Find the Roots (Zeros) 4x^3-12x^2-9x-1

$4 x^{3} - 12 x^{2} - 9 x - 1$

Step 1

Set $4 x^{3} - 12 x^{2} - 9 x - 1$ equal to $0$ .

$4 x^{3} - 12 x^{2} - 9 x - 1 = 0$

Step 2

Solve for $x$ .

Factor $4 x^{3} - 12 x^{2} - 9 x - 1$ 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.

$p = \pm 1 q = \pm 1, \pm 4, \pm 2$

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

$\pm 1, \pm 0.25, \pm 0.5$

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.

$4 {(- 0.5)}^{3} - 12 {(- 0.5)}^{2} - 9 \cdot - 0.5 - 1$

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

$4 \cdot - 0.125 - 12 {(- 0.5)}^{2} - 9 \cdot - 0.5 - 1$

Multiply $4$ by $- 0.125$ .

$- 0.5 - 12 {(- 0.5)}^{2} - 9 \cdot - 0.5 - 1$

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

$- 0.5 - 12 \cdot 0.25 - 9 \cdot - 0.5 - 1$

Multiply $- 12$ by $0.25$ .

$- 0.5 - 3 - 9 \cdot - 0.5 - 1$

Subtract $3$ from $- 0.5$ .

$- 3.5 - 9 \cdot - 0.5 - 1$

Multiply $- 9$ by $- 0.5$ .

$- 3.5 + 4.5 - 1$

Add $- 3.5$ and $4.5$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

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.

$\frac{4 x^{3} - 12 x^{2} - 9 x - 1}{2 x + 1}$

Divide $4 x^{3} - 12 x^{2} - 9 x - 1$ 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 $0$ .

$2 x$

$1$

$4 x^{3}$

$12 x^{2}$

$9 x$

$1$

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

					$2 x^{2}$
	$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			+	$4 x^{3}$	+	$2 x^{2}$

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

				$2 x^{2}$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$

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

				$2 x^{2}$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$

Divide the highest order term in the dividend $- 14 x^{2}$ by the highest order term in divisor $2 x$ .

				$2 x^{2}$	-	$7 x$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$7 x$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$
					-	$14 x^{2}$	-	$7 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 14 x^{2} - 7 x$

				$2 x^{2}$	-	$7 x$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$
					+	$14 x^{2}$	+	$7 x$

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

				$2 x^{2}$	-	$7 x$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$
					+	$14 x^{2}$	+	$7 x$
							-	$2 x$

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

				$2 x^{2}$	-	$7 x$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$
					+	$14 x^{2}$	+	$7 x$
							-	$2 x$	-	$1$

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

				$2 x^{2}$	-	$7 x$	-	$1$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$
					+	$14 x^{2}$	+	$7 x$
							-	$2 x$	-	$1$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$7 x$	-	$1$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$
					+	$14 x^{2}$	+	$7 x$
							-	$2 x$	-	$1$
							-	$2 x$	-	$1$

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x - 1$

				$2 x^{2}$	-	$7 x$	-	$1$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$
					+	$14 x^{2}$	+	$7 x$
							-	$2 x$	-	$1$
							+	$2 x$	+	$1$

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

				$2 x^{2}$	-	$7 x$	-	$1$
$2 x$	+	$1$		$4 x^{3}$	-	$12 x^{2}$	-	$9 x$	-	$1$
			-	$4 x^{3}$	-	$2 x^{2}$
					-	$14 x^{2}$	-	$9 x$
					+	$14 x^{2}$	+	$7 x$
							-	$2 x$	-	$1$
							+	$2 x$	+	$1$
										$0$

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

$2 x^{2} - 7 x - 1$

Write $4 x^{3} - 12 x^{2} - 9 x - 1$ as a set of factors.

$(2 x + 1) (2 x^{2} - 7 x - 1) = 0$

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

$2 x + 1 = 0$

$2 x^{2} - 7 x - 1 = 0$

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 $x$ .

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.

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

Set $2 x^{2} - 7 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x^{2} - 7 x - 1$ equal to $0$ .

$2 x^{2} - 7 x - 1 = 0$

Solve $2 x^{2} - 7 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = - 7$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (2 \cdot - 1)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 2 \cdot - 1}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 1$ .

Multiply $- 4$ by $2$ .

$x = \frac{7 \pm \sqrt{49 - 8 \cdot - 1}}{2 \cdot 2}$

Multiply $- 8$ by $- 1$ .

$x = \frac{7 \pm \sqrt{49 + 8}}{2 \cdot 2}$

Add $49$ and $8$ .

$x = \frac{7 \pm \sqrt{57}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{7 \pm \sqrt{57}}{4}$

The final answer is the combination of both solutions.

$x = \frac{7 + \sqrt{57}}{4}, \frac{7 - \sqrt{57}}{4}$

The final solution is all the values that make $(2 x + 1) (2 x^{2} - 7 x - 1) = 0$ true.

$x = - \frac{1}{2}, \frac{7 + \sqrt{57}}{4}, \frac{7 - \sqrt{57}}{4}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{2}, \frac{7 + \sqrt{57}}{4}, \frac{7 - \sqrt{57}}{4}$

Decimal Form:

$x = - 0.5, 3.63745860 \dots, - 0.13745860 \dots$

Step 4

Find the Roots (Zeros) 4x^3-12x^2-9x-1

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