Find the Roots (Zeros) f(x)=3x^3+12x^2+3x-18

$f (x) = 3 x^{3} + 12 x^{2} + 3 x - 18$

Step 1

Set $3 x^{3} + 12 x^{2} + 3 x - 18$ equal to $0$ .

$3 x^{3} + 12 x^{2} + 3 x - 18 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $3$ out of $3 x^{3} + 12 x^{2} + 3 x - 18$ .

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

$3 (x^{3}) + 12 x^{2} + 3 x - 18 = 0$

Factor $3$ out of $12 x^{2}$ .

$3 (x^{3}) + 3 (4 x^{2}) + 3 x - 18 = 0$

Factor $3$ out of $3 x$ .

$3 (x^{3}) + 3 (4 x^{2}) + 3 (x) - 18 = 0$

Factor $3$ out of $- 18$ .

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

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

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

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

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

Factor $3$ out of $3 (x^{3} + 4 x^{2} + x) + 3 \cdot - 6$ .

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

Factor $x^{3} + 4 x^{2} + x - 6$ 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.

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

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

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

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.

$1^{3} + 4 \cdot 1^{2} + 1 - 6$

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

$1 + 4 \cdot 1^{2} + 1 - 6$

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

$1 + 4 \cdot 1 + 1 - 6$

Multiply $4$ by $1$ .

$1 + 4 + 1 - 6$

Add $1$ and $4$ .

$5 + 1 - 6$

Add $5$ and $1$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

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.

$\frac{x^{3} + 4 x^{2} + x - 6}{x - 1}$

Divide $x^{3} + 4 x^{2} + x - 6$ 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 $0$ .

$x$

$1$

$x^{3}$

$4 x^{2}$

$x$

$6$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$

Divide the highest order term in the dividend $5 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$5 x$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$5 x$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$
					+	$5 x^{2}$	-	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{2} - 5 x$

				$x^{2}$	+	$5 x$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$
					-	$5 x^{2}$	+	$5 x$

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

				$x^{2}$	+	$5 x$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$
					-	$5 x^{2}$	+	$5 x$
							+	$6 x$

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

				$x^{2}$	+	$5 x$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$
					-	$5 x^{2}$	+	$5 x$
							+	$6 x$	-	$6$

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

				$x^{2}$	+	$5 x$	+	$6$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$
					-	$5 x^{2}$	+	$5 x$
							+	$6 x$	-	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$5 x$	+	$6$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$
					-	$5 x^{2}$	+	$5 x$
							+	$6 x$	-	$6$
							+	$6 x$	-	$6$

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

				$x^{2}$	+	$5 x$	+	$6$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$
					-	$5 x^{2}$	+	$5 x$
							+	$6 x$	-	$6$
							-	$6 x$	+	$6$

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

				$x^{2}$	+	$5 x$	+	$6$
$x$	-	$1$		$x^{3}$	+	$4 x^{2}$	+	$x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$5 x^{2}$	+	$x$
					-	$5 x^{2}$	+	$5 x$
							+	$6 x$	-	$6$
							-	$6 x$	+	$6$
										$0$

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

$x^{2} + 5 x + 6$

Write $x^{3} + 4 x^{2} + x - 6$ as a set of factors.

$3 ((x - 1) (x^{2} + 5 x + 6)) = 0$

Factor.

Factor $x^{2} + 5 x + 6$ 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 $5$ .

$2, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

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

$x - 1 = 0$

$x + 2 = 0$

$x + 3 = 0$

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.

$x = 1$

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.

$x = - 3$

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

$x = 1, - 2, - 3$

Step 3

Find the Roots (Zeros) f(x)=3x^3+12x^2+3x-18

Related Questions

Support LCMGCF.com by disabling your adblocker.