Precalculus Examples

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

Step 1

Set $x^{3} + 7 x^{2} + 7 x - 15$ equal to $0$ .

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

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

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

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Step 2.1.1.1

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.

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

$q = \pm 1$

Step 2.1.1.2

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

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

Step 2.1.1.3

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

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Step 2.1.1.3.1

Substitute $1$ into the polynomial.

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

Step 2.1.1.3.2

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

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

Step 2.1.1.3.3

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

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

Step 2.1.1.3.4

Multiply $7$ by $1$ .

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

Step 2.1.1.3.5

Add $1$ and $7$ .

$8 + 7 \cdot 1 - 15$

Step 2.1.1.3.6

Multiply $7$ by $1$ .

$8 + 7 - 15$

Step 2.1.1.3.7

Add $8$ and $7$ .

$15 - 15$

Step 2.1.1.3.8

Subtract $15$ from $15$ .

$0$

Step 2.1.1.4

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.

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

Step 2.1.1.5

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

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Step 2.1.1.5.1

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

$x$

$1$

$x^{3}$

$7 x^{2}$

$7 x$

$15$

Step 2.1.1.5.2

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

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

Step 2.1.1.5.3

Multiply the new quotient term by the divisor.

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

Step 2.1.1.5.4

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

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

Step 2.1.1.5.5

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

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

Step 2.1.1.5.6

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

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

Step 2.1.1.5.7

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

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

Step 2.1.1.5.8

Multiply the new quotient term by the divisor.

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

Step 2.1.1.5.9

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

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

Step 2.1.1.5.10

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

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

Step 2.1.1.5.11

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

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

Step 2.1.1.5.12

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

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

Step 2.1.1.5.13

Multiply the new quotient term by the divisor.

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

Step 2.1.1.5.14

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

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

Step 2.1.1.5.15

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

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

Step 2.1.1.5.16

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

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

Step 2.1.1.6

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

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

Step 2.1.2

Factor $x^{2} + 8 x + 15$ using the AC method.

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Step 2.1.2.1

Factor $x^{2} + 8 x + 15$ using the AC method.

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Step 2.1.2.1.1

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 $8$ .

$3, 5$

Step 2.1.2.1.2

Write the factored form using these integers.

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

Step 2.1.2.2

Remove unnecessary parentheses.

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

Step 2.2

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

$x - 1 = 0$

$x + 3 = 0$

$x + 5 = 0$

Step 2.3

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

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Step 2.3.1

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

$x - 1 = 0$

Step 2.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.4

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

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Step 2.4.1

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

$x + 3 = 0$

Step 2.4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.5

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

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Step 2.5.1

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

$x + 5 = 0$

Step 2.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.6

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

$x = 1, - 3, - 5$

Step 3