Find the Roots/Zeros Using the Rational Roots Test x^4+6x^3-9x^2-12x+14

$x^{4} + 6 x^{3} - 9 x^{2} - 12 x + 14$

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.

$p = \pm 1, \pm 2, \pm 7, \pm 14$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 7, \pm 14$

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.

${(1)}^{4} + 6 {(1)}^{3} - 9 {(1)}^{2} - 12 \cdot 1 + 14$

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.

$1 + 6 {(1)}^{3} - 9 {(1)}^{2} - 12 \cdot 1 + 14$

One to any power is one.

$1 + 6 \cdot 1 - 9 {(1)}^{2} - 12 \cdot 1 + 14$

Multiply $6$ by $1$ .

$1 + 6 - 9 {(1)}^{2} - 12 \cdot 1 + 14$

One to any power is one.

$1 + 6 - 9 \cdot 1 - 12 \cdot 1 + 14$

Multiply $- 9$ by $1$ .

$1 + 6 - 9 - 12 \cdot 1 + 14$

Multiply $- 12$ by $1$ .

$1 + 6 - 9 - 12 + 14$

Simplify by adding and subtracting.

Add $1$ and $6$ .

$7 - 9 - 12 + 14$

Subtract $9$ from $7$ .

$- 2 - 12 + 14$

Subtract $12$ from $- 2$ .

$- 14 + 14$

Add $- 14$ and $14$ .

$0$

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.

$\frac{x^{4} + 6 x^{3} - 9 x^{2} - 12 x + 14}{x - 1}$

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.

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\xa0				\xa0

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

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\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 $(6)$ .

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\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.

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\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 $(- 9)$ .

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\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.

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\xa0	$1$	$7$		\xa0
	$1$	$7$	$- 2$

Multiply the newest entry in the result $(- 2)$ by the divisor $(1)$ and place the result of $(- 2)$ under the next term in the dividend $(- 12)$ .

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\xa0	$1$	$7$	$- 2$	\xa0
	$1$	$7$	$- 2$

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

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\xa0	$1$	$7$	$- 2$	\xa0
	$1$	$7$	$- 2$	$- 14$

Multiply the newest entry in the result $(- 14)$ by the divisor $(1)$ and place the result of $(- 14)$ under the next term in the dividend $(14)$ .

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\xa0	$1$	$7$	$- 2$	$- 14$
	$1$	$7$	$- 2$	$- 14$

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

$1$	$1$	$6$	$- 9$	$- 12$	$14$
	\xa0	$1$	$7$	$- 2$	$- 14$
	$1$	$7$	$- 2$	$- 14$	$0$

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

$1 x^{3} + 7 x^{2} + (- 2) x - 14$

Simplify the quotient polynomial.

$x^{3} + 7 x^{2} - 2 x - 14$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (x^{3} + 7 x^{2}) - 2 x - 14$

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

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

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

Step 8

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

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

Step 9

Factor the left side of the equation.

Regroup terms.

$6 x^{3} - 12 x + x^{4} - 9 x^{2} + 14 = 0$

Factor $6 x$ out of $6 x^{3} - 12 x$ .

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

$6 x (x^{2}) - 12 x + x^{4} - 9 x^{2} + 14 = 0$

Factor $6 x$ out of $- 12 x$ .

$6 x (x^{2}) + 6 x (- 2) + x^{4} - 9 x^{2} + 14 = 0$

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

$6 x (x^{2} - 2) + x^{4} - 9 x^{2} + 14 = 0$

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

$6 x (x^{2} - 2) + {(x^{2})}^{2} - 9 x^{2} + 14 = 0$

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

$6 x (x^{2} - 2) + u^{2} - 9 u + 14 = 0$

Factor $u^{2} - 9 u + 14$ 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 $14$ and whose sum is $- 9$ .

$- 7, - 2$

Write the factored form using these integers.

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

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

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

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

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

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

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

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

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

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

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

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

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.

$(x^{2} - 2) ((u - 1) (u + 7)) = 0$

Factor.

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

$(x^{2} - 2) ((x - 1) (x + 7)) = 0$

Remove unnecessary parentheses.

$(x^{2} - 2) (x - 1) (x + 7) = 0$

Step 10

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

$x^{2} - 2 = 0$

$x - 1 = 0$

$x + 7 = 0$

Step 11

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

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

$x^{2} - 2 = 0$

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

Add $2$ to both sides of the equation.

$x^{2} = 2$

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

$x = \pm \sqrt{2}$

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.

$x = \sqrt{2}$

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

$x = - \sqrt{2}$

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

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

Step 12

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 13

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 14

The final solution is all the values that make $(x^{2} - 2) (x - 1) (x + 7) = 0$ true.

$x = \sqrt{2}, - \sqrt{2}, 1, - 7$

Step 15

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{2}, - \sqrt{2}, 1, - 7$

Decimal Form:

$x = 1.41421356 \dots, - 1.41421356 \dots, 1, - 7$

Step 16

Find the Roots/Zeros Using the Rational Roots Test x^4+6x^3-9x^2-12x+14

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