Algebra Examples

$27 y^{3} - 27 y^{2} + 9 y - 1 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $27 y^{3} - 27 y^{2} + 9 y - 1$ using the rational roots test.

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Step 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.

$p = \pm 1$

$q = \pm 1, \pm 27, \pm 3, \pm 9$

Step 1.1.2

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

$\pm 1, \pm 0.\overline{037}, \pm 0.\overline{3}, \pm 0.\overline{1}$

Step 1.1.3

Substitute $0.\overline{3}$ and simplify the expression. In this case, the expression is equal to $0$ so $0.\overline{3}$ is a root of the polynomial.

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Step 1.1.3.1

Substitute $0.\overline{3}$ into the polynomial.

$27 \cdot {0.\overline{3}}^{3} - 27 \cdot {0.\overline{3}}^{2} + 9 \cdot 0.\overline{3} - 1$

Step 1.1.3.2

Raise $0.\overline{3}$ to the power of $3$ .

$27 \cdot 0.\overline{037} - 27 \cdot {0.\overline{3}}^{2} + 9 \cdot 0.\overline{3} - 1$

Step 1.1.3.3

Multiply $27$ by $0.\overline{037}$ .

$1 - 27 \cdot {0.\overline{3}}^{2} + 9 \cdot 0.\overline{3} - 1$

Step 1.1.3.4

Raise $0.\overline{3}$ to the power of $2$ .

$1 - 27 \cdot 0.\overline{1} + 9 \cdot 0.\overline{3} - 1$

Step 1.1.3.5

Multiply $- 27$ by $0.\overline{1}$ .

$1 - 3 + 9 \cdot 0.\overline{3} - 1$

Step 1.1.3.6

Subtract $3$ from $1$ .

$- 2 + 9 \cdot 0.\overline{3} - 1$

Step 1.1.3.7

Multiply $9$ by $0.\overline{3}$ .

$- 2 + 3 - 1$

Step 1.1.3.8

Add $- 2$ and $3$ .

$1 - 1$

Step 1.1.3.9

Subtract $1$ from $1$ .

$0$

Step 1.1.4

Since $0.\overline{3}$ is a known root, divide the polynomial by $3 y - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{27 y^{3} - 27 y^{2} + 9 y - 1}{3 y - 1}$

Step 1.1.5

Divide $27 y^{3} - 27 y^{2} + 9 y - 1$ by $3 y - 1$ .

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Step 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 $0$ .

$3 y$

$1$

$27 y^{3}$

$27 y^{2}$

$9 y$

$1$

Step 1.1.5.2

Divide the highest order term in the dividend $27 y^{3}$ by the highest order term in divisor $3 y$ .

					$9 y^{2}$
	$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$

Step 1.1.5.3

Multiply the new quotient term by the divisor.

				$9 y^{2}$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			+	$27 y^{3}$	-	$9 y^{2}$

Step 1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $27 y^{3} - 9 y^{2}$

				$9 y^{2}$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$

Step 1.1.5.5

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

				$9 y^{2}$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$

Step 1.1.5.6

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

				$9 y^{2}$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$

Step 1.1.5.7

Divide the highest order term in the dividend $- 18 y^{2}$ by the highest order term in divisor $3 y$ .

				$9 y^{2}$	-	$6 y$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$

Step 1.1.5.8

Multiply the new quotient term by the divisor.

				$9 y^{2}$	-	$6 y$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$
					-	$18 y^{2}$	+	$6 y$

Step 1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 18 y^{2} + 6 y$

				$9 y^{2}$	-	$6 y$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$
					+	$18 y^{2}$	-	$6 y$

Step 1.1.5.10

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

				$9 y^{2}$	-	$6 y$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$
					+	$18 y^{2}$	-	$6 y$
							+	$3 y$

Step 1.1.5.11

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

				$9 y^{2}$	-	$6 y$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$
					+	$18 y^{2}$	-	$6 y$
							+	$3 y$	-	$1$

Step 1.1.5.12

Divide the highest order term in the dividend $3 y$ by the highest order term in divisor $3 y$ .

				$9 y^{2}$	-	$6 y$	+	$1$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$
					+	$18 y^{2}$	-	$6 y$
							+	$3 y$	-	$1$

Step 1.1.5.13

Multiply the new quotient term by the divisor.

				$9 y^{2}$	-	$6 y$	+	$1$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$
					+	$18 y^{2}$	-	$6 y$
							+	$3 y$	-	$1$
							+	$3 y$	-	$1$

Step 1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $3 y - 1$

				$9 y^{2}$	-	$6 y$	+	$1$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$
					+	$18 y^{2}$	-	$6 y$
							+	$3 y$	-	$1$
							-	$3 y$	+	$1$

Step 1.1.5.15

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

				$9 y^{2}$	-	$6 y$	+	$1$
$3 y$	-	$1$		$27 y^{3}$	-	$27 y^{2}$	+	$9 y$	-	$1$
			-	$27 y^{3}$	+	$9 y^{2}$
					-	$18 y^{2}$	+	$9 y$
					+	$18 y^{2}$	-	$6 y$
							+	$3 y$	-	$1$
							-	$3 y$	+	$1$
										$0$

Step 1.1.5.16

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

$9 y^{2} - 6 y + 1$

Step 1.1.6

Write $27 y^{3} - 27 y^{2} + 9 y - 1$ as a set of factors.

$(3 y - 1) (9 y^{2} - 6 y + 1) = 0$

Step 1.2

Factor using the perfect square rule.

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Step 1.2.1

Rewrite $9 y^{2}$ as ${(3 y)}^{2}$ .

$(3 y - 1) ({(3 y)}^{2} - 6 y + 1) = 0$

Step 1.2.2

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

$(3 y - 1) ({(3 y)}^{2} - 6 y + 1^{2}) = 0$

Step 1.2.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 y = 2 \cdot (3 y) \cdot 1$

Step 1.2.4

Rewrite the polynomial.

$(3 y - 1) ({(3 y)}^{2} - 2 \cdot (3 y) \cdot 1 + 1^{2}) = 0$

Step 1.2.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 y$ and $b = 1$ .

$(3 y - 1) {(3 y - 1)}^{2} = 0$

Step 1.3

Combine like factors.

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Step 1.3.1

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

$(3 y - 1) {(3 y - 1)}^{2} = 0$

Step 1.3.2

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

${(3 y - 1)}^{1 + 2} = 0$

Step 1.3.3

Add $1$ and $2$ .

${(3 y - 1)}^{3} = 0$

Step 2

Set the $3 y - 1$ equal to $0$ .

$3 y - 1 = 0$

Step 3

Solve for $y$ .

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Step 3.1

Add $1$ to both sides of the equation.

$3 y = 1$

Step 3.2

Divide each term in $3 y = 1$ by $3$ and simplify.

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Step 3.2.1

Divide each term in $3 y = 1$ by $3$ .

$\frac{3 y}{3} = \frac{1}{3}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $3$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{1}{3}$

Step 3.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{1}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$y = \frac{1}{3}$

Decimal Form:

$y = 0.\overline{3}$