Algebra Examples

$x^{4} + 5 x^{3} - x^{2} - 17 x + 12$

Step 1

Regroup terms.

$x^{4} - x^{2} + 5 x^{3} - 17 x + 12$

Step 2

Factor $x^{2}$ out of $x^{4} - x^{2}$ .

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Step 2.1

Factor $x^{2}$ out of $x^{4}$ .

$x^{2} x^{2} - x^{2} + 5 x^{3} - 17 x + 12$

Step 2.2

Factor $x^{2}$ out of $- x^{2}$ .

$x^{2} x^{2} + x^{2} \cdot - 1 + 5 x^{3} - 17 x + 12$

Step 2.3

Factor $x^{2}$ out of $x^{2} x^{2} + x^{2} \cdot - 1$ .

$x^{2} (x^{2} - 1) + 5 x^{3} - 17 x + 12$

Step 3

Rewrite $1$ as $1^{2}$ .

$x^{2} (x^{2} - 1^{2}) + 5 x^{3} - 17 x + 12$

Step 4

Factor.

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Step 4.1

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 $b = 1$ .

$x^{2} ((x + 1) (x - 1)) + 5 x^{3} - 17 x + 12$

Step 4.2

Remove unnecessary parentheses.

$x^{2} (x + 1) (x - 1) + 5 x^{3} - 17 x + 12$

Step 5

Factor $5 x^{3} - 17 x + 12$ using the rational roots test.

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Step 5.1

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.

$p = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

$q = \pm 1, \pm 5$

Step 5.2

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

$\pm 1, \pm 0.2, \pm 12, \pm 2.4, \pm 2, \pm 0.4, \pm 6, \pm 1.2, \pm 3, \pm 0.6, \pm 4, \pm 0.8$

Step 5.3

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 5.3.1

Substitute $1$ into the polynomial.

$5 \cdot 1^{3} - 17 \cdot 1 + 12$

Step 5.3.2

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

$5 \cdot 1 - 17 \cdot 1 + 12$

Step 5.3.3

Multiply $5$ by $1$ .

$5 - 17 \cdot 1 + 12$

Step 5.3.4

Multiply $- 17$ by $1$ .

$5 - 17 + 12$

Step 5.3.5

Subtract $17$ from $5$ .

$- 12 + 12$

Step 5.3.6

Add $- 12$ and $12$ .

$0$

Step 5.4

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.

$\frac{5 x^{3} - 17 x + 12}{x - 1}$

Step 5.5

Divide $5 x^{3} - 17 x + 12$ by $x - 1$ .

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Step 5.5.1

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

$x$

$1$

$5 x^{3}$

$0 x^{2}$

$17 x$

$12$

Step 5.5.2

Divide the highest order term in the dividend $5 x^{3}$ by the highest order term in divisor $x$ .

					$5 x^{2}$
	$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$

Step 5.5.3

Multiply the new quotient term by the divisor.

				$5 x^{2}$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			+	$5 x^{3}$	-	$5 x^{2}$

Step 5.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{3} - 5 x^{2}$

				$5 x^{2}$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$

Step 5.5.5

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

				$5 x^{2}$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$

Step 5.5.6

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

				$5 x^{2}$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$

Step 5.5.7

Divide the highest order term in the dividend $5 x^{2}$ by the highest order term in divisor $x$ .

				$5 x^{2}$	+	$5 x$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$

Step 5.5.8

Multiply the new quotient term by the divisor.

				$5 x^{2}$	+	$5 x$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$
					+	$5 x^{2}$	-	$5 x$

Step 5.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{2} - 5 x$

				$5 x^{2}$	+	$5 x$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$
					-	$5 x^{2}$	+	$5 x$

Step 5.5.10

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

				$5 x^{2}$	+	$5 x$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$
					-	$5 x^{2}$	+	$5 x$
							-	$12 x$

Step 5.5.11

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

				$5 x^{2}$	+	$5 x$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$
					-	$5 x^{2}$	+	$5 x$
							-	$12 x$	+	$12$

Step 5.5.12

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $x$ .

				$5 x^{2}$	+	$5 x$	-	$12$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$
					-	$5 x^{2}$	+	$5 x$
							-	$12 x$	+	$12$

Step 5.5.13

Multiply the new quotient term by the divisor.

				$5 x^{2}$	+	$5 x$	-	$12$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$
					-	$5 x^{2}$	+	$5 x$
							-	$12 x$	+	$12$
							-	$12 x$	+	$12$

Step 5.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 x + 12$

				$5 x^{2}$	+	$5 x$	-	$12$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$
					-	$5 x^{2}$	+	$5 x$
							-	$12 x$	+	$12$
							+	$12 x$	-	$12$

Step 5.5.15

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

				$5 x^{2}$	+	$5 x$	-	$12$
$x$	-	$1$		$5 x^{3}$	+	$0 x^{2}$	-	$17 x$	+	$12$
			-	$5 x^{3}$	+	$5 x^{2}$
					+	$5 x^{2}$	-	$17 x$
					-	$5 x^{2}$	+	$5 x$
							-	$12 x$	+	$12$
							+	$12 x$	-	$12$
										$0$

Step 5.5.16

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

$5 x^{2} + 5 x - 12$

Step 5.6

Write $5 x^{3} - 17 x + 12$ as a set of factors.

$x^{2} (x + 1) (x - 1) + (x - 1) (5 x^{2} + 5 x - 12)$

Step 6

Factor $x - 1$ out of $x^{2} (x + 1) (x - 1) + (x - 1) (5 x^{2} + 5 x - 12)$ .

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Step 6.1

Factor $x - 1$ out of $x^{2} (x + 1) (x - 1)$ .

$(x - 1) (x^{2} (x + 1)) + (x - 1) (5 x^{2} + 5 x - 12)$

Step 6.2

Factor $x - 1$ out of $(x - 1) (x^{2} (x + 1)) + (x - 1) (5 x^{2} + 5 x - 12)$ .

$(x - 1) (x^{2} (x + 1) + 5 x^{2} + 5 x - 12)$

Step 7

Apply the distributive property.

$(x - 1) (x^{2} x + x^{2} \cdot 1 + 5 x^{2} + 5 x - 12)$

Step 8

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 8.1

Multiply $x^{2}$ by $x$ .

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Step 8.1.1

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

$(x - 1) (x^{2} x^{1} + x^{2} \cdot 1 + 5 x^{2} + 5 x - 12)$

Step 8.1.2

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

$(x - 1) (x^{2 + 1} + x^{2} \cdot 1 + 5 x^{2} + 5 x - 12)$

Step 8.2

Add $2$ and $1$ .

$(x - 1) (x^{3} + x^{2} \cdot 1 + 5 x^{2} + 5 x - 12)$

Step 9

Multiply $x^{2}$ by $1$ .

$(x - 1) (x^{3} + x^{2} + 5 x^{2} + 5 x - 12)$

Step 10

Add $x^{2}$ and $5 x^{2}$ .

$(x - 1) (x^{3} + 6 x^{2} + 5 x - 12)$

Step 11

Factor.

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Step 11.1

Rewrite $x^{3} + 6 x^{2} + 5 x - 12$ in a factored form.

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Step 11.1.1

Factor $x^{3} + 6 x^{2} + 5 x - 12$ using the rational roots test.

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Step 11.1.1.1

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.

$p = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

$q = \pm 1$

Step 11.1.1.2

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

$\pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

Step 11.1.1.3

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 11.1.1.3.1

Substitute $1$ into the polynomial.

$1^{3} + 6 \cdot 1^{2} + 5 \cdot 1 - 12$

Step 11.1.1.3.2

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

$1 + 6 \cdot 1^{2} + 5 \cdot 1 - 12$

Step 11.1.1.3.3

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

$1 + 6 \cdot 1 + 5 \cdot 1 - 12$

Step 11.1.1.3.4

Multiply $6$ by $1$ .

$1 + 6 + 5 \cdot 1 - 12$

Step 11.1.1.3.5

Add $1$ and $6$ .

$7 + 5 \cdot 1 - 12$

Step 11.1.1.3.6

Multiply $5$ by $1$ .

$7 + 5 - 12$

Step 11.1.1.3.7

Add $7$ and $5$ .

$12 - 12$

Step 11.1.1.3.8

Subtract $12$ from $12$ .

$0$

Step 11.1.1.4

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.

$\frac{x^{3} + 6 x^{2} + 5 x - 12}{x - 1}$

Step 11.1.1.5

Divide $x^{3} + 6 x^{2} + 5 x - 12$ by $x - 1$ .

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Step 11.1.1.5.1

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

$x$

$1$

$x^{3}$

$6 x^{2}$

$5 x$

$12$

Step 11.1.1.5.2

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$

Step 11.1.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			+	$x^{3}$	-	$x^{2}$

Step 11.1.1.5.4

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$

Step 11.1.1.5.5

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$

Step 11.1.1.5.6

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$

Step 11.1.1.5.7

Divide the highest order term in the dividend $7 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$7 x$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$

Step 11.1.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$7 x$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$
					+	$7 x^{2}$	-	$7 x$

Step 11.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{2} - 7 x$

				$x^{2}$	+	$7 x$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$
					-	$7 x^{2}$	+	$7 x$

Step 11.1.1.5.10

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

				$x^{2}$	+	$7 x$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$
					-	$7 x^{2}$	+	$7 x$
							+	$12 x$

Step 11.1.1.5.11

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

				$x^{2}$	+	$7 x$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$
					-	$7 x^{2}$	+	$7 x$
							+	$12 x$	-	$12$

Step 11.1.1.5.12

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$7 x$	+	$12$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$
					-	$7 x^{2}$	+	$7 x$
							+	$12 x$	-	$12$

Step 11.1.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$7 x$	+	$12$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$
					-	$7 x^{2}$	+	$7 x$
							+	$12 x$	-	$12$
							+	$12 x$	-	$12$

Step 11.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $12 x - 12$

				$x^{2}$	+	$7 x$	+	$12$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$
					-	$7 x^{2}$	+	$7 x$
							+	$12 x$	-	$12$
							-	$12 x$	+	$12$

Step 11.1.1.5.15

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

				$x^{2}$	+	$7 x$	+	$12$
$x$	-	$1$		$x^{3}$	+	$6 x^{2}$	+	$5 x$	-	$12$
			-	$x^{3}$	+	$x^{2}$
					+	$7 x^{2}$	+	$5 x$
					-	$7 x^{2}$	+	$7 x$
							+	$12 x$	-	$12$
							-	$12 x$	+	$12$
										$0$

Step 11.1.1.5.16

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

$x^{2} + 7 x + 12$

Step 11.1.1.6

Write $x^{3} + 6 x^{2} + 5 x - 12$ as a set of factors.

$(x - 1) ((x - 1) (x^{2} + 7 x + 12))$

Step 11.1.2

Factor $x^{2} + 7 x + 12$ using the AC method.

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Step 11.1.2.1

Factor $x^{2} + 7 x + 12$ using the AC method.

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Step 11.1.2.1.1

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 $12$ and whose sum is $7$ .

$3, 4$

Step 11.1.2.1.2

Write the factored form using these integers.

$(x - 1) ((x - 1) ((x + 3) (x + 4)))$

Step 11.1.2.2

Remove unnecessary parentheses.

$(x - 1) ((x - 1) (x + 3) (x + 4))$

Step 11.2

Remove unnecessary parentheses.

$(x - 1) (x - 1) (x + 3) (x + 4)$

Step 12

Combine exponents.

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Step 12.1

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

${(x - 1)}^{1} (x - 1) (x + 3) (x + 4)$

Step 12.2

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

${(x - 1)}^{1} {(x - 1)}^{1} (x + 3) (x + 4)$

Step 12.3

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

${(x - 1)}^{1 + 1} (x + 3) (x + 4)$

Step 12.4

Add $1$ and $1$ .

${(x - 1)}^{2} (x + 3) (x + 4)$