Find the Roots/Zeros Using the Rational Roots Test f(x)=9x^4-9x^3-58x^2+4x+24

$f (x) = 9 x^{4} - 9 x^{3} - 58 x^{2} + 4 x + 24$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

$q = \pm 1, \pm 3, \pm 9$

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm \frac{1}{9}, \pm 2, \pm \frac{2}{3}, \pm \frac{2}{9}, \pm 3, \pm 4, \pm \frac{4}{3}, \pm \frac{4}{9}, \pm 6, \pm 8, \pm \frac{8}{3}, \pm \frac{8}{9}, \pm 12, \pm 24$

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.

$9 {(- 2)}^{4} - 9 {(- 2)}^{3} - 58 {(- 2)}^{2} + 4 (- 2) + 24$

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 $4$ .

$9 \cdot 16 - 9 {(- 2)}^{3} - 58 {(- 2)}^{2} + 4 (- 2) + 24$

Multiply $9$ by $16$ .

$144 - 9 {(- 2)}^{3} - 58 {(- 2)}^{2} + 4 (- 2) + 24$

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

$144 - 9 \cdot - 8 - 58 {(- 2)}^{2} + 4 (- 2) + 24$

Multiply $- 9$ by $- 8$ .

$144 + 72 - 58 {(- 2)}^{2} + 4 (- 2) + 24$

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

$144 + 72 - 58 \cdot 4 + 4 (- 2) + 24$

Multiply $- 58$ by $4$ .

$144 + 72 - 232 + 4 (- 2) + 24$

Multiply $4$ by $- 2$ .

$144 + 72 - 232 - 8 + 24$

Simplify by adding and subtracting.

Add $144$ and $72$ .

$216 - 232 - 8 + 24$

Subtract $232$ from $216$ .

$- 16 - 8 + 24$

Subtract $8$ from $- 16$ .

$- 24 + 24$

Add $- 24$ and $24$ .

$0$

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.

$\frac{9 x^{4} - 9 x^{3} - 58 x^{2} + 4 x + 24}{x + 2}$

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.

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0				\xa0

The first number in the dividend $(9)$ is put into the first position of the result area (below the horizontal line).

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0				\xa0
	$9$

Multiply the newest entry in the result $(9)$ by the divisor $(- 2)$ and place the result of $(- 18)$ under the next term in the dividend $(- 9)$ .

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0	$- 18$			\xa0
	$9$

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

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0	$- 18$			\xa0
	$9$	$- 27$

Multiply the newest entry in the result $(- 27)$ by the divisor $(- 2)$ and place the result of $(54)$ under the next term in the dividend $(- 58)$ .

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0	$- 18$	$54$		\xa0
	$9$	$- 27$

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

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0	$- 18$	$54$		\xa0
	$9$	$- 27$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(- 2)$ and place the result of $(8)$ under the next term in the dividend $(4)$ .

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0	$- 18$	$54$	$8$	\xa0
	$9$	$- 27$	$- 4$

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

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0	$- 18$	$54$	$8$	\xa0
	$9$	$- 27$	$- 4$	$12$

Multiply the newest entry in the result $(12)$ by the divisor $(- 2)$ and place the result of $(- 24)$ under the next term in the dividend $(24)$ .

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0	$- 18$	$54$	$8$	$- 24$
	$9$	$- 27$	$- 4$	$12$

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

$- 2$	$9$	$- 9$	$- 58$	$4$	$24$
	\xa0	$- 18$	$54$	$8$	$- 24$
	$9$	$- 27$	$- 4$	$12$	$0$

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

$9 x^{3} + - 27 x^{2} + (- 4) x + 12$

Simplify the quotient polynomial.

$9 x^{3} - 27 x^{2} - 4 x + 12$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 2) (9 x^{3} - 27 x^{2}) - 4 x + 12$

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

$(x + 2) + 9 x^{2} (x - 3) - 4 (x - 3)$

$(x + 2) \cdot 9 x^{2} (x - 3) - 4 (x - 3)$

Step 8

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

$(x + 2) (x - 3) (9 x^{2} - 4)$

Step 9

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$(x + 2) (x - 3) ({(3 x)}^{2} - 4)$

Step 10

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

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

Step 11

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

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

Step 12

Factor the left side of the equation.

Regroup terms.

$- 9 x^{3} + 4 x + 9 x^{4} - 58 x^{2} + 24 = 0$

Factor $- x$ out of $- 9 x^{3} + 4 x$ .

Factor $- x$ out of $- 9 x^{3}$ .

$- x (9 x^{2}) + 4 x + 9 x^{4} - 58 x^{2} + 24 = 0$

Factor $- x$ out of $4 x$ .

$- x (9 x^{2}) - x \cdot - 4 + 9 x^{4} - 58 x^{2} + 24 = 0$

Factor $- x$ out of $- x (9 x^{2}) - x (- 4)$ .

$- x (9 x^{2} - 4) + 9 x^{4} - 58 x^{2} + 24 = 0$

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$- x ({(3 x)}^{2} - 4) + 9 x^{4} - 58 x^{2} + 24 = 0$

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

$- x ({(3 x)}^{2} - 2^{2}) + 9 x^{4} - 58 x^{2} + 24 = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 2$ .

$- x ((3 x + 2) (3 x - 2)) + 9 x^{4} - 58 x^{2} + 24 = 0$

Remove unnecessary parentheses.

$- x (3 x + 2) (3 x - 2) + 9 x^{4} - 58 x^{2} + 24 = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- x (3 x + 2) (3 x - 2) + 9 {(x^{2})}^{2} - 58 x^{2} + 24 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- x (3 x + 2) (3 x - 2) + 9 u^{2} - 58 u + 24 = 0$

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 $a \cdot c = 9 \cdot 24 = 216$ and whose sum is $b = - 58$ .

Factor $- 58$ out of $- 58 u$ .

$- x (3 x + 2) (3 x - 2) + 9 u^{2} - 58 u + 24 = 0$

Rewrite $- 58$ as $- 4$ plus $- 54$

$- x (3 x + 2) (3 x - 2) + 9 u^{2} + (- 4 - 54) u + 24 = 0$

Apply the distributive property.

$- x (3 x + 2) (3 x - 2) + 9 u^{2} - 4 u - 54 u + 24 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- x (3 x + 2) (3 x - 2) + 9 u^{2} - 4 u - 54 u + 24 = 0$

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

$- x (3 x + 2) (3 x - 2) + u (9 u - 4) - 6 (9 u - 4) = 0$

Factor the polynomial by factoring out the greatest common factor, $9 u - 4$ .

$- x (3 x + 2) (3 x - 2) + (9 u - 4) (u - 6) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

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

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 2$ .

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

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

Factor $(3 x + 2) (3 x - 2)$ out of $- x (3 x + 2) (3 x - 2)$ .

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

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

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

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

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

Factor $- u + u^{2} - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Factor.

Replace all occurrences of $u$ with $x$ .

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

Remove unnecessary parentheses.

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

Step 13

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

$3 x + 2 = 0$

$3 x - 2 = 0$

$x - 3 = 0$

$x + 2 = 0$

Step 14

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 $x$ .

Subtract $2$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 15

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 16

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 17

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 18

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

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

Step 19

Find the Roots/Zeros Using the Rational Roots Test f(x)=9x^4-9x^3-58x^2+4x+24

Related Questions

Support LCMGCF.com by disabling your adblocker.