Find the Roots/Zeros Using the Rational Roots Test x^3+7x^2+7x-15

$x^{3} + 7 x^{2} + 7 x - 15$

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.

$p = \pm 1, \pm 3, \pm 5, \pm 15$

$q = \pm 1$

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

$\pm 1, \pm 3, \pm 5, \pm 15$

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.

${(1)}^{3} + 7 {(1)}^{2} + 7 (1) - 15$

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.

$1 + 7 {(1)}^{2} + 7 (1) - 15$

One to any power is one.

$1 + 7 \cdot 1 + 7 (1) - 15$

Multiply $7$ by $1$ .

$1 + 7 + 7 (1) - 15$

Multiply $7$ by $1$ .

$1 + 7 + 7 - 15$

Simplify by adding and subtracting.

Add $1$ and $7$ .

$8 + 7 - 15$

Add $8$ and $7$ .

$15 - 15$

Subtract $15$ from $15$ .

$0$

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.

$\frac{x^{3} + 7 x^{2} + 7 x - 15}{x - 1}$

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.

$1$	$1$	$7$	$7$	$- 15$
	\xa0			\xa0

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

$1$	$1$	$7$	$7$	$- 15$
	\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 $(7)$ .

$1$	$1$	$7$	$7$	$- 15$
	\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.

$1$	$1$	$7$	$7$	$- 15$
	\xa0	$1$		\xa0
	$1$	$8$

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 $(7)$ .

$1$	$1$	$7$	$7$	$- 15$
	\xa0	$1$	$8$	\xa0
	$1$	$8$

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

$1$	$1$	$7$	$7$	$- 15$
	\xa0	$1$	$8$	\xa0
	$1$	$8$	$15$

Multiply the newest entry in the result $(15)$ by the divisor $(1)$ and place the result of $(15)$ under the next term in the dividend $(- 15)$ .

$1$	$1$	$7$	$7$	$- 15$
	\xa0	$1$	$8$	$15$
	$1$	$8$	$15$

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

$1$	$1$	$7$	$7$	$- 15$
	\xa0	$1$	$8$	$15$
	$1$	$8$	$15$	$0$

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

$1 x^{2} + 8 x + 15$

Simplify the quotient polynomial.

$x^{2} + 8 x + 15$

Factor $x^{2} + 8 x + 15$ 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 $15$ and whose sum is $8$ .

$3, 5$

Write the factored form using these integers.

$(x - 1) + (x + 3) (x + 5)$

$(x - 1) (x + 3) (x + 5)$

Factor the left side of the equation.

Factor $x^{3} + 7 x^{2} + 7 x - 15$ using the rational roots test.

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.

$p = \pm 1, \pm 15, \pm 3, \pm 5 q = \pm 1$

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

$\pm 1, \pm 15, \pm 3, \pm 5$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$1^{3} + 7 \cdot 1^{2} + 7 \cdot 1 - 15$

Raise $1$ to the power of $3$ .

$1 + 7 \cdot 1^{2} + 7 \cdot 1 - 15$

Raise $1$ to the power of $2$ .

$1 + 7 \cdot 1 + 7 \cdot 1 - 15$

Multiply $7$ by $1$ .

$1 + 7 + 7 \cdot 1 - 15$

Add $1$ and $7$ .

$8 + 7 \cdot 1 - 15$

Multiply $7$ by $1$ .

$8 + 7 - 15$

Add $8$ and $7$ .

$15 - 15$

Subtract $15$ from $15$ .

$0$

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.

$\frac{x^{3} + 7 x^{2} + 7 x - 15}{x - 1}$

Divide $x^{3} + 7 x^{2} + 7 x - 15$ by $x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{3}$

$7 x^{2}$

$7 x$

$15$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			+	$x^{3}$	-	$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - x^{2}$

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$

Divide the highest order term in the dividend $8 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$8 x$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$8 x$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$
					+	$8 x^{2}$	-	$8 x$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x^{2} - 8 x$

				$x^{2}$	+	$8 x$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$
					-	$8 x^{2}$	+	$8 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	+	$8 x$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$
					-	$8 x^{2}$	+	$8 x$
							+	$15 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	+	$8 x$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$
					-	$8 x^{2}$	+	$8 x$
							+	$15 x$	-	$15$

Divide the highest order term in the dividend $15 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$8 x$	+	$15$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$
					-	$8 x^{2}$	+	$8 x$
							+	$15 x$	-	$15$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$8 x$	+	$15$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$
					-	$8 x^{2}$	+	$8 x$
							+	$15 x$	-	$15$
							+	$15 x$	-	$15$

The expression needs to be subtracted from the dividend, so change all the signs in $15 x - 15$

				$x^{2}$	+	$8 x$	+	$15$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$
					-	$8 x^{2}$	+	$8 x$
							+	$15 x$	-	$15$
							-	$15 x$	+	$15$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	+	$8 x$	+	$15$
$x$	-	$1$		$x^{3}$	+	$7 x^{2}$	+	$7 x$	-	$15$
			-	$x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	+	$7 x$
					-	$8 x^{2}$	+	$8 x$
							+	$15 x$	-	$15$
							-	$15 x$	+	$15$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} + 8 x + 15$

Write $x^{3} + 7 x^{2} + 7 x - 15$ as a set of factors.

$(x - 1) (x^{2} + 8 x + 15) = 0$

Factor $x^{2} + 8 x + 15$ 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 $15$ and whose sum is $8$ .

$3, 5$

Write the factored form using these integers.

$(x - 1) ((x + 3) (x + 5)) = 0$

Remove unnecessary parentheses.

$(x - 1) (x + 3) (x + 5) = 0$

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

$x - 1 = 0$

$x + 3 = 0$

$x + 5 = 0$

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.

$x = 1$

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 $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 1) (x + 3) (x + 5) = 0$ true.

$x = 1, - 3, - 5$

Find the Roots/Zeros Using the Rational Roots Test x^3+7x^2+7x-15

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