = 3$ (Multiplicity of $1$ )

$x = - 2$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=2x^3-3x^2-11x+6

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