What is (2x^2+x)/6*8/(4x+2) ?

The solution to (2x^2+x)/6*8/(4x+2) is (2x)/3

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simplify (2x^2+x)/6*8/(4x+2)

simplify

\frac{2 x ^{2} + x}{6} \cdot \frac{8}{4 x + 2}

\frac{2 x}{3}

Solution steps

\frac{2 x ^{2} + x}{6} \cdot \frac{8}{4 x + 2}

Cancel

\frac{8}{4 x + 2} : \frac{4}{2 x + 1}

= \frac{2 x ^{2} + x}{6} \cdot \frac{4}{2 x + 1}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( 2 x ^{2} + x ) \cdot 4}{6 ( 2 x + 1 )}

Factor the number:

4 = 2 \cdot 2

= \frac{( 2 x ^{2} + x ) \cdot 2 \cdot 2}{6 ( 2 x + 1 )}

Factor the number:

6 = 2 \cdot 3

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

Cancel the common factor:

2

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

Factor

2 x^{2} + x : x (2 x + 1)

= \frac{x ( 2 x + 1 ) \cdot 2}{3 ( 2 x + 1 )}

Cancel the common factor:

2 x + 1

= \frac{x \cdot 2}{3}

= \frac{2 x}{3}

\frac{x ^{2}}{4 x}

8 c^{3} - 27

t^{2} - 8 t + 16

\frac{5}{12} + \frac{5}{6}

8 \frac{1}{5}