Solve for y 27y^3-27y^2+9y-1=0

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

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.

$p = \pm 1$

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

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm \frac{1}{9}, \pm \frac{1}{27}$

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.

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

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $y = \frac{1}{3}$ is a root of the polynomial.

Simplify each term.

Apply the product rule to $\frac{1}{3}$ .

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

One to any power is one.

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

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

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

Cancel the common factor of $27$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the product rule to $\frac{1}{3}$ .

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

One to any power is one.

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

Raise $3$ to the power of $2$ .

$1 - 27 (\frac{1}{9}) + 9 (\frac{1}{3}) - 1$

Cancel the common factor of $9$ .

Factor $9$ out of $- 27$ .

$1 + 9 (- 3) \frac{1}{9} + 9 (\frac{1}{3}) - 1$

Cancel the common factor.

$1 + 9 \cdot - 3 \frac{1}{9} + 9 (\frac{1}{3}) - 1$

Rewrite the expression.

$1 - 3 + 9 (\frac{1}{3}) - 1$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$1 - 3 + 3 (3) \frac{1}{3} - 1$

Cancel the common factor.

$1 - 3 + 3 \cdot 3 \frac{1}{3} - 1$

Rewrite the expression.

$1 - 3 + 3 - 1$

Simplify by adding and subtracting.

Subtract $3$ from $1$ .

$- 2 + 3 - 1$

Add $- 2$ and $3$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

Step 5

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

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

Step 6

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

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

$\frac{1}{3}$	$27$	$- 27$	$9$	$- 1$
	\xa0			\xa0

The first number in the dividend $(27)$ is put into the first position of the result area (below the horizontal line).

$\frac{1}{3}$	$27$	$- 27$	$9$	$- 1$
	\xa0			\xa0
	$27$

Multiply the newest entry in the result $(27)$ by the divisor $(\frac{1}{3})$ and place the result of $(9)$ under the next term in the dividend $(- 27)$ .

$\frac{1}{3}$	$27$	$- 27$	$9$	$- 1$
	\xa0	$9$		\xa0
	$27$

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

$\frac{1}{3}$	$27$	$- 27$	$9$	$- 1$
	\xa0	$9$		\xa0
	$27$	$- 18$

Multiply the newest entry in the result $(- 18)$ by the divisor $(\frac{1}{3})$ and place the result of $(- 6)$ under the next term in the dividend $(9)$ .

$\frac{1}{3}$	$27$	$- 27$	$9$	$- 1$
	\xa0	$9$	$- 6$	\xa0
	$27$	$- 18$

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

$\frac{1}{3}$	$27$	$- 27$	$9$	$- 1$
	\xa0	$9$	$- 6$	\xa0
	$27$	$- 18$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(\frac{1}{3})$ and place the result of $(1)$ under the next term in the dividend $(- 1)$ .

$\frac{1}{3}$	$27$	$- 27$	$9$	$- 1$
	\xa0	$9$	$- 6$	$1$
	$27$	$- 18$	$3$

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

$\frac{1}{3}$	$27$	$- 27$	$9$	$- 1$
	\xa0	$9$	$- 6$	$1$
	$27$	$- 18$	$3$	$0$

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

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

Simplify the quotient polynomial.

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

Step 7

Factor $3$ out of $27 y^{2} - 18 y + 3$ .

Factor $3$ out of $27 y^{2}$ .

$(y - \frac{1}{3}) (3 (9 y^{2}) - 18 y + 3)$

Factor $3$ out of $- 18 y$ .

$(y - \frac{1}{3}) (3 (9 y^{2}) + 3 (- 6 y) + 3)$

Factor $3$ out of $3$ .

$(y - \frac{1}{3}) (3 (9 y^{2}) + 3 (- 6 y) + 3 (1))$

Factor $3$ out of $3 (9 y^{2}) + 3 (- 6 y)$ .

$(y - \frac{1}{3}) (3 (9 y^{2} - 6 y) + 3 (1))$

Factor $3$ out of $3 (9 y^{2} - 6 y) + 3 (1)$ .

$(y - \frac{1}{3}) (3 (9 y^{2} - 6 y + 1))$

Step 8

Factor using the perfect square rule.

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

$(y - \frac{1}{3}) (3 ({(3 y)}^{2} - 6 y + 1))$

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

$(y - \frac{1}{3}) (3 ({(3 y)}^{2} - 6 y + 1^{2}))$

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

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

Rewrite the polynomial.

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

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

$(y - \frac{1}{3}) (3 {(3 y - 1)}^{2})$

Step 9

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

$y - \frac{1}{3} = 0$

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

Step 10

Set $y - \frac{1}{3}$ equal to $0$ and solve for $y$ .

Set $y - \frac{1}{3}$ equal to $0$ .

$y - \frac{1}{3} = 0$

Add $\frac{1}{3}$ to both sides of the equation.

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

Step 11

The final solution is all the values that make $(y - \frac{1}{3}) (3 {(3 y - 1)}^{2}) = 0$ true.

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

Step 12

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$y = 0.\overline{3}$

Solve for y 27y^3-27y^2+9y-1=0

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