Find the Roots/Zeros Using the Rational Roots Test x^4+6x^3-16x^2-54x+63

$x^{4} + 6 x^{3} - 16 x^{2} - 54 x + 63$

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.

$p = \pm 1, \pm 3, \pm 7, \pm 9, \pm 21, \pm 63$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 3, \pm 7, \pm 9, \pm 21, \pm 63$

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.

${(1)}^{4} + 6 {(1)}^{3} - 16 {(1)}^{2} - 54 \cdot 1 + 63$

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.

$1 + 6 {(1)}^{3} - 16 {(1)}^{2} - 54 \cdot 1 + 63$

One to any power is one.

$1 + 6 \cdot 1 - 16 {(1)}^{2} - 54 \cdot 1 + 63$

Multiply $6$ by $1$ .

$1 + 6 - 16 {(1)}^{2} - 54 \cdot 1 + 63$

One to any power is one.

$1 + 6 - 16 \cdot 1 - 54 \cdot 1 + 63$

Multiply $- 16$ by $1$ .

$1 + 6 - 16 - 54 \cdot 1 + 63$

Multiply $- 54$ by $1$ .

$1 + 6 - 16 - 54 + 63$

Simplify by adding and subtracting.

Add $1$ and $6$ .

$7 - 16 - 54 + 63$

Subtract $16$ from $7$ .

$- 9 - 54 + 63$

Subtract $54$ from $- 9$ .

$- 63 + 63$

Add $- 63$ and $63$ .

$0$

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.

$\frac{x^{4} + 6 x^{3} - 16 x^{2} - 54 x + 63}{x - 1}$

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.

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\xa0				\xa0

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

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\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 $(6)$ .

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\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.

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\xa0	$1$			\xa0
	$1$	$7$

Multiply the newest entry in the result $(7)$ by the divisor $(1)$ and place the result of $(7)$ under the next term in the dividend $(- 16)$ .

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\xa0	$1$	$7$		\xa0
	$1$	$7$

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

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\xa0	$1$	$7$		\xa0
	$1$	$7$	$- 9$

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 $(- 54)$ .

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\xa0	$1$	$7$	$- 9$	\xa0
	$1$	$7$	$- 9$

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

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\xa0	$1$	$7$	$- 9$	\xa0
	$1$	$7$	$- 9$	$- 63$

Multiply the newest entry in the result $(- 63)$ by the divisor $(1)$ and place the result of $(- 63)$ under the next term in the dividend $(63)$ .

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\xa0	$1$	$7$	$- 9$	$- 63$
	$1$	$7$	$- 9$	$- 63$

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

$1$	$1$	$6$	$- 16$	$- 54$	$63$
	\xa0	$1$	$7$	$- 9$	$- 63$
	$1$	$7$	$- 9$	$- 63$	$0$

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

$1 x^{3} + 7 x^{2} + (- 9) x - 63$

Simplify the quotient polynomial.

$x^{3} + 7 x^{2} - 9 x - 63$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (x^{3} + 7 x^{2}) - 9 x - 63$

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

$(x - 1) + x^{2} (x + 7) - 9 (x + 7)$

$(x - 1) x^{2} (x + 7) - 9 (x + 7)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 7$ .

$(x - 1) (x + 7) (x^{2} - 9)$

Step 9

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

$(x - 1) (x + 7) (x^{2} - 3^{2})$

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 $b = 3$ .

$(x - 1) + (x + 7) ((x + 3) (x - 3))$

Remove unnecessary parentheses.

$(x - 1) + (x + 7) (x + 3) (x - 3)$

$(x - 1) (x + 7) (x + 3) (x - 3)$

Step 11

Factor the left side of the equation.

Regroup terms.

$6 x^{3} - 54 x + x^{4} - 16 x^{2} + 63 = 0$

Factor $6 x$ out of $6 x^{3} - 54 x$ .

Factor $6 x$ out of $6 x^{3}$ .

$6 x (x^{2}) - 54 x + x^{4} - 16 x^{2} + 63 = 0$

Factor $6 x$ out of $- 54 x$ .

$6 x (x^{2}) + 6 x (- 9) + x^{4} - 16 x^{2} + 63 = 0$

Factor $6 x$ out of $6 x (x^{2}) + 6 x (- 9)$ .

$6 x (x^{2} - 9) + x^{4} - 16 x^{2} + 63 = 0$

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

$6 x (x^{2} - 3^{2}) + x^{4} - 16 x^{2} + 63 = 0$

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 $b = 3$ .

$6 x ((x + 3) (x - 3)) + x^{4} - 16 x^{2} + 63 = 0$

Remove unnecessary parentheses.

$6 x (x + 3) (x - 3) + x^{4} - 16 x^{2} + 63 = 0$

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

$6 x (x + 3) (x - 3) + {(x^{2})}^{2} - 16 x^{2} + 63 = 0$

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

$6 x (x + 3) (x - 3) + u^{2} - 16 u + 63 = 0$

Factor $u^{2} - 16 u + 63$ 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 $63$ and whose sum is $- 16$ .

$- 9, - 7$

Write the factored form using these integers.

$6 x (x + 3) (x - 3) + (u - 9) (u - 7) = 0$

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

$6 x (x + 3) (x - 3) + (x^{2} - 9) (x^{2} - 7) = 0$

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

$6 x (x + 3) (x - 3) + (x^{2} - 3^{2}) (x^{2} - 7) = 0$

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 $b = 3$ .

$6 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} - 7) = 0$

Factor $(x + 3) (x - 3)$ out of $6 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} - 7)$ .

Factor $(x + 3) (x - 3)$ out of $6 x (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (6 x) + (x + 3) (x - 3) (x^{2} - 7) = 0$

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (6 x) + (x + 3) (x - 3) (x^{2} - 7)$ .

$(x + 3) (x - 3) (6 x + x^{2} - 7) = 0$

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

$(x + 3) (x - 3) (6 u + u^{2} - 7) = 0$

Factor $6 u + u^{2} - 7$ 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 $- 7$ and whose sum is $6$ .

$- 1, 7$

Write the factored form using these integers.

$(x + 3) (x - 3) ((u - 1) (u + 7)) = 0$

Factor.

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

$(x + 3) (x - 3) ((x - 1) (x + 7)) = 0$

Remove unnecessary parentheses.

$(x + 3) (x - 3) (x - 1) (x + 7) = 0$

Step 12

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

$x + 3 = 0$

$x - 3 = 0$

$x - 1 = 0$

$x + 7 = 0$

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 $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 17

The final solution is all the values that make $(x + 3) (x - 3) (x - 1) (x + 7) = 0$ true.

$x = - 3, 3, 1, - 7$

Step 18

Find the Roots/Zeros Using the Rational Roots Test x^4+6x^3-16x^2-54x+63

Related Questions

Support LCMGCF.com by disabling your adblocker.