What is ((18n^2-9n+1))/((3n-1)) ?

The solution to ((18n^2-9n+1))/((3n-1)) is 6n-1

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simplify ((18n^2-9n+1))/((3n-1))

simplify

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

6 n - 1

Solution steps

\frac{18 n ^{2} - 9 n + 1}{3 n - 1}

Factor

18 n^{2} - 9 n + 1 : (6 n - 1) (3 n - 1)

= \frac{( 6 n - 1 ) ( 3 n - 1 )}{3 n - 1}

Cancel the common factor:

3 n - 1

= 6 n - 1

\frac{csc ( x )}{sec ( x )}

\frac{3}{11} - \frac{2}{3}

sin (θ) + cos (θ) cot (θ)

\frac{1}{cos ( θ )}

(3 x^{2})^{2} \cdot (\frac{1}{3} x)^{4}

What is ((18n^2-9n+1))/((3n-1)) ?
The solution to ((18n^2-9n+1))/((3n-1)) is 6n-1