What is expand (2y^2(3y^3+1))/(19) ?

The solution to expand (2y^2(3y^3+1))/(19) is (6y^5)/(19)+(2y^2)/(19)

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expand (2y^2(3y^3+1))/(19)

expand

\frac{2 y ^{2} ( 3 y ^{3} + 1 )}{19}

\frac{6 y ^{5}}{19} + \frac{2 y ^{2}}{19}

Solution steps

\frac{2 y ^{2} ( 3 y ^{3} + 1 )}{19}

Expand

2 y^{2} (3 y^{3} + 1) : 6 y^{5} + 2 y^{2}

= \frac{6 y ^{5} + 2 y ^{2}}{19}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{6 y ^{5} + 2 y ^{2}}{19} = \frac{6 y ^{5}}{19} + \frac{2 y ^{2}}{19}

= \frac{6 y ^{5}}{19} + \frac{2 y ^{2}}{19}

\frac{d ^{3}}{d x ^{3}} ((y) (2 x^{4} + 3 x^{2} + 12))

\frac{0.07}{2} (100 + \frac{L}{2})^{2}

\frac{s + 1}{s ( s - 1 ) ( s ^{2} + 4 s + 16 )}

\frac{5 ( x - 140 ) ( x - 220 ) ( x - 240 )}{64}

\frac{( 2 a - b ) - 3}{4 - 2 ( 2 a - b ) ^{2}}

What is expand (2y^2(3y^3+1))/(19) ?
The solution to expand (2y^2(3y^3+1))/(19) is (6y^5)/(19)+(2y^2)/(19)