Find the Roots/Zeros Using the Rational Roots Test 3x^3-5x^2-12x+20

$3 x^{3} - 5 x^{2} - 12 x + 20$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$

$q = \pm 1, \pm 3$

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 5, \pm \frac{5}{3}, \pm 10, \pm \frac{10}{3}, \pm 20, \pm \frac{20}{3}$

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.

$3 {(2)}^{3} - 5 {(2)}^{2} - 12 \cdot 2 + 20$

Step 4

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

Simplify each term.

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

$3 \cdot 8 - 5 {(2)}^{2} - 12 \cdot 2 + 20$

Multiply $3$ by $8$ .

$24 - 5 {(2)}^{2} - 12 \cdot 2 + 20$

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

$24 - 5 \cdot 4 - 12 \cdot 2 + 20$

Multiply $- 5$ by $4$ .

$24 - 20 - 12 \cdot 2 + 20$

Multiply $- 12$ by $2$ .

$24 - 20 - 24 + 20$

Simplify by adding and subtracting.

Subtract $20$ from $24$ .

$4 - 24 + 20$

Subtract $24$ from $4$ .

$- 20 + 20$

Add $- 20$ and $20$ .

$0$

Step 5

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

$\frac{3 x^{3} - 5 x^{2} - 12 x + 20}{x - 2}$

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.

$2$	$3$	$- 5$	$- 12$	$20$
	\xa0			\xa0

The first number in the dividend $(3)$ is put into the first position of the result area (below the horizontal line).

$2$	$3$	$- 5$	$- 12$	$20$
	\xa0			\xa0
	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(2)$ and place the result of $(6)$ under the next term in the dividend $(- 5)$ .

$2$	$3$	$- 5$	$- 12$	$20$
	\xa0	$6$		\xa0
	$3$

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

$2$	$3$	$- 5$	$- 12$	$20$
	\xa0	$6$		\xa0
	$3$	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(2)$ and place the result of $(2)$ under the next term in the dividend $(- 12)$ .

$2$	$3$	$- 5$	$- 12$	$20$
	\xa0	$6$	$2$	\xa0
	$3$	$1$

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

$2$	$3$	$- 5$	$- 12$	$20$
	\xa0	$6$	$2$	\xa0
	$3$	$1$	$- 10$

Multiply the newest entry in the result $(- 10)$ by the divisor $(2)$ and place the result of $(- 20)$ under the next term in the dividend $(20)$ .

$2$	$3$	$- 5$	$- 12$	$20$
	\xa0	$6$	$2$	$- 20$
	$3$	$1$	$- 10$

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

$2$	$3$	$- 5$	$- 12$	$20$
	\xa0	$6$	$2$	$- 20$
	$3$	$1$	$- 10$	$0$

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

$3 x^{2} + 1 x - 10$

Simplify the quotient polynomial.

$3 x^{2} + x - 10$

Step 7

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 $a \cdot c = 3 \cdot - 10 = - 30$ and whose sum is $b = 1$ .

Multiply by $1$ .

$(x - 2) + 3 x^{2} + 1 x - 10$

Rewrite $1$ as $- 5$ plus $6$

$(x - 2) + 3 x^{2} + (- 5 + 6) x - 10$

Apply the distributive property.

$(x - 2) + 3 x^{2} - 5 x + 6 x - 10$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 2) (3 x^{2} - 5 x) + 6 x - 10$

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

$(x - 2) + x (3 x - 5) + 2 (3 x - 5)$

Factor the polynomial by factoring out the greatest common factor, $3 x - 5$ .

$(x - 2) + (3 x - 5) (x + 2)$

$(x - 2) (3 x - 5) (x + 2)$

Step 8

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{3} - 5 x^{2}) - 12 x + 20 = 0$

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

$x^{2} (3 x - 5) - 4 (3 x - 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 5$ .

$(3 x - 5) (x^{2} - 4) = 0$

Rewrite $4$ as $2^{2}$ .

$(3 x - 5) (x^{2} - 2^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

Step 9

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

$3 x - 5 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 10

Set $3 x - 5$ equal to $0$ and solve for $x$ .

Set $3 x - 5$ equal to $0$ .

$3 x - 5 = 0$

Solve $3 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$3 x = 5$

Divide each term in $3 x = 5$ by $3$ and simplify.

Divide each term in $3 x = 5$ by $3$ .

$\frac{3 x}{3} = \frac{5}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{5}{3}$

Divide $x$ by $1$ .

$x = \frac{5}{3}$

Step 11

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 12

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 13

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

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

Step 14

Find the Roots/Zeros Using the Rational Roots Test 3x^3-5x^2-12x+20

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