Find the Roots (Zeros) x^3-5x^2+3x+9=0

$x^{3} - 5 x^{2} + 3 x + 9 = 0$

Step 1

Factor the left side of the equation.

Factor $x^{3} - 5 x^{2} + 3 x + 9$ 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.

$p = \pm 1, \pm 9, \pm 3 q = \pm 1$

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

$\pm 1, \pm 9, \pm 3$

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.

${(- 1)}^{3} - 5 {(- 1)}^{2} + 3 \cdot - 1 + 9$

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

$- 1 - 5 {(- 1)}^{2} + 3 \cdot - 1 + 9$

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

$- 1 - 5 \cdot 1 + 3 \cdot - 1 + 9$

Multiply $- 5$ by $1$ .

$- 1 - 5 + 3 \cdot - 1 + 9$

Subtract $5$ from $- 1$ .

$- 6 + 3 \cdot - 1 + 9$

Multiply $3$ by $- 1$ .

$- 6 - 3 + 9$

Subtract $3$ from $- 6$ .

$- 9 + 9$

Add $- 9$ and $9$ .

$0$

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.

$\frac{x^{3} - 5 x^{2} + 3 x + 9}{x + 1}$

Divide $x^{3} - 5 x^{2} + 3 x + 9$ 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 $0$ .

$x$

$1$

$x^{3}$

$5 x^{2}$

$3 x$

$9$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$
					-	$6 x^{2}$	-	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$
					+	$6 x^{2}$	+	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$
					+	$6 x^{2}$	+	$6 x$
							+	$9 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$
					+	$6 x^{2}$	+	$6 x$
							+	$9 x$	+	$9$

Divide the highest order term in the dividend $9 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$	+	$9$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$
					+	$6 x^{2}$	+	$6 x$
							+	$9 x$	+	$9$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$9$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$
					+	$6 x^{2}$	+	$6 x$
							+	$9 x$	+	$9$
							+	$9 x$	+	$9$

The expression needs to be subtracted from the dividend, so change all the signs in $9 x + 9$

				$x^{2}$	-	$6 x$	+	$9$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$
					+	$6 x^{2}$	+	$6 x$
							+	$9 x$	+	$9$
							-	$9 x$	-	$9$

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

				$x^{2}$	-	$6 x$	+	$9$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$9$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$3 x$
					+	$6 x^{2}$	+	$6 x$
							+	$9 x$	+	$9$
							-	$9 x$	-	$9$
										$0$

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

$x^{2} - 6 x + 9$

Write $x^{3} - 5 x^{2} + 3 x + 9$ as a set of factors.

$(x + 1) (x^{2} - 6 x + 9) = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

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

Step 2

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

$x + 1 = 0$

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

Step 3

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4

Set ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 3)}^{2}$ equal to $0$ .

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

Solve ${(x - 3)}^{2} = 0$ for $x$ .

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

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

$x = - 1, 3$

Step 6

Find the Roots (Zeros) x^3-5x^2+3x+9=0

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