Find the Roots (Zeros) 2x^3+9x^2+4x-15=0

$2 x^{3} + 9 x^{2} + 4 x - 15 = 0$

Step 1

Factor the left side of the equation.

Factor $2 x^{3} + 9 x^{2} + 4 x - 15$ 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.

$p = \pm 1, \pm 15, \pm 3, \pm 5 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 15, \pm 7.5, \pm 3, \pm 1.5, \pm 5, \pm 2.5$

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.

$2 \cdot 1^{3} + 9 \cdot 1^{2} + 4 \cdot 1 - 15$

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

$2 \cdot 1 + 9 \cdot 1^{2} + 4 \cdot 1 - 15$

Multiply $2$ by $1$ .

$2 + 9 \cdot 1^{2} + 4 \cdot 1 - 15$

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

$2 + 9 \cdot 1 + 4 \cdot 1 - 15$

Multiply $9$ by $1$ .

$2 + 9 + 4 \cdot 1 - 15$

Add $2$ and $9$ .

$11 + 4 \cdot 1 - 15$

Multiply $4$ by $1$ .

$11 + 4 - 15$

Add $11$ and $4$ .

$15 - 15$

Subtract $15$ from $15$ .

$0$

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.

$\frac{2 x^{3} + 9 x^{2} + 4 x - 15}{x - 1}$

Divide $2 x^{3} + 9 x^{2} + 4 x - 15$ 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 $0$ .

$x$

$1$

$2 x^{3}$

$9 x^{2}$

$4 x$

$15$

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

					$2 x^{2}$
	$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			+	$2 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$

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

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$

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

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$

Divide the highest order term in the dividend $11 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	+	$11 x$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$11 x$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$
					+	$11 x^{2}$	-	$11 x$

The expression needs to be subtracted from the dividend, so change all the signs in $11 x^{2} - 11 x$

				$2 x^{2}$	+	$11 x$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$
					-	$11 x^{2}$	+	$11 x$

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

				$2 x^{2}$	+	$11 x$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$
					-	$11 x^{2}$	+	$11 x$
							+	$15 x$

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

				$2 x^{2}$	+	$11 x$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$
					-	$11 x^{2}$	+	$11 x$
							+	$15 x$	-	$15$

Divide the highest order term in the dividend $15 x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	+	$11 x$	+	$15$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$
					-	$11 x^{2}$	+	$11 x$
							+	$15 x$	-	$15$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$11 x$	+	$15$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$
					-	$11 x^{2}$	+	$11 x$
							+	$15 x$	-	$15$
							+	$15 x$	-	$15$

The expression needs to be subtracted from the dividend, so change all the signs in $15 x - 15$

				$2 x^{2}$	+	$11 x$	+	$15$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$
					-	$11 x^{2}$	+	$11 x$
							+	$15 x$	-	$15$
							-	$15 x$	+	$15$

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

				$2 x^{2}$	+	$11 x$	+	$15$
$x$	-	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$4 x$	-	$15$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$11 x^{2}$	+	$4 x$
					-	$11 x^{2}$	+	$11 x$
							+	$15 x$	-	$15$
							-	$15 x$	+	$15$
										$0$

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

$2 x^{2} + 11 x + 15$

Write $2 x^{3} + 9 x^{2} + 4 x - 15$ as a set of factors.

$(x - 1) (2 x^{2} + 11 x + 15) = 0$

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 $a \cdot c = 2 \cdot 15 = 30$ and whose sum is $b = 11$ .

Factor $11$ out of $11 x$ .

$(x - 1) (2 x^{2} + 11 (x) + 15) = 0$

Rewrite $11$ as $5$ plus $6$

$(x - 1) (2 x^{2} + (5 + 6) x + 15) = 0$

Apply the distributive property.

$(x - 1) (2 x^{2} + 5 x + 6 x + 15) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) ((2 x^{2} + 5 x) + 6 x + 15) = 0$

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

$(x - 1) (x (2 x + 5) + 3 (2 x + 5)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 5$ .

$(x - 1) ((2 x + 5) (x + 3)) = 0$

Remove unnecessary parentheses.

$(x - 1) (2 x + 5) (x + 3) = 0$

Step 2

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

$x - 1 = 0$

$2 x + 5 = 0$

$x + 3 = 0$

Step 3

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.

$x = 1$

Step 4

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 5

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.

$x = - 3$

Step 6

The final solution is all the values that make $(x - 1) (2 x + 5) (x + 3) = 0$ true.

$x = 1, - \frac{5}{2}, - 3$

Step 7

Find the Roots (Zeros) 2x^3+9x^2+4x-15=0

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