Find the Roots/Zeros Using the Rational Roots Test H(x)=x^3-7x^2-x+7

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

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.

$p = \pm 1, \pm 7$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 7$

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.

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

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.

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

One to any power is one.

$1 - 7 \cdot 1 - (1) + 7$

Multiply $- 7$ by $1$ .

$1 - 7 - (1) + 7$

Multiply $- 1$ by $1$ .

$1 - 7 - 1 + 7$

Simplify by adding and subtracting.

Subtract $7$ from $1$ .

$- 6 - 1 + 7$

Subtract $1$ from $- 6$ .

$- 7 + 7$

Add $- 7$ and $7$ .

$0$

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.

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

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.

$1$	$1$	$- 7$	$- 1$	$7$
	\xa0			\xa0

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

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

$1$	$1$	$- 7$	$- 1$	$7$
	\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.

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

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 $(- 1)$ .

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

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

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

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 $(7)$ .

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

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

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

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

$1 x^{2} + - 6 x - 7$

Simplify the quotient polynomial.

$x^{2} - 6 x - 7$

Step 7

Factor $x^{2} - 6 x - 7$ 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 $- 6$ .

$- 7, 1$

Write the factored form using these integers.

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

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

Step 8

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Rewrite $1$ as $1^{2}$ .

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

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 $b = 1$ .

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

Remove unnecessary parentheses.

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

Step 9

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

$x - 7 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 10

Set $x - 7$ equal to $0$ and solve for $x$ .

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Step 11

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.

$x = - 1$

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

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

$x = 7, - 1, 1$

Step 14

Find the Roots/Zeros Using the Rational Roots Test H(x)=x^3-7x^2-x+7

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