Find the Roots (Zeros) f(x)=x^3-3x^2-5x+39

$f (x) = x^{3} - 3 x^{2} - 5 x + 39$

Step 1

Set $x^{3} - 3 x^{2} - 5 x + 39$ equal to $0$ .

$x^{3} - 3 x^{2} - 5 x + 39 = 0$

Step 2

Solve for $x$ .

Factor $x^{3} - 3 x^{2} - 5 x + 39$ 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.

$p = \pm 1, \pm 39, \pm 3, \pm 13 q = \pm 1$

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

$\pm 1, \pm 39, \pm 3, \pm 13$

Substitute $- 3$ and simplify the expression. In this case, the expression is equal to $0$ so $- 3$ is a root of the polynomial.

Substitute $- 3$ into the polynomial.

${(- 3)}^{3} - 3 {(- 3)}^{2} - 5 \cdot - 3 + 39$

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

$- 27 - 3 {(- 3)}^{2} - 5 \cdot - 3 + 39$

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

$- 27 - 3 \cdot 9 - 5 \cdot - 3 + 39$

Multiply $- 3$ by $9$ .

$- 27 - 27 - 5 \cdot - 3 + 39$

Subtract $27$ from $- 27$ .

$- 54 - 5 \cdot - 3 + 39$

Multiply $- 5$ by $- 3$ .

$- 54 + 15 + 39$

Add $- 54$ and $15$ .

$- 39 + 39$

Add $- 39$ and $39$ .

$0$

Since $- 3$ is a known root, divide the polynomial by $x + 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 3 x^{2} - 5 x + 39}{x + 3}$

Divide $x^{3} - 3 x^{2} - 5 x + 39$ by $x + 3$ .

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

$x$

$3$

$x^{3}$

$3 x^{2}$

$5 x$

$39$

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

					$x^{2}$
	$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			+	$x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$
					-	$6 x^{2}$	-	$18 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} - 18 x$

				$x^{2}$	-	$6 x$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$18 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$18 x$
							+	$13 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$18 x$
							+	$13 x$	+	$39$

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

				$x^{2}$	-	$6 x$	+	$13$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$18 x$
							+	$13 x$	+	$39$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$13$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$18 x$
							+	$13 x$	+	$39$
							+	$13 x$	+	$39$

The expression needs to be subtracted from the dividend, so change all the signs in $13 x + 39$

				$x^{2}$	-	$6 x$	+	$13$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$18 x$
							+	$13 x$	+	$39$
							-	$13 x$	-	$39$

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

				$x^{2}$	-	$6 x$	+	$13$
$x$	+	$3$		$x^{3}$	-	$3 x^{2}$	-	$5 x$	+	$39$
			-	$x^{3}$	-	$3 x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$18 x$
							+	$13 x$	+	$39$
							-	$13 x$	-	$39$
										$0$

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

$x^{2} - 6 x + 13$

Write $x^{3} - 3 x^{2} - 5 x + 39$ as a set of factors.

$(x + 3) (x^{2} - 6 x + 13) = 0$

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

$x + 3 = 0$

$x^{2} - 6 x + 13 = 0$

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.

$x = - 3$

Set $x^{2} - 6 x + 13$ equal to $0$ and solve for $x$ .

Set $x^{2} - 6 x + 13$ equal to $0$ .

$x^{2} - 6 x + 13 = 0$

Solve $x^{2} - 6 x + 13 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 13}}{2 \cdot 1}$

Multiply $- 4$ by $13$ .

$x = \frac{6 \pm \sqrt{36 - 52}}{2 \cdot 1}$

Subtract $52$ from $36$ .

$x = \frac{6 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$x = \frac{6 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{6 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

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

$x = \frac{6 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

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

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

Move $4$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

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

$x = 3 \pm 2 i$

The final answer is the combination of both solutions.

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

The final solution is all the values that make $(x + 3) (x^{2} - 6 x + 13) = 0$ true.

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

Step 3

Find the Roots (Zeros) f(x)=x^3-3x^2-5x+39

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