Precalculus Examples

$x^{3} - x^{2} - 37 x - 35$

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.

$p = \pm 1, \pm 5, \pm 7, \pm 35$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 5, \pm 7, \pm 35$

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.

${(- 1)}^{3} - {(- 1)}^{2} - 37 \cdot - 1 - 35$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

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

$- 1 - {(- 1)}^{2} - 37 \cdot - 1 - 35$

Step 4.1.2

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 4.1.2.1

Multiply $- 1$ by ${(- 1)}^{2}$ .

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Step 4.1.2.1.1

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

$- 1 + {(- 1)}^{1} {(- 1)}^{2} - 37 \cdot - 1 - 35$

Step 4.1.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 1 + {(- 1)}^{1 + 2} - 37 \cdot - 1 - 35$

Step 4.1.2.2

Add $1$ and $2$ .

$- 1 + {(- 1)}^{3} - 37 \cdot - 1 - 35$

Step 4.1.3

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

$- 1 - 1 - 37 \cdot - 1 - 35$

Step 4.1.4

Multiply $- 37$ by $- 1$ .

$- 1 - 1 + 37 - 35$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $1$ from $- 1$ .

$- 2 + 37 - 35$

Step 4.2.2

Add $- 2$ and $37$ .

$35 - 35$

Step 4.2.3

Subtract $35$ from $35$ .

$0$

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.

$\frac{x^{3} - x^{2} - 37 x - 35}{x + 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Step 6.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$- 1$	$1$	$- 1$	$- 37$	$- 35$

Step 6.2

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

$- 1$	$1$	$- 1$	$- 37$	$- 35$

	$1$

Step 6.3

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 $(- 1)$ .

$- 1$	$1$	$- 1$	$- 37$	$- 35$
		$- 1$
	$1$

Step 6.4

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

$- 1$	$1$	$- 1$	$- 37$	$- 35$
		$- 1$
	$1$	$- 2$

Step 6.5

Multiply the newest entry in the result $(- 2)$ by the divisor $(- 1)$ and place the result of $(2)$ under the next term in the dividend $(- 37)$ .

$- 1$	$1$	$- 1$	$- 37$	$- 35$
		$- 1$	$2$
	$1$	$- 2$

Step 6.6

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

$- 1$	$1$	$- 1$	$- 37$	$- 35$
		$- 1$	$2$
	$1$	$- 2$	$- 35$

Step 6.7

Multiply the newest entry in the result $(- 35)$ by the divisor $(- 1)$ and place the result of $(35)$ under the next term in the dividend $(- 35)$ .

$- 1$	$1$	$- 1$	$- 37$	$- 35$
		$- 1$	$2$	$35$
	$1$	$- 2$	$- 35$

Step 6.8

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

$- 1$	$1$	$- 1$	$- 37$	$- 35$
		$- 1$	$2$	$35$
	$1$	$- 2$	$- 35$	$0$

Step 6.9

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

$1 x^{2} + - 2 x - 35$

Step 6.10

Simplify the quotient polynomial.

$x^{2} - 2 x - 35$

Step 7

Factor $x^{2} - 2 x - 35$ using the AC method.

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Step 7.1

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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Step 7.2

Write the factored form using these integers.

$(x + 1) + (x - 7) (x + 5)$

$(x + 1) (x - 7) (x + 5)$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $x^{3} - x^{2} - 37 x - 35$ using the rational roots test.

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Step 8.1.1

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.

$p = \pm 1, \pm 35, \pm 5, \pm 7$

$q = \pm 1$

Step 8.1.2

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

$\pm 1, \pm 35, \pm 5, \pm 7$

Step 8.1.3

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 8.1.3.1

Substitute $- 1$ into the polynomial.

${(- 1)}^{3} - {(- 1)}^{2} - 37 \cdot - 1 - 35$

Step 8.1.3.2

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

$- 1 - {(- 1)}^{2} - 37 \cdot - 1 - 35$

Step 8.1.3.3

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

$- 1 - 1 \cdot 1 - 37 \cdot - 1 - 35$

Step 8.1.3.4

Multiply $- 1$ by $1$ .

$- 1 - 1 - 37 \cdot - 1 - 35$

Step 8.1.3.5

Subtract $1$ from $- 1$ .

$- 2 - 37 \cdot - 1 - 35$

Step 8.1.3.6

Multiply $- 37$ by $- 1$ .

$- 2 + 37 - 35$

Step 8.1.3.7

Add $- 2$ and $37$ .

$35 - 35$

Step 8.1.3.8

Subtract $35$ from $35$ .

$0$

Step 8.1.4

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.

$\frac{x^{3} - x^{2} - 37 x - 35}{x + 1}$

Step 8.1.5

Divide $x^{3} - x^{2} - 37 x - 35$ by $x + 1$ .

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Step 8.1.5.1

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

$x$

$1$

$x^{3}$

$x^{2}$

$37 x$

$35$

Step 8.1.5.2

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$

Step 8.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			+	$x^{3}$	+	$x^{2}$

Step 8.1.5.4

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$

Step 8.1.5.5

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$

Step 8.1.5.6

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$

Step 8.1.5.7

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

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$

Step 8.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$
					-	$2 x^{2}$	-	$2 x$

Step 8.1.5.9

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

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$
					+	$2 x^{2}$	+	$2 x$

Step 8.1.5.10

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

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$
					+	$2 x^{2}$	+	$2 x$
							-	$35 x$

Step 8.1.5.11

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

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$
					+	$2 x^{2}$	+	$2 x$
							-	$35 x$	-	$35$

Step 8.1.5.12

Divide the highest order term in the dividend $- 35 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$2 x$	-	$35$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$
					+	$2 x^{2}$	+	$2 x$
							-	$35 x$	-	$35$

Step 8.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	-	$35$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$
					+	$2 x^{2}$	+	$2 x$
							-	$35 x$	-	$35$
							-	$35 x$	-	$35$

Step 8.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 35 x - 35$

				$x^{2}$	-	$2 x$	-	$35$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$
					+	$2 x^{2}$	+	$2 x$
							-	$35 x$	-	$35$
							+	$35 x$	+	$35$

Step 8.1.5.15

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

				$x^{2}$	-	$2 x$	-	$35$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	-	$37 x$	-	$35$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	-	$37 x$
					+	$2 x^{2}$	+	$2 x$
							-	$35 x$	-	$35$
							+	$35 x$	+	$35$
										$0$

Step 8.1.5.16

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

$x^{2} - 2 x - 35$

Step 8.1.6

Write $x^{3} - x^{2} - 37 x - 35$ as a set of factors.

$(x + 1) (x^{2} - 2 x - 35) = 0$

Step 8.2

Factor $x^{2} - 2 x - 35$ using the AC method.

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Step 8.2.1

Factor $x^{2} - 2 x - 35$ using the AC method.

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Step 8.2.1.1

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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Step 8.2.1.2

Write the factored form using these integers.

$(x + 1) ((x - 7) (x + 5)) = 0$

Step 8.2.2

Remove unnecessary parentheses.

$(x + 1) (x - 7) (x + 5) = 0$

Step 9

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

$x + 1 = 0$

$x - 7 = 0$

$x + 5 = 0$

Step 10

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

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Step 10.1

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

$x + 1 = 0$

Step 10.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

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

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Step 11.1

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

$x - 7 = 0$

Step 11.2

Add $7$ to both sides of the equation.

$x = 7$

Step 12

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

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Step 12.1

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

$x + 5 = 0$

Step 12.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 13

The final solution is all the values that make $(x + 1) (x - 7) (x + 5) = 0$ true.

$x = - 1, 7, - 5$

Step 14