Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+10x^3-20x^2-90x+99

$f (x) = x^{4} + 10 x^{3} - 20 x^{2} - 90 x + 99$

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.

$p = \pm 1, \pm 3, \pm 9, \pm 11, \pm 33, \pm 99$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 3, \pm 9, \pm 11, \pm 33, \pm 99$

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.

${(1)}^{4} + 10 {(1)}^{3} - 20 {(1)}^{2} - 90 \cdot 1 + 99$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

Simplify each term.

One to any power is one.

$1 + 10 {(1)}^{3} - 20 {(1)}^{2} - 90 \cdot 1 + 99$

One to any power is one.

$1 + 10 \cdot 1 - 20 {(1)}^{2} - 90 \cdot 1 + 99$

Multiply $10$ by $1$ .

$1 + 10 - 20 {(1)}^{2} - 90 \cdot 1 + 99$

One to any power is one.

$1 + 10 - 20 \cdot 1 - 90 \cdot 1 + 99$

Multiply $- 20$ by $1$ .

$1 + 10 - 20 - 90 \cdot 1 + 99$

Multiply $- 90$ by $1$ .

$1 + 10 - 20 - 90 + 99$

Simplify by adding and subtracting.

Add $1$ and $10$ .

$11 - 20 - 90 + 99$

Subtract $20$ from $11$ .

$- 9 - 90 + 99$

Subtract $90$ from $- 9$ .

$- 99 + 99$

Add $- 99$ and $99$ .

$0$

Step 5

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

$\frac{x^{4} + 10 x^{3} - 20 x^{2} - 90 x + 99}{x - 1}$

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.

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0				\xa0

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

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0				\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(10)$ .

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0	$1$			\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.

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0	$1$			\xa0
	$1$	$11$

Multiply the newest entry in the result $(11)$ by the divisor $(1)$ and place the result of $(11)$ under the next term in the dividend $(- 20)$ .

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0	$1$	$11$		\xa0
	$1$	$11$

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

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0	$1$	$11$		\xa0
	$1$	$11$	$- 9$

Multiply the newest entry in the result $(- 9)$ by the divisor $(1)$ and place the result of $(- 9)$ under the next term in the dividend $(- 90)$ .

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0	$1$	$11$	$- 9$	\xa0
	$1$	$11$	$- 9$

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

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0	$1$	$11$	$- 9$	\xa0
	$1$	$11$	$- 9$	$- 99$

Multiply the newest entry in the result $(- 99)$ by the divisor $(1)$ and place the result of $(- 99)$ under the next term in the dividend $(99)$ .

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0	$1$	$11$	$- 9$	$- 99$
	$1$	$11$	$- 9$	$- 99$

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

$1$	$1$	$10$	$- 20$	$- 90$	$99$
	\xa0	$1$	$11$	$- 9$	$- 99$
	$1$	$11$	$- 9$	$- 99$	$0$

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

$1 x^{3} + 11 x^{2} + (- 9) x - 99$

Simplify the quotient polynomial.

$x^{3} + 11 x^{2} - 9 x - 99$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (x^{3} + 11 x^{2}) - 9 x - 99$

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

$(x - 1) + x^{2} (x + 11) - 9 (x + 11)$

$(x - 1) x^{2} (x + 11) - 9 (x + 11)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 11$ .

$(x - 1) (x + 11) (x^{2} - 9)$

Step 9

Rewrite $9$ as $3^{2}$ .

$(x - 1) (x + 11) (x^{2} - 3^{2})$

Step 10

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 $b = 3$ .

$(x - 1) + (x + 11) ((x + 3) (x - 3))$

Remove unnecessary parentheses.

$(x - 1) + (x + 11) (x + 3) (x - 3)$

$(x - 1) (x + 11) (x + 3) (x - 3)$

Step 11

Factor the left side of the equation.

Regroup terms.

$10 x^{3} - 90 x + x^{4} - 20 x^{2} + 99 = 0$

Factor $10 x$ out of $10 x^{3} - 90 x$ .

Factor $10 x$ out of $10 x^{3}$ .

$10 x (x^{2}) - 90 x + x^{4} - 20 x^{2} + 99 = 0$

Factor $10 x$ out of $- 90 x$ .

$10 x (x^{2}) + 10 x (- 9) + x^{4} - 20 x^{2} + 99 = 0$

Factor $10 x$ out of $10 x (x^{2}) + 10 x (- 9)$ .

$10 x (x^{2} - 9) + x^{4} - 20 x^{2} + 99 = 0$

Rewrite $9$ as $3^{2}$ .

$10 x (x^{2} - 3^{2}) + x^{4} - 20 x^{2} + 99 = 0$

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 $b = 3$ .

$10 x ((x + 3) (x - 3)) + x^{4} - 20 x^{2} + 99 = 0$

Remove unnecessary parentheses.

$10 x (x + 3) (x - 3) + x^{4} - 20 x^{2} + 99 = 0$

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

$10 x (x + 3) (x - 3) + {(x^{2})}^{2} - 20 x^{2} + 99 = 0$

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

$10 x (x + 3) (x - 3) + u^{2} - 20 u + 99 = 0$

Factor $u^{2} - 20 u + 99$ 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 $99$ and whose sum is $- 20$ .

$- 11, - 9$

Write the factored form using these integers.

$10 x (x + 3) (x - 3) + (u - 11) (u - 9) = 0$

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

$10 x (x + 3) (x - 3) + (x^{2} - 11) (x^{2} - 9) = 0$

Rewrite $9$ as $3^{2}$ .

$10 x (x + 3) (x - 3) + (x^{2} - 11) (x^{2} - 3^{2}) = 0$

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 $b = 3$ .

$10 x (x + 3) (x - 3) + (x^{2} - 11) ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$10 x (x + 3) (x - 3) + (x^{2} - 11) (x + 3) (x - 3) = 0$

Factor $(x + 3) (x - 3)$ out of $10 x (x + 3) (x - 3) + (x^{2} - 11) (x + 3) (x - 3)$ .

Factor $(x + 3) (x - 3)$ out of $10 x (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (10 x) + (x^{2} - 11) (x + 3) (x - 3) = 0$

Factor $(x + 3) (x - 3)$ out of $(x^{2} - 11) (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (10 x) + (x + 3) (x - 3) (x^{2} - 11) = 0$

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (10 x) + (x + 3) (x - 3) (x^{2} - 11)$ .

$(x + 3) (x - 3) (10 x + x^{2} - 11) = 0$

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

$(x + 3) (x - 3) (10 u + u^{2} - 11) = 0$

Factor $10 u + u^{2} - 11$ 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 $- 11$ and whose sum is $10$ .

$- 1, 11$

Write the factored form using these integers.

$(x + 3) (x - 3) ((u - 1) (u + 11)) = 0$

Factor.

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

$(x + 3) (x - 3) ((x - 1) (x + 11)) = 0$

Remove unnecessary parentheses.

$(x + 3) (x - 3) (x - 1) (x + 11) = 0$

Step 12

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

$x + 3 = 0$

$x - 3 = 0$

$x - 1 = 0$

$x + 11 = 0$

Step 13

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 14

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 15

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 16

Set $x + 11$ equal to $0$ and solve for $x$ .

Set $x + 11$ equal to $0$ .

$x + 11 = 0$

Subtract $11$ from both sides of the equation.

$x = - 11$

Step 17

The final solution is all the values that make $(x + 3) (x - 3) (x - 1) (x + 11) = 0$ true.

$x = - 3, 3, 1, - 11$

Step 18

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+10x^3-20x^2-90x+99

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