Find the Roots (Zeros) 0=x^3-3x-52

$0 = x^{3} - 3 x - 52$

Step 1

Rewrite the equation as $x^{3} - 3 x - 52 = 0$ .

$x^{3} - 3 x - 52 = 0$

Step 2

Factor $x^{3} - 3 x - 52$ 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.

$p = \pm 1, \pm 52, \pm 2, \pm 26, \pm 4, \pm 13 q = \pm 1$

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

$\pm 1, \pm 52, \pm 2, \pm 26, \pm 4, \pm 13$

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

Substitute $4$ into the polynomial.

$4^{3} - 3 \cdot 4 - 52$

Raise $4$ to the power of $3$ .

$64 - 3 \cdot 4 - 52$

Multiply $- 3$ by $4$ .

$64 - 12 - 52$

Subtract $12$ from $64$ .

$52 - 52$

Subtract $52$ from $52$ .

$0$

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 3 x - 52}{x - 4}$

Divide $x^{3} - 3 x - 52$ by $x - 4$ .

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

$x$

$4$

$x^{3}$

$0 x^{2}$

$3 x$

$52$

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

					$x^{2}$
	$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			+	$x^{3}$	-	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$

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

				$x^{2}$	+	$4 x$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$4 x$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$
					+	$4 x^{2}$	-	$16 x$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{2} - 16 x$

				$x^{2}$	+	$4 x$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$
					-	$4 x^{2}$	+	$16 x$

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

				$x^{2}$	+	$4 x$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$
					-	$4 x^{2}$	+	$16 x$
							+	$13 x$

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

				$x^{2}$	+	$4 x$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$
					-	$4 x^{2}$	+	$16 x$
							+	$13 x$	-	$52$

Divide the highest order term in the dividend $13 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$4 x$	+	$13$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$
					-	$4 x^{2}$	+	$16 x$
							+	$13 x$	-	$52$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$4 x$	+	$13$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$
					-	$4 x^{2}$	+	$16 x$
							+	$13 x$	-	$52$
							+	$13 x$	-	$52$

The expression needs to be subtracted from the dividend, so change all the signs in $13 x - 52$

				$x^{2}$	+	$4 x$	+	$13$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$
					-	$4 x^{2}$	+	$16 x$
							+	$13 x$	-	$52$
							-	$13 x$	+	$52$

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

				$x^{2}$	+	$4 x$	+	$13$
$x$	-	$4$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$52$
			-	$x^{3}$	+	$4 x^{2}$
					+	$4 x^{2}$	-	$3 x$
					-	$4 x^{2}$	+	$16 x$
							+	$13 x$	-	$52$
							-	$13 x$	+	$52$
										$0$

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

$x^{2} + 4 x + 13$

Write $x^{3} - 3 x - 52$ as a set of factors.

$(x - 4) (x^{2} + 4 x + 13) = 0$

Step 3

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

$x - 4 = 0$

$x^{2} + 4 x + 13 = 0$

Step 4

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 5

Set $x^{2} + 4 x + 13$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4 x + 13$ equal to $0$ .

$x^{2} + 4 x + 13 = 0$

Solve $x^{2} + 4 x + 13 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 4$ , and $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 13$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 13}}{2 \cdot 1}$

Multiply $- 4$ by $13$ .

$x = \frac{- 4 \pm \sqrt{16 - 52}}{2 \cdot 1}$

Subtract $52$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 36}}{2 \cdot 1}$

Rewrite $- 36$ as $- 1 (36)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 4 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 4 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 4 \pm i \cdot 6}{2 \cdot 1}$

Move $6$ to the left of $i$ .

$x = \frac{- 4 \pm 6 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 6 i}{2}$

Simplify $\frac{- 4 \pm 6 i}{2}$ .

$x = - 2 \pm 3 i$

The final answer is the combination of both solutions.

$x = - 2 + 3 i, - 2 - 3 i$

Step 6

The final solution is all the values that make $(x - 4) (x^{2} + 4 x + 13) = 0$ true.

$x = 4, - 2 + 3 i, - 2 - 3 i$

Step 7

Find the Roots (Zeros) 0=x^3-3x-52

Related Questions

Support LCMGCF.com by disabling your adblocker.