Algebra Examples

$f (x) = x^{3} - 111 x + 110$

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.

$p = \pm 1, \pm 2, \pm 5, \pm 10, \pm 11, \pm 22, \pm 55, \pm 110$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 5, \pm 10, \pm 11, \pm 22, \pm 55, \pm 110$

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.

${(1)}^{3} - 111 \cdot 1 + 110$

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

One to any power is one.

$1 - 111 \cdot 1 + 110$

Step 4.1.2

Multiply $- 111$ by $1$ .

$1 - 111 + 110$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $111$ from $1$ .

$- 110 + 110$

Step 4.2.2

Add $- 110$ and $110$ .

$0$

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.

$\frac{x^{3} - 111 x + 110}{x - 1}$

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.

$1$	$1$	$0$	$- 111$	$110$

Step 6.2

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

$1$	$1$	$0$	$- 111$	$110$

	$1$

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 $(0)$ .

$1$	$1$	$0$	$- 111$	$110$
		$1$
	$1$

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.

$1$	$1$	$0$	$- 111$	$110$
		$1$
	$1$	$1$

Step 6.5

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 $(- 111)$ .

$1$	$1$	$0$	$- 111$	$110$
		$1$	$1$
	$1$	$1$

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.

$1$	$1$	$0$	$- 111$	$110$
		$1$	$1$
	$1$	$1$	$- 110$

Step 6.7

Multiply the newest entry in the result $(- 110)$ by the divisor $(1)$ and place the result of $(- 110)$ under the next term in the dividend $(110)$ .

$1$	$1$	$0$	$- 111$	$110$
		$1$	$1$	$- 110$
	$1$	$1$	$- 110$

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.

$1$	$1$	$0$	$- 111$	$110$
		$1$	$1$	$- 110$
	$1$	$1$	$- 110$	$0$

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.

$1 x^{2} + 1 x - 110$

Step 6.10

Simplify the quotient polynomial.

$x^{2} + x - 110$

Step 7

Factor $x^{2} + x - 110$ 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 $- 110$ and whose sum is $1$ .

$- 10, 11$

Step 7.2

Write the factored form using these integers.

$(x - 1) + (x - 10) (x + 11)$

$(x - 1) (x - 10) (x + 11)$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $x^{3} - 111 x + 110$ 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.

$p = \pm 1, \pm 110, \pm 2, \pm 55, \pm 5, \pm 22, \pm 10, \pm 11$

$q = \pm 1$

Step 8.1.2

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

$\pm 1, \pm 110, \pm 2, \pm 55, \pm 5, \pm 22, \pm 10, \pm 11$

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.

$1^{3} - 111 \cdot 1 + 110$

Step 8.1.3.2

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

$1 - 111 \cdot 1 + 110$

Step 8.1.3.3

Multiply $- 111$ by $1$ .

$1 - 111 + 110$

Step 8.1.3.4

Subtract $111$ from $1$ .

$- 110 + 110$

Step 8.1.3.5

Add $- 110$ and $110$ .

$0$

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.

$\frac{x^{3} - 111 x + 110}{x - 1}$

Step 8.1.5

Divide $x^{3} - 111 x + 110$ 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 $0$ .

$x$

$1$

$x^{3}$

$0 x^{2}$

$111 x$

$110$

Step 8.1.5.2

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$

Step 8.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			+	$x^{3}$	-	$x^{2}$

Step 8.1.5.4

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$

Step 8.1.5.5

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$

Step 8.1.5.6

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$

Step 8.1.5.7

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$

Step 8.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$
					+	$x^{2}$	-	$x$

Step 8.1.5.9

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$
					-	$x^{2}$	+	$x$

Step 8.1.5.10

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$
					-	$x^{2}$	+	$x$
							-	$110 x$

Step 8.1.5.11

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$
					-	$x^{2}$	+	$x$
							-	$110 x$	+	$110$

Step 8.1.5.12

Divide the highest order term in the dividend $- 110 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$x$	-	$110$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$
					-	$x^{2}$	+	$x$
							-	$110 x$	+	$110$

Step 8.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	-	$110$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$
					-	$x^{2}$	+	$x$
							-	$110 x$	+	$110$
							-	$110 x$	+	$110$

Step 8.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 110 x + 110$

				$x^{2}$	+	$x$	-	$110$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$
					-	$x^{2}$	+	$x$
							-	$110 x$	+	$110$
							+	$110 x$	-	$110$

Step 8.1.5.15

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

				$x^{2}$	+	$x$	-	$110$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$111 x$	+	$110$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$111 x$
					-	$x^{2}$	+	$x$
							-	$110 x$	+	$110$
							+	$110 x$	-	$110$
										$0$

Step 8.1.5.16

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

$x^{2} + x - 110$

Step 8.1.6

Write $x^{3} - 111 x + 110$ as a set of factors.

$(x - 1) (x^{2} + x - 110) = 0$

Step 8.2

Factor $x^{2} + x - 110$ using the AC method.

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Step 8.2.1

Factor $x^{2} + x - 110$ 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 $- 110$ and whose sum is $1$ .

$- 10, 11$

Step 8.2.1.2

Write the factored form using these integers.

$(x - 1) ((x - 10) (x + 11)) = 0$

Step 8.2.2

Remove unnecessary parentheses.

$(x - 1) (x - 10) (x + 11) = 0$

Step 9

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

$x - 1 = 0$

$x - 10 = 0$

$x + 11 = 0$

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

Add $1$ to both sides of the equation.

$x = 1$

Step 11

Set $x - 10$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Step 11.2

Add $10$ to both sides of the equation.

$x = 10$

Step 12

Set $x + 11$ equal to $0$ and solve for $x$ .

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Step 12.1

Set $x + 11$ equal to $0$ .

$x + 11 = 0$

Step 12.2

Subtract $11$ from both sides of the equation.

$x = - 11$

Step 13

The final solution is all the values that make $(x - 1) (x - 10) (x + 11) = 0$ true.

$x = 1, 10, - 11$

Step 14