Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+7x^3-22x^2+56x-240

$f (x) = x^{4} + 7 x^{3} - 22 x^{2} + 56 x - 240$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 16, \pm 20, \pm 24, \pm 30, \pm 40, \pm 48, \pm 60, \pm 80, \pm 120, \pm 240$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 16, \pm 20, \pm 24, \pm 30, \pm 40, \pm 48, \pm 60, \pm 80, \pm 120, \pm 240$

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.

${(3)}^{4} + 7 {(3)}^{3} - 22 {(3)}^{2} + 56 (3) - 240$

Step 4

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

Simplify each term.

Raise $3$ to the power of $4$ .

$81 + 7 {(3)}^{3} - 22 {(3)}^{2} + 56 (3) - 240$

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

$81 + 7 \cdot 27 - 22 {(3)}^{2} + 56 (3) - 240$

Multiply $7$ by $27$ .

$81 + 189 - 22 {(3)}^{2} + 56 (3) - 240$

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

$81 + 189 - 22 \cdot 9 + 56 (3) - 240$

Multiply $- 22$ by $9$ .

$81 + 189 - 198 + 56 (3) - 240$

Multiply $56$ by $3$ .

$81 + 189 - 198 + 168 - 240$

Simplify by adding and subtracting.

Add $81$ and $189$ .

$270 - 198 + 168 - 240$

Subtract $198$ from $270$ .

$72 + 168 - 240$

Add $72$ and $168$ .

$240 - 240$

Subtract $240$ from $240$ .

$0$

Step 5

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

$\frac{x^{4} + 7 x^{3} - 22 x^{2} + 56 x - 240}{x - 3}$

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.

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0				\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0				\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(3)$ and place the result of $(3)$ under the next term in the dividend $(7)$ .

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0	$3$			\xa0
	$1$

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

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0	$3$			\xa0
	$1$	$10$

Multiply the newest entry in the result $(10)$ by the divisor $(3)$ and place the result of $(30)$ under the next term in the dividend $(- 22)$ .

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0	$3$	$30$		\xa0
	$1$	$10$

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

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0	$3$	$30$		\xa0
	$1$	$10$	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(3)$ and place the result of $(24)$ under the next term in the dividend $(56)$ .

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0	$3$	$30$	$24$	\xa0
	$1$	$10$	$8$

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

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0	$3$	$30$	$24$	\xa0
	$1$	$10$	$8$	$80$

Multiply the newest entry in the result $(80)$ by the divisor $(3)$ and place the result of $(240)$ under the next term in the dividend $(- 240)$ .

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0	$3$	$30$	$24$	$240$
	$1$	$10$	$8$	$80$

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

$3$	$1$	$7$	$- 22$	$56$	$- 240$
	\xa0	$3$	$30$	$24$	$240$
	$1$	$10$	$8$	$80$	$0$

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

$1 x^{3} + 10 x^{2} + (8) x + 80$

Simplify the quotient polynomial.

$x^{3} + 10 x^{2} + 8 x + 80$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 3) (x^{3} + 10 x^{2}) + 8 x + 80$

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

$(x - 3) + x^{2} (x + 10) + 8 (x + 10)$

$(x - 3) x^{2} (x + 10) + 8 (x + 10)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 10$ .

$(x - 3) (x + 10) (x^{2} + 8)$

Step 9

Factor the left side of the equation.

Regroup terms.

$7 x^{3} + 56 x + x^{4} - 22 x^{2} - 240 = 0$

Factor $7 x$ out of $7 x^{3} + 56 x$ .

Factor $7 x$ out of $7 x^{3}$ .

$7 x (x^{2}) + 56 x + x^{4} - 22 x^{2} - 240 = 0$

Factor $7 x$ out of $56 x$ .

$7 x (x^{2}) + 7 x (8) + x^{4} - 22 x^{2} - 240 = 0$

Factor $7 x$ out of $7 x (x^{2}) + 7 x (8)$ .

$7 x (x^{2} + 8) + x^{4} - 22 x^{2} - 240 = 0$

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

$7 x (x^{2} + 8) + {(x^{2})}^{2} - 22 x^{2} - 240 = 0$

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

$7 x (x^{2} + 8) + u^{2} - 22 u - 240 = 0$

Factor $u^{2} - 22 u - 240$ 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 $- 240$ and whose sum is $- 22$ .

$- 30, 8$

Write the factored form using these integers.

$7 x (x^{2} + 8) + (u - 30) (u + 8) = 0$

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

$7 x (x^{2} + 8) + (x^{2} - 30) (x^{2} + 8) = 0$

Factor $x^{2} + 8$ out of $7 x (x^{2} + 8) + (x^{2} - 30) (x^{2} + 8)$ .

Factor $x^{2} + 8$ out of $7 x (x^{2} + 8)$ .

$(x^{2} + 8) (7 x) + (x^{2} - 30) (x^{2} + 8) = 0$

Factor $x^{2} + 8$ out of $(x^{2} - 30) (x^{2} + 8)$ .

$(x^{2} + 8) (7 x) + (x^{2} + 8) (x^{2} - 30) = 0$

Factor $x^{2} + 8$ out of $(x^{2} + 8) (7 x) + (x^{2} + 8) (x^{2} - 30)$ .

$(x^{2} + 8) (7 x + x^{2} - 30) = 0$

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

$(x^{2} + 8) (7 u + u^{2} - 30) = 0$

Factor $7 u + u^{2} - 30$ 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 $- 30$ and whose sum is $7$ .

$- 3, 10$

Write the factored form using these integers.

$(x^{2} + 8) ((u - 3) (u + 10)) = 0$

Factor.

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

$(x^{2} + 8) ((x - 3) (x + 10)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 8) (x - 3) (x + 10) = 0$

Step 10

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

$x^{2} + 8 = 0$

$x - 3 = 0$

$x + 10 = 0$

Step 11

Set $x^{2} + 8$ equal to $0$ and solve for $x$ .

Set $x^{2} + 8$ equal to $0$ .

$x^{2} + 8 = 0$

Solve $x^{2} + 8 = 0$ for $x$ .

Subtract $8$ from both sides of the equation.

$x^{2} = - 8$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 8}$

Simplify $\pm \sqrt{- 8}$ .

Rewrite $- 8$ as $- 1 (8)$ .

$x = \pm \sqrt{- 1 (8)}$

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{8}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{8}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \pm i \cdot \sqrt{4 (2)}$

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

$x = \pm i \cdot \sqrt{2^{2} \cdot 2}$

Pull terms out from under the radical.

$x = \pm i \cdot (2 \sqrt{2})$

Move $2$ to the left of $i$ .

$x = \pm 2 i \sqrt{2}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2 i \sqrt{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2 i \sqrt{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2 i \sqrt{2}, - 2 i \sqrt{2}$

Step 12

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 13

Set $x + 10$ equal to $0$ and solve for $x$ .

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

Step 14

The final solution is all the values that make $(x^{2} + 8) (x - 3) (x + 10) = 0$ true.

$x = 2 i \sqrt{2}, - 2 i \sqrt{2}, 3, - 10$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+7x^3-22x^2+56x-240

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