Find the Roots/Zeros Using the Rational Roots Test 60x^2-57x-18

$60 x^{2} - 57 x - 18$

Step 1

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.

$p = \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$

$q = \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{5}, \pm \frac{1}{6}, \pm \frac{1}{10}, \pm \frac{1}{12}, \pm \frac{1}{15}, \pm \frac{1}{20}, \pm \frac{1}{30}, \pm \frac{1}{60}, \pm 2, \pm \frac{2}{3}, \pm \frac{2}{5}, \pm \frac{2}{15}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{5}, \pm \frac{3}{10}, \pm \frac{3}{20}, \pm 6, \pm \frac{6}{5}, \pm 9, \pm \frac{9}{2}, \pm \frac{9}{4}, \pm \frac{9}{5}, \pm \frac{9}{10}, \pm \frac{9}{20}, \pm 18, \pm \frac{18}{5}$

Step 3

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.

$60 {(- \frac{1}{4})}^{2} - 57 (- \frac{1}{4}) - 18$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - \frac{1}{4}$ is a root of the polynomial.

Simplify each term.

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

Apply the product rule to $- \frac{1}{4}$ .

$60 ({(- 1)}^{2} {(\frac{1}{4})}^{2}) - 57 (- \frac{1}{4}) - 18$

Apply the product rule to $\frac{1}{4}$ .

$60 ({(- 1)}^{2} \frac{1^{2}}{4^{2}}) - 57 (- \frac{1}{4}) - 18$

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

$60 (1 \frac{1^{2}}{4^{2}}) - 57 (- \frac{1}{4}) - 18$

Multiply $\frac{1^{2}}{4^{2}}$ by $1$ .

$60 \frac{1^{2}}{4^{2}} - 57 (- \frac{1}{4}) - 18$

One to any power is one.

$60 \frac{1}{4^{2}} - 57 (- \frac{1}{4}) - 18$

Raise $4$ to the power of $2$ .

$60 (\frac{1}{16}) - 57 (- \frac{1}{4}) - 18$

Cancel the common factor of $4$ .

Factor $4$ out of $60$ .

$4 (15) \frac{1}{16} - 57 (- \frac{1}{4}) - 18$

Factor $4$ out of $16$ .

$4 \cdot 15 \frac{1}{4 \cdot 4} - 57 (- \frac{1}{4}) - 18$

Cancel the common factor.

$4 \cdot 15 \frac{1}{4 \cdot 4} - 57 (- \frac{1}{4}) - 18$

Rewrite the expression.

$15 (\frac{1}{4}) - 57 (- \frac{1}{4}) - 18$

Combine $15$ and $\frac{1}{4}$ .

$\frac{15}{4} - 57 (- \frac{1}{4}) - 18$

Multiply $- 57 (- \frac{1}{4})$ .

Multiply $- 1$ by $- 57$ .

$\frac{15}{4} + 57 (\frac{1}{4}) - 18$

Combine $57$ and $\frac{1}{4}$ .

$\frac{15}{4} + \frac{57}{4} - 18$

Combine fractions.

Combine the numerators over the common denominator.

$- 18 + \frac{15 + 57}{4}$

Simplify the expression.

Add $15$ and $57$ .

$- 18 + \frac{72}{4}$

Divide $72$ by $4$ .

$- 18 + 18$

Add $- 18$ and $18$ .

$0$

Step 5

Since $- \frac{1}{4}$ is a known root, divide the polynomial by $x + \frac{1}{4}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{60 x^{2} - 57 x - 18}{x + \frac{1}{4}}$

Step 6

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

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

$- \frac{1}{4}$	$60$	$- 57$	$- 18$
	\xa0		\xa0

The first number in the dividend $(60)$ is put into the first position of the result area (below the horizontal line).

$- \frac{1}{4}$	$60$	$- 57$	$- 18$
	\xa0		\xa0
	$60$

Multiply the newest entry in the result $(60)$ by the divisor $(- \frac{1}{4})$ and place the result of $(- 15)$ under the next term in the dividend $(- 57)$ .

$- \frac{1}{4}$	$60$	$- 57$	$- 18$
	\xa0	$- 15$	\xa0
	$60$

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

$- \frac{1}{4}$	$60$	$- 57$	$- 18$
	\xa0	$- 15$	\xa0
	$60$	$- 72$

Multiply the newest entry in the result $(- 72)$ by the divisor $(- \frac{1}{4})$ and place the result of $(18)$ under the next term in the dividend $(- 18)$ .

$- \frac{1}{4}$	$60$	$- 57$	$- 18$
	\xa0	$- 15$	$18$
	$60$	$- 72$

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

$- \frac{1}{4}$	$60$	$- 57$	$- 18$
	\xa0	$- 15$	$18$
	$60$	$- 72$	$0$

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

$(60) x - 72$

Simplify the quotient polynomial.

$60 x - 72$

Step 7

Factor $12$ out of $60 x - 72$ .

Factor $12$ out of $60 x$ .

$(x + \frac{1}{4}) (12 (5 x) - 72)$

Factor $12$ out of $- 72$ .

$(x + \frac{1}{4}) (12 (5 x) + 12 (- 6))$

Factor $12$ out of $12 (5 x) + 12 (- 6)$ .

$(x + \frac{1}{4}) (12 (5 x - 6))$

Step 8

Factor the left side of the equation.

Factor $3$ out of $60 x^{2} - 57 x - 18$ .

Factor $3$ out of $60 x^{2}$ .

$3 (20 x^{2}) - 57 x - 18 = 0$

Factor $3$ out of $- 57 x$ .

$3 (20 x^{2}) + 3 (- 19 x) - 18 = 0$

Factor $3$ out of $- 18$ .

$3 (20 x^{2}) + 3 (- 19 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (20 x^{2}) + 3 (- 19 x)$ .

$3 (20 x^{2} - 19 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (20 x^{2} - 19 x) + 3 (- 6)$ .

$3 (20 x^{2} - 19 x - 6) = 0$

Factor.

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 $a \cdot c = 20 \cdot - 6 = - 120$ and whose sum is $b = - 19$ .

Factor $- 19$ out of $- 19 x$ .

$3 (20 x^{2} - 19 x - 6) = 0$

Rewrite $- 19$ as $5$ plus $- 24$

$3 (20 x^{2} + (5 - 24) x - 6) = 0$

Apply the distributive property.

$3 (20 x^{2} + 5 x - 24 x - 6) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$3 ((20 x^{2} + 5 x) - 24 x - 6) = 0$

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

$3 (5 x (4 x + 1) - 6 (4 x + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x + 1$ .

$3 ((4 x + 1) (5 x - 6)) = 0$

Remove unnecessary parentheses.

$3 (4 x + 1) (5 x - 6) = 0$

Step 9

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

$4 x + 1 = 0$

$5 x - 6 = 0$

Step 10

Set $4 x + 1$ equal to $0$ and solve for $x$ .

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Solve $4 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Divide each term in $4 x = - 1$ by $4$ and simplify.

Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Step 11

Set $5 x - 6$ equal to $0$ and solve for $x$ .

Set $5 x - 6$ equal to $0$ .

$5 x - 6 = 0$

Solve $5 x - 6 = 0$ for $x$ .

Add $6$ to both sides of the equation.

$5 x = 6$

Divide each term in $5 x = 6$ by $5$ and simplify.

Divide each term in $5 x = 6$ by $5$ .

$\frac{5 x}{5} = \frac{6}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{6}{5}$

Divide $x$ by $1$ .

$x = \frac{6}{5}$

Step 12

The final solution is all the values that make $3 (4 x + 1) (5 x - 6) = 0$ true.

$x = - \frac{1}{4}, \frac{6}{5}$

Step 13

Find the Roots/Zeros Using the Rational Roots Test 60x^2-57x-18

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