Find the Roots (Zeros) -x^3+2x^2+5x-6

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

Step 1

Set $- x^{3} + 2 x^{2} + 5 x - 6$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $- x^{3} + 2 x^{2} + 5 x - 6$ 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.

$- 1^{3} + 2 \cdot 1^{2} + 5 \cdot 1 - 6$

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

$- 1 \cdot 1 + 2 \cdot 1^{2} + 5 \cdot 1 - 6$

Multiply $- 1$ by $1$ .

$- 1 + 2 \cdot 1^{2} + 5 \cdot 1 - 6$

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

$- 1 + 2 \cdot 1 + 5 \cdot 1 - 6$

Multiply $2$ by $1$ .

$- 1 + 2 + 5 \cdot 1 - 6$

Add $- 1$ and $2$ .

$1 + 5 \cdot 1 - 6$

Multiply $5$ by $1$ .

$1 + 5 - 6$

Add $1$ and $5$ .

$6 - 6$

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.

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

Divide $- x^{3} + 2 x^{2} + 5 x - 6$ 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 $0$ .

$x$

$1$

$x^{3}$

$2 x^{2}$

$5 x$

$6$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

Write $- x^{3} + 2 x^{2} + 5 x - 6$ as a set of factors.

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

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 6 = - 6$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

Rewrite $1$ as $- 2$ plus $3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- x - 2$ .

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

Remove unnecessary parentheses.

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

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 $0$ .

$- x - 2 = 0$

Solve $- x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$- x = 2$

Divide each term in $- x = 2$ by $- 1$ and simplify.

Divide each term in $- x = 2$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $- 1$ .

$x = - 2$

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.

$x = 3$

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

$x = 1, - 2, 3$

Step 3

Find the Roots (Zeros) -x^3+2x^2+5x-6

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