Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+2x^3-5x^2-4x+6

$f (x) = x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 6$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 6$

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.

${(1)}^{4} + 2 {(1)}^{3} - 5 {(1)}^{2} - 4 \cdot 1 + 6$

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.

$1 + 2 {(1)}^{3} - 5 {(1)}^{2} - 4 \cdot 1 + 6$

One to any power is one.

$1 + 2 \cdot 1 - 5 {(1)}^{2} - 4 \cdot 1 + 6$

Multiply $2$ by $1$ .

$1 + 2 - 5 {(1)}^{2} - 4 \cdot 1 + 6$

One to any power is one.

$1 + 2 - 5 \cdot 1 - 4 \cdot 1 + 6$

Multiply $- 5$ by $1$ .

$1 + 2 - 5 - 4 \cdot 1 + 6$

Multiply $- 4$ by $1$ .

$1 + 2 - 5 - 4 + 6$

Simplify by adding and subtracting.

Add $1$ and $2$ .

$3 - 5 - 4 + 6$

Subtract $5$ from $3$ .

$- 2 - 4 + 6$

Subtract $4$ from $- 2$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

$\frac{x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6}{x - 1}$

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.

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\xa0				\xa0

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

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\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 $(2)$ .

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\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.

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\xa0	$1$			\xa0
	$1$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(1)$ and place the result of $(3)$ under the next term in the dividend $(- 5)$ .

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\xa0	$1$	$3$		\xa0
	$1$	$3$

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

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\xa0	$1$	$3$		\xa0
	$1$	$3$	$- 2$

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 $(- 4)$ .

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\xa0	$1$	$3$	$- 2$	\xa0
	$1$	$3$	$- 2$

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

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\xa0	$1$	$3$	$- 2$	\xa0
	$1$	$3$	$- 2$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(1)$ and place the result of $(- 6)$ under the next term in the dividend $(6)$ .

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\xa0	$1$	$3$	$- 2$	$- 6$
	$1$	$3$	$- 2$	$- 6$

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

$1$	$1$	$2$	$- 5$	$- 4$	$6$
	\xa0	$1$	$3$	$- 2$	$- 6$
	$1$	$3$	$- 2$	$- 6$	$0$

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

$1 x^{3} + 3 x^{2} + (- 2) x - 6$

Simplify the quotient polynomial.

$x^{3} + 3 x^{2} - 2 x - 6$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (x^{3} + 3 x^{2}) - 2 x - 6$

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

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

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

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

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

Step 9

Factor the left side of the equation.

Regroup terms.

$2 x^{3} - 4 x + x^{4} - 5 x^{2} + 6 = 0$

Factor $2 x$ out of $2 x^{3} - 4 x$ .

Factor $2 x$ out of $2 x^{3}$ .

$2 x (x^{2}) - 4 x + x^{4} - 5 x^{2} + 6 = 0$

Factor $2 x$ out of $- 4 x$ .

$2 x (x^{2}) + 2 x (- 2) + x^{4} - 5 x^{2} + 6 = 0$

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

$2 x (x^{2} - 2) + x^{4} - 5 x^{2} + 6 = 0$

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

$2 x (x^{2} - 2) + {(x^{2})}^{2} - 5 x^{2} + 6 = 0$

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

$2 x (x^{2} - 2) + u^{2} - 5 u + 6 = 0$

Factor $u^{2} - 5 u + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

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

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

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

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

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

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

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

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

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

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

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

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

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

$- 1, 3$

Write the factored form using these integers.

$(x^{2} - 2) ((u - 1) (u + 3)) = 0$

Factor.

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

$(x^{2} - 2) ((x - 1) (x + 3)) = 0$

Remove unnecessary parentheses.

$(x^{2} - 2) (x - 1) (x + 3) = 0$

Step 10

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

$x^{2} - 2 = 0$

$x - 1 = 0$

$x + 3 = 0$

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 $x + 3$ 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

The final solution is all the values that make $(x^{2} - 2) (x - 1) (x + 3) = 0$ true.

$x = \sqrt{2}, - \sqrt{2}, 1, - 3$

Step 15

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{2}, - \sqrt{2}, 1, - 3$

Decimal Form:

$x = 1.41421356 \dots, - 1.41421356 \dots, 1, - 3$

Step 16

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+2x^3-5x^2-4x+6

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