Find the Roots/Zeros Using the Rational Roots Test t(x)=x^4-7x^3+3x^2+63x-108

$t (x) = x^{4} - 7 x^{3} + 3 x^{2} + 63 x - 108$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 27, \pm 36, \pm 54, \pm 108$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 27, \pm 36, \pm 54, \pm 108$

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.

${(3)}^{4} - 7 {(3)}^{3} + 3 {(3)}^{2} + 63 (3) - 108$

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 $4$ .

$81 - 7 {(3)}^{3} + 3 {(3)}^{2} + 63 (3) - 108$

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

$81 - 7 \cdot 27 + 3 {(3)}^{2} + 63 (3) - 108$

Multiply $- 7$ by $27$ .

$81 - 189 + 3 {(3)}^{2} + 63 (3) - 108$

Multiply $3$ by ${(3)}^{2}$ by adding the exponents.

Multiply $3$ by ${(3)}^{2}$ .

Raise $3$ to the power of $1$ .

$81 - 189 + 3^{1} {(3)}^{2} + 63 (3) - 108$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$81 - 189 + 3^{1 + 2} + 63 (3) - 108$

Add $1$ and $2$ .

$81 - 189 + 3^{3} + 63 (3) - 108$

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

$81 - 189 + 27 + 63 (3) - 108$

Multiply $63$ by $3$ .

$81 - 189 + 27 + 189 - 108$

Simplify by adding and subtracting.

Subtract $189$ from $81$ .

$- 108 + 27 + 189 - 108$

Add $- 108$ and $27$ .

$- 81 + 189 - 108$

Add $- 81$ and $189$ .

$108 - 108$

Subtract $108$ from $108$ .

$0$

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.

$\frac{x^{4} - 7 x^{3} + 3 x^{2} + 63 x - 108}{x - 3}$

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.

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\xa0				\xa0

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

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\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 $(- 7)$ .

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\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.

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\xa0	$3$			\xa0
	$1$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(3)$ and place the result of $(- 12)$ under the next term in the dividend $(3)$ .

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\xa0	$3$	$- 12$		\xa0
	$1$	$- 4$

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

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\xa0	$3$	$- 12$		\xa0
	$1$	$- 4$	$- 9$

Multiply the newest entry in the result $(- 9)$ by the divisor $(3)$ and place the result of $(- 27)$ under the next term in the dividend $(63)$ .

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\xa0	$3$	$- 12$	$- 27$	\xa0
	$1$	$- 4$	$- 9$

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

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\xa0	$3$	$- 12$	$- 27$	\xa0
	$1$	$- 4$	$- 9$	$36$

Multiply the newest entry in the result $(36)$ by the divisor $(3)$ and place the result of $(108)$ under the next term in the dividend $(- 108)$ .

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\xa0	$3$	$- 12$	$- 27$	$108$
	$1$	$- 4$	$- 9$	$36$

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

$3$	$1$	$- 7$	$3$	$63$	$- 108$
	\xa0	$3$	$- 12$	$- 27$	$108$
	$1$	$- 4$	$- 9$	$36$	$0$

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

$1 x^{3} + - 4 x^{2} + (- 9) x + 36$

Simplify the quotient polynomial.

$x^{3} - 4 x^{2} - 9 x + 36$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 3) (x^{3} - 4 x^{2}) - 9 x + 36$

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

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

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

Step 8

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

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

Step 9

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

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

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 $b = 3$ .

$(x - 3) + (x - 4) ((x + 3) (x - 3))$

Remove unnecessary parentheses.

$(x - 3) + (x - 4) (x + 3) (x - 3)$

$(x - 3) (x - 4) (x + 3) (x - 3)$

Step 11

Factor the left side of the equation.

Regroup terms.

$- 7 x^{3} + 63 x + x^{4} + 3 x^{2} - 108 = 0$

Factor $- 7 x$ out of $- 7 x^{3} + 63 x$ .

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

$- 7 x \cdot x^{2} + 63 x + x^{4} + 3 x^{2} - 108 = 0$

Factor $- 7 x$ out of $63 x$ .

$- 7 x \cdot x^{2} - 7 x \cdot - 9 + x^{4} + 3 x^{2} - 108 = 0$

Factor $- 7 x$ out of $- 7 x (x^{2}) - 7 x (- 9)$ .

$- 7 x (x^{2} - 9) + x^{4} + 3 x^{2} - 108 = 0$

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

$- 7 x (x^{2} - 3^{2}) + x^{4} + 3 x^{2} - 108 = 0$

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 $b = 3$ .

$- 7 x ((x + 3) (x - 3)) + x^{4} + 3 x^{2} - 108 = 0$

Remove unnecessary parentheses.

$- 7 x (x + 3) (x - 3) + x^{4} + 3 x^{2} - 108 = 0$

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

$- 7 x (x + 3) (x - 3) + {(x^{2})}^{2} + 3 x^{2} - 108 = 0$

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

$- 7 x (x + 3) (x - 3) + u^{2} + 3 u - 108 = 0$

Factor $u^{2} + 3 u - 108$ 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 $- 108$ and whose sum is $3$ .

$- 9, 12$

Write the factored form using these integers.

$- 7 x (x + 3) (x - 3) + (u - 9) (u + 12) = 0$

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

$- 7 x (x + 3) (x - 3) + (x^{2} - 9) (x^{2} + 12) = 0$

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

$- 7 x (x + 3) (x - 3) + (x^{2} - 3^{2}) (x^{2} + 12) = 0$

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 $b = 3$ .

$- 7 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} + 12) = 0$

Factor $(x + 3) (x - 3)$ out of $- 7 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} + 12)$ .

Factor $(x + 3) (x - 3)$ out of $- 7 x (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (- 7 x) + (x + 3) (x - 3) (x^{2} + 12) = 0$

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (- 7 x) + (x + 3) (x - 3) (x^{2} + 12)$ .

$(x + 3) (x - 3) (- 7 x + x^{2} + 12) = 0$

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

$(x + 3) (x - 3) (- 7 u + u^{2} + 12) = 0$

Factor $- 7 u + u^{2} + 12$ 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 $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$(x + 3) (x - 3) ((u - 4) (u - 3)) = 0$

Factor.

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

$(x + 3) (x - 3) ((x - 4) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x + 3) (x - 3) (x - 4) (x - 3) = 0$

Combine exponents.

Raise $x - 3$ to the power of $1$ .

$(x + 3) ((x - 3) (x - 3)) (x - 4) = 0$

Raise $x - 3$ to the power of $1$ .

$(x + 3) ((x - 3) (x - 3)) (x - 4) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Step 12

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

$x + 3 = 0$

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

$x - 4 = 0$

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 ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 3)}^{2}$ equal to $0$ .

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

Solve ${(x - 3)}^{2} = 0$ for $x$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 15

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 16

The final solution is all the values that make $(x + 3) {(x - 3)}^{2} (x - 4) = 0$ true.

$x = - 3, 3, 4$

Step 17

Find the Roots/Zeros Using the Rational Roots Test t(x)=x^4-7x^3+3x^2+63x-108

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