Find the Roots/Zeros Using the Rational Roots Test x^4+6x^3-11x^2-24x+28

$x^{4} + 6 x^{3} - 11 x^{2} - 24 x + 28$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28$

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.

${(1)}^{4} + 6 {(1)}^{3} - 11 {(1)}^{2} - 24 \cdot 1 + 28$

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.

$1 + 6 {(1)}^{3} - 11 {(1)}^{2} - 24 \cdot 1 + 28$

One to any power is one.

$1 + 6 \cdot 1 - 11 {(1)}^{2} - 24 \cdot 1 + 28$

Multiply $6$ by $1$ .

$1 + 6 - 11 {(1)}^{2} - 24 \cdot 1 + 28$

One to any power is one.

$1 + 6 - 11 \cdot 1 - 24 \cdot 1 + 28$

Multiply $- 11$ by $1$ .

$1 + 6 - 11 - 24 \cdot 1 + 28$

Multiply $- 24$ by $1$ .

$1 + 6 - 11 - 24 + 28$

Simplify by adding and subtracting.

Add $1$ and $6$ .

$7 - 11 - 24 + 28$

Subtract $11$ from $7$ .

$- 4 - 24 + 28$

Subtract $24$ from $- 4$ .

$- 28 + 28$

Add $- 28$ and $28$ .

$0$

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.

$\frac{x^{4} + 6 x^{3} - 11 x^{2} - 24 x + 28}{x - 1}$

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.

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\xa0				\xa0

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

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\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 $(6)$ .

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\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.

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\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 $(- 11)$ .

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\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.

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\xa0	$1$	$7$		\xa0
	$1$	$7$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(1)$ and place the result of $(- 4)$ under the next term in the dividend $(- 24)$ .

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\xa0	$1$	$7$	$- 4$	\xa0
	$1$	$7$	$- 4$

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

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\xa0	$1$	$7$	$- 4$	\xa0
	$1$	$7$	$- 4$	$- 28$

Multiply the newest entry in the result $(- 28)$ by the divisor $(1)$ and place the result of $(- 28)$ under the next term in the dividend $(28)$ .

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\xa0	$1$	$7$	$- 4$	$- 28$
	$1$	$7$	$- 4$	$- 28$

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

$1$	$1$	$6$	$- 11$	$- 24$	$28$
	\xa0	$1$	$7$	$- 4$	$- 28$
	$1$	$7$	$- 4$	$- 28$	$0$

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

$1 x^{3} + 7 x^{2} + (- 4) x - 28$

Simplify the quotient polynomial.

$x^{3} + 7 x^{2} - 4 x - 28$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (x^{3} + 7 x^{2}) - 4 x - 28$

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

$(x - 1) + x^{2} (x + 7) - 4 (x + 7)$

$(x - 1) x^{2} (x + 7) - 4 (x + 7)$

Step 8

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

$(x - 1) (x + 7) (x^{2} - 4)$

Step 9

Rewrite $4$ as $2^{2}$ .

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

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 $b = 2$ .

$(x - 1) + (x + 7) ((x + 2) (x - 2))$

Remove unnecessary parentheses.

$(x - 1) + (x + 7) (x + 2) (x - 2)$

$(x - 1) (x + 7) (x + 2) (x - 2)$

Step 11

Factor the left side of the equation.

Regroup terms.

$6 x^{3} - 24 x + x^{4} - 11 x^{2} + 28 = 0$

Factor $6 x$ out of $6 x^{3} - 24 x$ .

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

$6 x (x^{2}) - 24 x + x^{4} - 11 x^{2} + 28 = 0$

Factor $6 x$ out of $- 24 x$ .

$6 x (x^{2}) + 6 x (- 4) + x^{4} - 11 x^{2} + 28 = 0$

Factor $6 x$ out of $6 x (x^{2}) + 6 x (- 4)$ .

$6 x (x^{2} - 4) + x^{4} - 11 x^{2} + 28 = 0$

Rewrite $4$ as $2^{2}$ .

$6 x (x^{2} - 2^{2}) + x^{4} - 11 x^{2} + 28 = 0$

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 $b = 2$ .

$6 x ((x + 2) (x - 2)) + x^{4} - 11 x^{2} + 28 = 0$

Remove unnecessary parentheses.

$6 x (x + 2) (x - 2) + x^{4} - 11 x^{2} + 28 = 0$

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

$6 x (x + 2) (x - 2) + {(x^{2})}^{2} - 11 x^{2} + 28 = 0$

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

$6 x (x + 2) (x - 2) + u^{2} - 11 u + 28 = 0$

Factor $u^{2} - 11 u + 28$ 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 $28$ and whose sum is $- 11$ .

$- 7, - 4$

Write the factored form using these integers.

$6 x (x + 2) (x - 2) + (u - 7) (u - 4) = 0$

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

$6 x (x + 2) (x - 2) + (x^{2} - 7) (x^{2} - 4) = 0$

Rewrite $4$ as $2^{2}$ .

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

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

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

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

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

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

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

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

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

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

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.

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

Factor.

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

$(x + 2) (x - 2) ((x - 1) (x + 7)) = 0$

Remove unnecessary parentheses.

$(x + 2) (x - 2) (x - 1) (x + 7) = 0$

Step 12

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

$x + 2 = 0$

$x - 2 = 0$

$x - 1 = 0$

$x + 7 = 0$

Step 13

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 14

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

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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 $(x + 2) (x - 2) (x - 1) (x + 7) = 0$ true.

$x = - 2, 2, 1, - 7$

Step 18

Find the Roots/Zeros Using the Rational Roots Test x^4+6x^3-11x^2-24x+28

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