Find the Roots (Zeros) 2x^3+23x^2+58x-35

$2 x^{3} + 23 x^{2} + 58 x - 35$

Step 1

Set $2 x^{3} + 23 x^{2} + 58 x - 35$ equal to $0$ .

$2 x^{3} + 23 x^{2} + 58 x - 35 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2 x^{3} + 23 x^{2} + 58 x - 35$ 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.

$p = \pm 1, \pm 35, \pm 5, \pm 7 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 35, \pm 17.5, \pm 5, \pm 2.5, \pm 7, \pm 3.5$

Substitute $0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $0.5$ is a root of the polynomial.

Substitute $0.5$ into the polynomial.

$2 \cdot {0.5}^{3} + 23 \cdot {0.5}^{2} + 58 \cdot 0.5 - 35$

Raise $0.5$ to the power of $3$ .

$2 \cdot 0.125 + 23 \cdot {0.5}^{2} + 58 \cdot 0.5 - 35$

Multiply $2$ by $0.125$ .

$0.25 + 23 \cdot {0.5}^{2} + 58 \cdot 0.5 - 35$

Raise $0.5$ to the power of $2$ .

$0.25 + 23 \cdot 0.25 + 58 \cdot 0.5 - 35$

Multiply $23$ by $0.25$ .

$0.25 + 5.75 + 58 \cdot 0.5 - 35$

Add $0.25$ and $5.75$ .

$6 + 58 \cdot 0.5 - 35$

Multiply $58$ by $0.5$ .

$6 + 29 - 35$

Add $6$ and $29$ .

$35 - 35$

Subtract $35$ from $35$ .

$0$

Since $0.5$ is a known root, divide the polynomial by $2 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} + 23 x^{2} + 58 x - 35}{2 x - 1}$

Divide $2 x^{3} + 23 x^{2} + 58 x - 35$ by $2 x - 1$ .

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

$2 x$

$1$

$2 x^{3}$

$23 x^{2}$

$58 x$

$35$

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

					$x^{2}$
	$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			+	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$

Divide the highest order term in the dividend $24 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					+	$24 x^{2}$	-	$12 x$

The expression needs to be subtracted from the dividend, so change all the signs in $24 x^{2} - 12 x$

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$

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

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$

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

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$

Divide the highest order term in the dividend $70 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$12 x$	+	$35$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$12 x$	+	$35$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$
							+	$70 x$	-	$35$

The expression needs to be subtracted from the dividend, so change all the signs in $70 x - 35$

				$x^{2}$	+	$12 x$	+	$35$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$
							-	$70 x$	+	$35$

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

				$x^{2}$	+	$12 x$	+	$35$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$
							-	$70 x$	+	$35$
										$0$

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

$x^{2} + 12 x + 35$

Write $2 x^{3} + 23 x^{2} + 58 x - 35$ as a set of factors.

$(2 x - 1) (x^{2} + 12 x + 35) = 0$

Factor $x^{2} + 12 x + 35$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $35$ and whose sum is $12$ .

$5, 7$

Write the factored form using these integers.

$(2 x - 1) ((x + 5) (x + 7)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x + 5) (x + 7) = 0$

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

$2 x - 1 = 0$

$x + 5 = 0$

$x + 7 = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(2 x - 1) (x + 5) (x + 7) = 0$ true.

$x = \frac{1}{2}, - 5, - 7$

Step 3

Find the Roots (Zeros) 2x^3+23x^2+58x-35

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