Find the Roots/Zeros Using the Rational Roots Test f(x)=54x^3-135x^2+93x-18

$f (x) = 54 x^{3} - 135 x^{2} + 93 x - 18$

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.

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

$q = \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm 27, \pm 54$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{1}{9}, \pm \frac{1}{18}, \pm \frac{1}{27}, \pm \frac{1}{54}, \pm 2, \pm \frac{2}{3}, \pm \frac{2}{9}, \pm \frac{2}{27}, \pm 3, \pm \frac{3}{2}, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18$

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.

$54 {(\frac{3}{2})}^{3} - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{3}{2}$ is a root of the polynomial.

Simplify each term.

Apply the product rule to $\frac{3}{2}$ .

$54 \frac{3^{3}}{2^{3}} - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

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

$54 \frac{27}{2^{3}} - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

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

$54 (\frac{27}{8}) - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

Cancel the common factor of $2$ .

Factor $2$ out of $54$ .

$2 (27) \frac{27}{8} - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

Factor $2$ out of $8$ .

$2 \cdot 27 \frac{27}{2 \cdot 4} - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

Cancel the common factor.

$2 \cdot 27 \frac{27}{2 \cdot 4} - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

Rewrite the expression.

$27 (\frac{27}{4}) - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

Combine $27$ and $\frac{27}{4}$ .

$\frac{27 \cdot 27}{4} - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

Multiply $27$ by $27$ .

$\frac{729}{4} - 135 {(\frac{3}{2})}^{2} + 93 (\frac{3}{2}) - 18$

Apply the product rule to $\frac{3}{2}$ .

$\frac{729}{4} - 135 \frac{3^{2}}{2^{2}} + 93 (\frac{3}{2}) - 18$

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

$\frac{729}{4} - 135 \frac{9}{2^{2}} + 93 (\frac{3}{2}) - 18$

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

$\frac{729}{4} - 135 (\frac{9}{4}) + 93 (\frac{3}{2}) - 18$

Multiply $- 135 (\frac{9}{4})$ .

Combine $- 135$ and $\frac{9}{4}$ .

$\frac{729}{4} + \frac{- 135 \cdot 9}{4} + 93 (\frac{3}{2}) - 18$

Multiply $- 135$ by $9$ .

$\frac{729}{4} + \frac{- 1215}{4} + 93 (\frac{3}{2}) - 18$

Move the negative in front of the fraction.

$\frac{729}{4} - \frac{1215}{4} + 93 (\frac{3}{2}) - 18$

Multiply $93 (\frac{3}{2})$ .

Combine $93$ and $\frac{3}{2}$ .

$\frac{729}{4} - \frac{1215}{4} + \frac{93 \cdot 3}{2} - 18$

Multiply $93$ by $3$ .

$\frac{729}{4} - \frac{1215}{4} + \frac{279}{2} - 18$

Combine fractions.

Combine the numerators over the common denominator.

$- 18 + \frac{729 - 1215}{4} + \frac{279}{2}$

Subtract $1215$ from $729$ .

$- 18 + \frac{- 486}{4} + \frac{279}{2}$

Find the common denominator.

Write $- 18$ as a fraction with denominator $1$ .

$\frac{- 18}{1} + \frac{- 486}{4} + \frac{279}{2}$

Multiply $\frac{- 18}{1}$ by $\frac{4}{4}$ .

$\frac{- 18}{1} \cdot \frac{4}{4} + \frac{- 486}{4} + \frac{279}{2}$

Multiply $\frac{- 18}{1}$ by $\frac{4}{4}$ .

$\frac{- 18 \cdot 4}{4} + \frac{- 486}{4} + \frac{279}{2}$

Multiply $\frac{279}{2}$ by $\frac{2}{2}$ .

$\frac{- 18 \cdot 4}{4} + \frac{- 486}{4} + \frac{279}{2} \cdot \frac{2}{2}$

Multiply $\frac{279}{2}$ by $\frac{2}{2}$ .

$\frac{- 18 \cdot 4}{4} + \frac{- 486}{4} + \frac{279 \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{- 18 \cdot 4}{4} + \frac{- 486}{4} + \frac{279 \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{- 18 \cdot 4 - 486 + 279 \cdot 2}{4}$

Simplify each term.

Multiply $- 18$ by $4$ .

$\frac{- 72 - 486 + 279 \cdot 2}{4}$

Multiply $279$ by $2$ .

$\frac{- 72 - 486 + 558}{4}$

Simplify the expression.

Subtract $486$ from $- 72$ .

$\frac{- 558 + 558}{4}$

Add $- 558$ and $558$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

Step 5

Since $\frac{3}{2}$ is a known root, divide the polynomial by $x - \frac{3}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{54 x^{3} - 135 x^{2} + 93 x - 18}{x - \frac{3}{2}}$

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.

$\frac{3}{2}$	$54$	$- 135$	$93$	$- 18$
	\xa0			\xa0

The first number in the dividend $(54)$ is put into the first position of the result area (below the horizontal line).

$\frac{3}{2}$	$54$	$- 135$	$93$	$- 18$
	\xa0			\xa0
	$54$

Multiply the newest entry in the result $(54)$ by the divisor $(\frac{3}{2})$ and place the result of $(81)$ under the next term in the dividend $(- 135)$ .

$\frac{3}{2}$	$54$	$- 135$	$93$	$- 18$
	\xa0	$81$		\xa0
	$54$

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

$\frac{3}{2}$	$54$	$- 135$	$93$	$- 18$
	\xa0	$81$		\xa0
	$54$	$- 54$

Multiply the newest entry in the result $(- 54)$ by the divisor $(\frac{3}{2})$ and place the result of $(- 81)$ under the next term in the dividend $(93)$ .

$\frac{3}{2}$	$54$	$- 135$	$93$	$- 18$
	\xa0	$81$	$- 81$	\xa0
	$54$	$- 54$

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

$\frac{3}{2}$	$54$	$- 135$	$93$	$- 18$
	\xa0	$81$	$- 81$	\xa0
	$54$	$- 54$	$12$

Multiply the newest entry in the result $(12)$ by the divisor $(\frac{3}{2})$ and place the result of $(18)$ under the next term in the dividend $(- 18)$ .

$\frac{3}{2}$	$54$	$- 135$	$93$	$- 18$
	\xa0	$81$	$- 81$	$18$
	$54$	$- 54$	$12$

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

$\frac{3}{2}$	$54$	$- 135$	$93$	$- 18$
	\xa0	$81$	$- 81$	$18$
	$54$	$- 54$	$12$	$0$

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

$54 x^{2} + - 54 x + 12$

Simplify the quotient polynomial.

$54 x^{2} - 54 x + 12$

Step 7

Factor $6$ out of $54 x^{2} - 54 x + 12$ .

Factor $6$ out of $54 x^{2}$ .

$(x - \frac{3}{2}) (6 (9 x^{2}) - 54 x + 12)$

Factor $6$ out of $- 54 x$ .

$(x - \frac{3}{2}) (6 (9 x^{2}) + 6 (- 9 x) + 12)$

Factor $6$ out of $12$ .

$(x - \frac{3}{2}) (6 (9 x^{2}) + 6 (- 9 x) + 6 (2))$

Factor $6$ out of $6 (9 x^{2}) + 6 (- 9 x)$ .

$(x - \frac{3}{2}) (6 (9 x^{2} - 9 x) + 6 (2))$

Factor $6$ out of $6 (9 x^{2} - 9 x) + 6 (2)$ .

$(x - \frac{3}{2}) (6 (9 x^{2} - 9 x + 2))$

Step 8

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 $a \cdot c = 9 \cdot 2 = 18$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

$(x - \frac{3}{2}) (6 (9 x^{2} - 9 x + 2))$

Rewrite $- 9$ as $- 3$ plus $- 6$

$(x - \frac{3}{2}) (6 (9 x^{2} + (- 3 - 6) x + 2))$

Apply the distributive property.

$(x - \frac{3}{2}) (6 (9 x^{2} - 3 x - 6 x + 2))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - \frac{3}{2}) (6 ((9 x^{2} - 3 x) - 6 x + 2))$

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

$(x - \frac{3}{2}) (6 (3 x (3 x - 1) - 2 (3 x - 1)))$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(x - \frac{3}{2}) (6 ((3 x - 1) (3 x - 2)))$

Remove unnecessary parentheses.

$(x - \frac{3}{2}) (6 (3 x - 1) (3 x - 2))$

Step 9

Factor the left side of the equation.

Factor $3$ out of $54 x^{3} - 135 x^{2} + 93 x - 18$ .

Factor $3$ out of $54 x^{3}$ .

$3 (18 x^{3}) - 135 x^{2} + 93 x - 18 = 0$

Factor $3$ out of $- 135 x^{2}$ .

$3 (18 x^{3}) + 3 (- 45 x^{2}) + 93 x - 18 = 0$

Factor $3$ out of $93 x$ .

$3 (18 x^{3}) + 3 (- 45 x^{2}) + 3 (31 x) - 18 = 0$

Factor $3$ out of $- 18$ .

$3 (18 x^{3}) + 3 (- 45 x^{2}) + 3 (31 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (18 x^{3}) + 3 (- 45 x^{2})$ .

$3 (18 x^{3} - 45 x^{2}) + 3 (31 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (18 x^{3} - 45 x^{2}) + 3 (31 x)$ .

$3 (18 x^{3} - 45 x^{2} + 31 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (18 x^{3} - 45 x^{2} + 31 x) + 3 (- 6)$ .

$3 (18 x^{3} - 45 x^{2} + 31 x - 6) = 0$

Factor $18 x^{3} - 45 x^{2} + 31 x - 6$ 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.

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

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

$\pm 1, \pm 0.0\overline{5}, \pm 0.5, \pm 0.\overline{1}, \pm 0.\overline{3}, \pm 0.1\overline{6}, \pm 6, \pm 3, \pm 0.\overline{6}, \pm 2, \pm 0.\overline{2}, \pm 1.5$

Substitute $0.\overline{3}$ and simplify the expression. In this case, the expression is equal to $0$ so $0.\overline{3}$ is a root of the polynomial.

Substitute $0.\overline{3}$ into the polynomial.

$18 \cdot {0.\overline{3}}^{3} - 45 \cdot {0.\overline{3}}^{2} + 31 \cdot 0.\overline{3} - 6$

Raise $0.\overline{3}$ to the power of $3$ .

$18 \cdot 0.\overline{037} - 45 \cdot {0.\overline{3}}^{2} + 31 \cdot 0.\overline{3} - 6$

Multiply $18$ by $0.\overline{037}$ .

$0.\overline{6} - 45 \cdot {0.\overline{3}}^{2} + 31 \cdot 0.\overline{3} - 6$

Raise $0.\overline{3}$ to the power of $2$ .

$0.\overline{6} - 45 \cdot 0.\overline{1} + 31 \cdot 0.\overline{3} - 6$

Multiply $- 45$ by $0.\overline{1}$ .

$0.\overline{6} - 5 + 31 \cdot 0.\overline{3} - 6$

Subtract $5$ from $0.\overline{6}$ .

$- 4.33333333 + 31 \cdot 0.\overline{3} - 6$

Multiply $31$ by $0.\overline{3}$ .

$- 4.33333333 + 10.\overline{3} - 6$

Add $- 4.33333333$ and $10.\overline{3}$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

Since $0.\overline{3}$ is a known root, divide the polynomial by $3 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{18 x^{3} - 45 x^{2} + 31 x - 6}{3 x - 1}$

Divide $18 x^{3} - 45 x^{2} + 31 x - 6$ by $3 x - 1$ .

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

$3 x$

$1$

$18 x^{3}$

$45 x^{2}$

$31 x$

$6$

Divide the highest order term in the dividend $18 x^{3}$ by the highest order term in divisor $3 x$ .

					$6 x^{2}$
	$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$

Multiply the new quotient term by the divisor.

				$6 x^{2}$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			+	$18 x^{3}$	-	$6 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $18 x^{3} - 6 x^{2}$

				$6 x^{2}$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$

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

				$6 x^{2}$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$

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

				$6 x^{2}$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$

Divide the highest order term in the dividend $- 39 x^{2}$ by the highest order term in divisor $3 x$ .

				$6 x^{2}$	-	$13 x$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	-	$13 x$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$
					-	$39 x^{2}$	+	$13 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 39 x^{2} + 13 x$

				$6 x^{2}$	-	$13 x$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$
					+	$39 x^{2}$	-	$13 x$

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

				$6 x^{2}$	-	$13 x$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$
					+	$39 x^{2}$	-	$13 x$
							+	$18 x$

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

				$6 x^{2}$	-	$13 x$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$
					+	$39 x^{2}$	-	$13 x$
							+	$18 x$	-	$6$

Divide the highest order term in the dividend $18 x$ by the highest order term in divisor $3 x$ .

				$6 x^{2}$	-	$13 x$	+	$6$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$
					+	$39 x^{2}$	-	$13 x$
							+	$18 x$	-	$6$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	-	$13 x$	+	$6$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$
					+	$39 x^{2}$	-	$13 x$
							+	$18 x$	-	$6$
							+	$18 x$	-	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $18 x - 6$

				$6 x^{2}$	-	$13 x$	+	$6$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$
					+	$39 x^{2}$	-	$13 x$
							+	$18 x$	-	$6$
							-	$18 x$	+	$6$

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

				$6 x^{2}$	-	$13 x$	+	$6$
$3 x$	-	$1$		$18 x^{3}$	-	$45 x^{2}$	+	$31 x$	-	$6$
			-	$18 x^{3}$	+	$6 x^{2}$
					-	$39 x^{2}$	+	$31 x$
					+	$39 x^{2}$	-	$13 x$
							+	$18 x$	-	$6$
							-	$18 x$	+	$6$
										$0$

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

$6 x^{2} - 13 x + 6$

Write $18 x^{3} - 45 x^{2} + 31 x - 6$ as a set of factors.

$3 ((3 x - 1) (6 x^{2} - 13 x + 6)) = 0$

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 $a \cdot c = 6 \cdot 6 = 36$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 x$ .

$3 ((3 x - 1) (6 x^{2} - 13 x + 6)) = 0$

Rewrite $- 13$ as $- 4$ plus $- 9$

$3 ((3 x - 1) (6 x^{2} + (- 4 - 9) x + 6)) = 0$

Apply the distributive property.

$3 ((3 x - 1) (6 x^{2} - 4 x - 9 x + 6)) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$3 ((3 x - 1) ((6 x^{2} - 4 x) - 9 x + 6)) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Step 10

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

$3 x - 1 = 0$

$3 x - 2 = 0$

$2 x - 3 = 0$

Step 11

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 12

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve $3 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 13

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 14

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

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

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=54x^3-135x^2+93x-18

Related Questions

Support LCMGCF.com by disabling your adblocker.