Find the Roots/Zeros Using the Rational Roots Test x^4+6x^3+7x^2-6x-8

$x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 8$

$q = \pm 1$

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

$\pm 1, \pm 2, \pm 4, \pm 8$

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.

${(1)}^{4} + 6 {(1)}^{3} + 7 {(1)}^{2} - 6 \cdot 1 - 8$

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.

$1 + 6 {(1)}^{3} + 7 {(1)}^{2} - 6 \cdot 1 - 8$

One to any power is one.

$1 + 6 \cdot 1 + 7 {(1)}^{2} - 6 \cdot 1 - 8$

Multiply $6$ by $1$ .

$1 + 6 + 7 {(1)}^{2} - 6 \cdot 1 - 8$

One to any power is one.

$1 + 6 + 7 \cdot 1 - 6 \cdot 1 - 8$

Multiply $7$ by $1$ .

$1 + 6 + 7 - 6 \cdot 1 - 8$

Multiply $- 6$ by $1$ .

$1 + 6 + 7 - 6 - 8$

Simplify by adding and subtracting.

Add $1$ and $6$ .

$7 + 7 - 6 - 8$

Add $7$ and $7$ .

$14 - 6 - 8$

Subtract $6$ from $14$ .

$8 - 8$

Subtract $8$ from $8$ .

$0$

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.

$\frac{x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8}{x - 1}$

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.

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\xa0				\xa0

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

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\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 $(6)$ .

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\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.

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\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 $(7)$ .

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\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.

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\xa0	$1$	$7$		\xa0
	$1$	$7$	$14$

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 $(- 6)$ .

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\xa0	$1$	$7$	$14$	\xa0
	$1$	$7$	$14$

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

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\xa0	$1$	$7$	$14$	\xa0
	$1$	$7$	$14$	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(1)$ and place the result of $(8)$ under the next term in the dividend $(- 8)$ .

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\xa0	$1$	$7$	$14$	$8$
	$1$	$7$	$14$	$8$

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

$1$	$1$	$6$	$7$	$- 6$	$- 8$
	\xa0	$1$	$7$	$14$	$8$
	$1$	$7$	$14$	$8$	$0$

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

$1 x^{3} + 7 x^{2} + (14) x + 8$

Simplify the quotient polynomial.

$x^{3} + 7 x^{2} + 14 x + 8$

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Regroup terms.

$x^{3} + 8 + 7 x^{2} + 14 x = 0$

Rewrite $8$ as $2^{3}$ .

$x^{3} + 2^{3} + 7 x^{2} + 14 x = 0$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 2$ .

$(x + 2) (x^{2} - x \cdot 2 + 2^{2}) + 7 x^{2} + 14 x = 0$

Simplify.

Multiply $2$ by $- 1$ .

$(x + 2) (x^{2} - 2 x + 2^{2}) + 7 x^{2} + 14 x = 0$

Raise $2$ to the power of $2$ .

$(x + 2) (x^{2} - 2 x + 4) + 7 x^{2} + 14 x = 0$

Factor $7 x$ out of $7 x^{2} + 14 x$ .

Factor $7 x$ out of $7 x^{2}$ .

$(x + 2) (x^{2} - 2 x + 4) + 7 x (x) + 14 x = 0$

Factor $7 x$ out of $14 x$ .

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

Factor $7 x$ out of $7 x (x) + 7 x (2)$ .

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

Factor $x + 2$ out of $(x + 2) (x^{2} - 2 x + 4) + 7 x (x + 2)$ .

Factor $x + 2$ out of $7 x (x + 2)$ .

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

Factor $x + 2$ out of $(x + 2) (x^{2} - 2 x + 4) + (x + 2) (7 x)$ .

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

Add $- 2 x$ and $7 x$ .

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

Factor.

Factor $x^{2} + 5 x + 4$ 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 $4$ and whose sum is $5$ .

$1, 4$

Write the factored form using these integers.

$(x + 2) ((x + 1) (x + 4)) = 0$

Remove unnecessary parentheses.

$(x + 2) (x + 1) (x + 4) = 0$

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

$x + 2 = 0$

$x + 1 = 0$

$x + 4 = 0$

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.

$x = - 2$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x + 2) (x + 1) (x + 4) = 0$ true.

$x = - 2, - 1, - 4$

The polynomial can be written as a set of linear factors.

$(x - 1) (x + 2) (x + 1) (x + 4)$

These are the roots (zeros) of the polynomial $x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8$ .

$x = 1, - 2, - 1, - 4$

Find the Roots/Zeros Using the Rational Roots Test x^4+6x^3+7x^2-6x-8

Related Questions

Support LCMGCF.com by disabling your adblocker.