Find the Roots/Zeros Using the Rational Roots Test g(x)=2x^3-23x^2+58x+35

$g (x) = 2 x^{3} - 23 x^{2} + 58 x + 35$

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.

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

$q = \pm 1, \pm 2$

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

$\pm 1, \pm \frac{1}{2}, \pm 5, \pm \frac{5}{2}, \pm 7, \pm \frac{7}{2}, \pm 35, \pm \frac{35}{2}$

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

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

Simplify the expression. In this case, the expression is equal to $0$ so $x = - \frac{1}{2}$ is a root of the polynomial.

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Simplify each term.

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Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- \frac{1}{2}$ .

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

Apply the product rule to $\frac{1}{2}$ .

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

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

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

One to any power is one.

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

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

$2 (- \frac{1}{8}) - 23 {(- \frac{1}{2})}^{2} + 58 (- \frac{1}{2}) + 35$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{8}$ into the numerator.

$2 (\frac{- 1}{8}) - 23 {(- \frac{1}{2})}^{2} + 58 (- \frac{1}{2}) + 35$

Factor $2$ out of $8$ .

$2 \frac{- 1}{2 (4)} - 23 {(- \frac{1}{2})}^{2} + 58 (- \frac{1}{2}) + 35$

Cancel the common factor.

$2 \frac{- 1}{2 \cdot 4} - 23 {(- \frac{1}{2})}^{2} + 58 (- \frac{1}{2}) + 35$

Rewrite the expression.

$\frac{- 1}{4} - 23 {(- \frac{1}{2})}^{2} + 58 (- \frac{1}{2}) + 35$

Move the negative in front of the fraction.

$- \frac{1}{4} - 23 {(- \frac{1}{2})}^{2} + 58 (- \frac{1}{2}) + 35$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- \frac{1}{2}$ .

$- \frac{1}{4} - 23 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) + 58 (- \frac{1}{2}) + 35$

Apply the product rule to $\frac{1}{2}$ .

$- \frac{1}{4} - 23 ({(- 1)}^{2} \frac{1^{2}}{2^{2}}) + 58 (- \frac{1}{2}) + 35$

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

$- \frac{1}{4} - 23 (1 \frac{1^{2}}{2^{2}}) + 58 (- \frac{1}{2}) + 35$

Multiply $\frac{1^{2}}{2^{2}}$ by $1$ .

$- \frac{1}{4} - 23 \frac{1^{2}}{2^{2}} + 58 (- \frac{1}{2}) + 35$

One to any power is one.

$- \frac{1}{4} - 23 \frac{1}{2^{2}} + 58 (- \frac{1}{2}) + 35$

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

$- \frac{1}{4} - 23 (\frac{1}{4}) + 58 (- \frac{1}{2}) + 35$

Combine $- 23$ and $\frac{1}{4}$ .

$- \frac{1}{4} + \frac{- 23}{4} + 58 (- \frac{1}{2}) + 35$

Move the negative in front of the fraction.

$- \frac{1}{4} - \frac{23}{4} + 58 (- \frac{1}{2}) + 35$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{2}$ into the numerator.

$- \frac{1}{4} - \frac{23}{4} + 58 (\frac{- 1}{2}) + 35$

Factor $2$ out of $58$ .

$- \frac{1}{4} - \frac{23}{4} + 2 (29) \frac{- 1}{2} + 35$

Cancel the common factor.

$- \frac{1}{4} - \frac{23}{4} + 2 \cdot 29 \frac{- 1}{2} + 35$

Rewrite the expression.

$- \frac{1}{4} - \frac{23}{4} + 29 \cdot - 1 + 35$

Multiply $29$ by $- 1$ .

$- \frac{1}{4} - \frac{23}{4} - 29 + 35$

Combine fractions.

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Combine fractions with similar denominators.

$- 29 + 35 + \frac{- 1 - 23}{4}$

Simplify the expression.

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Subtract $23$ from $- 1$ .

$- 29 + 35 + \frac{- 24}{4}$

Divide $- 24$ by $4$ .

$- 29 + 35 - 6$

Add $- 29$ and $35$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

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

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

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Place the numbers representing the divisor and the dividend into a division-like configuration.

$- \frac{1}{2}$	$2$	$- 23$	$58$	$35$

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$- \frac{1}{2}$	$2$	$- 23$	$58$	$35$

	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(- \frac{1}{2})$ and place the result of $(- 1)$ under the next term in the dividend $(- 23)$ .

$- \frac{1}{2}$	$2$	$- 23$	$58$	$35$
		$- 1$
	$2$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- \frac{1}{2}$	$2$	$- 23$	$58$	$35$
		$- 1$
	$2$	$- 24$

Multiply the newest entry in the result $(- 24)$ by the divisor $(- \frac{1}{2})$ and place the result of $(12)$ under the next term in the dividend $(58)$ .

$- \frac{1}{2}$	$2$	$- 23$	$58$	$35$
		$- 1$	$12$
	$2$	$- 24$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- \frac{1}{2}$	$2$	$- 23$	$58$	$35$
		$- 1$	$12$
	$2$	$- 24$	$70$

Multiply the newest entry in the result $(70)$ by the divisor $(- \frac{1}{2})$ and place the result of $(- 35)$ under the next term in the dividend $(35)$ .

$- \frac{1}{2}$	$2$	$- 23$	$58$	$35$
		$- 1$	$12$	$- 35$
	$2$	$- 24$	$70$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- \frac{1}{2}$	$2$	$- 23$	$58$	$35$
		$- 1$	$12$	$- 35$
	$2$	$- 24$	$70$	$0$

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$2 x^{2} + - 24 x + 70$

Simplify the quotient polynomial.

$2 x^{2} - 24 x + 70$

Factor $2$ out of $2 x^{2} - 24 x + 70$ .

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Factor $2$ out of $2 x^{2}$ .

$(x + \frac{1}{2}) (2 (x^{2}) - 24 x + 70)$

Factor $2$ out of $- 24 x$ .

$(x + \frac{1}{2}) (2 (x^{2}) + 2 (- 12 x) + 70)$

Factor $2$ out of $70$ .

$(x + \frac{1}{2}) (2 x^{2} + 2 (- 12 x) + 2 \cdot 35)$

Factor $2$ out of $2 x^{2} + 2 (- 12 x)$ .

$(x + \frac{1}{2}) (2 (x^{2} - 12 x) + 2 \cdot 35)$

Factor $2$ out of $2 (x^{2} - 12 x) + 2 \cdot 35$ .

$(x + \frac{1}{2}) (2 (x^{2} - 12 x + 35))$

Factor.

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

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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 $- 12$ .

$- 7, - 5$

Write the factored form using these integers.

$(x + \frac{1}{2}) (2 ((x - 7) (x - 5)))$

Remove unnecessary parentheses.

$(x + \frac{1}{2}) (2 (x - 7) (x - 5))$

Factor the left side of the equation.

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Factor $2 x^{3} - 23 x^{2} + 58 x + 35$ using the rational roots test.

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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.

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Substitute $- 0.5$ into the polynomial.

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

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

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

Multiply $2$ by $- 0.125$ .

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

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

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

Multiply $- 23$ by $0.25$ .

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

Subtract $5.75$ from $- 0.25$ .

$- 6 + 58 \cdot - 0.5 + 35$

Multiply $58$ by $- 0.5$ .

$- 6 - 29 + 35$

Subtract $29$ from $- 6$ .

$- 35 + 35$

Add $- 35$ and $35$ .

$0$

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 $2 x + 1$ .

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

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.

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

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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 $- 12$ .

$- 7, - 5$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$2 x + 1 = 0$

$x - 7 = 0$

$x - 5 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$2 x + 1 = 0$

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term by $2$ and simplify.

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

Move the negative in front of the fraction.

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

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

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

Find the Roots/Zeros Using the Rational Roots Test g(x)=2x^3-23x^2+58x+35

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