Find the Roots/Zeros Using the Rational Roots Test x^3-7x^2-9x+63

$x^{3} - 7 x^{2} - 9 x + 63$

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.

$p = \pm 1, \pm 3, \pm 7, \pm 9, \pm 21, \pm 63$

$q = \pm 1$

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

$\pm 1, \pm 3, \pm 7, \pm 9, \pm 21, \pm 63$

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.

${(3)}^{3} - 7 {(3)}^{2} - 9 \cdot 3 + 63$

Simplify the expression. In this case, the expression is equal to $0$ so $x = 3$ is a root of the polynomial.

Simplify each term.

Raise $3$ to the power of $3$ .

$27 - 7 {(3)}^{2} - 9 \cdot 3 + 63$

Raise $3$ to the power of $2$ .

$27 - 7 \cdot 9 - 9 \cdot 3 + 63$

Multiply $- 7$ by $9$ .

$27 - 63 - 9 \cdot 3 + 63$

Multiply $- 9$ by $3$ .

$27 - 63 - 27 + 63$

Simplify by adding and subtracting.

Subtract $63$ from $27$ .

$- 36 - 27 + 63$

Subtract $27$ from $- 36$ .

$- 63 + 63$

Add $- 63$ and $63$ .

$0$

Since $3$ is a known root, divide the polynomial by $x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 7 x^{2} - 9 x + 63}{x - 3}$

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.

$3$	$1$	$- 7$	$- 9$	$63$
	\xa0			\xa0

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

$3$	$1$	$- 7$	$- 9$	$63$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(3)$ and place the result of $(3)$ under the next term in the dividend $(- 7)$ .

$3$	$1$	$- 7$	$- 9$	$63$
	\xa0	$3$		\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.

$3$	$1$	$- 7$	$- 9$	$63$
	\xa0	$3$		\xa0
	$1$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(3)$ and place the result of $(- 12)$ under the next term in the dividend $(- 9)$ .

$3$	$1$	$- 7$	$- 9$	$63$
	\xa0	$3$	$- 12$	\xa0
	$1$	$- 4$

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

$3$	$1$	$- 7$	$- 9$	$63$
	\xa0	$3$	$- 12$	\xa0
	$1$	$- 4$	$- 21$

Multiply the newest entry in the result $(- 21)$ by the divisor $(3)$ and place the result of $(- 63)$ under the next term in the dividend $(63)$ .

$3$	$1$	$- 7$	$- 9$	$63$
	\xa0	$3$	$- 12$	$- 63$
	$1$	$- 4$	$- 21$

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

$3$	$1$	$- 7$	$- 9$	$63$
	\xa0	$3$	$- 12$	$- 63$
	$1$	$- 4$	$- 21$	$0$

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

$1 x^{2} + - 4 x - 21$

Simplify the quotient polynomial.

$x^{2} - 4 x - 21$

Factor $x^{2} - 4 x - 21$ 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 $- 21$ and whose sum is $- 4$ .

$- 7, 3$

Write the factored form using these integers.

$(x - 3) + (x - 7) (x + 3)$

$(x - 3) (x - 7) (x + 3)$

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.

$(x^{3} - 7 x^{2}) - 9 x + 63 = 0$

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

$x^{2} (x - 7) - 9 (x - 7) = 0$

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

$(x - 7) (x^{2} - 9) = 0$

Rewrite $9$ as $3^{2}$ .

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

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 $b = 3$ .

$(x - 7) ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x - 7) (x + 3) (x - 3) = 0$

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

$x - 7 = 0$

$x + 3 = 0$

$x - 3 = 0$

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.

$x = 7$

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.

$x = - 3$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x - 7) (x + 3) (x - 3) = 0$ true.

$x = 7, - 3, 3$

Find the Roots/Zeros Using the Rational Roots Test x^3-7x^2-9x+63

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