What is 2(-2k^3+2k+k^2-k^4+(-8k+5k^4+8k^3)/6) ?

The solution to 2(-2k^3+2k+k^2-k^4+(-8k+5k^4+8k^3)/6) is (-k^4-4k^3+6k^2+4k)/3

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simplify 2(-2k^3+2k+k^2-k^4+(-8k+5k^4+8k^3)/6)

simplify

2 (- 2 k^{3} + 2 k + k^{2} - k^{4} + \frac{- 8 k + 5 k ^{4} + 8 k ^{3}}{6})

\frac{- k ^{4} - 4 k ^{3} + 6 k ^{2} + 4 k}{3}

Solution steps

2 (- 2 k^{3} + 2 k + k^{2} - k^{4} + \frac{- 8 k + 5 k ^{4} + 8 k ^{3}}{6})

Join

- 2 k^{3} + 2 k + k^{2} - k^{4} + \frac{- 8 k + 5 k ^{4} + 8 k ^{3}}{6} : \frac{- k ^{4} - 4 k ^{3} + k ^{2} \cdot 6 + 4 k}{6}

= 2 \cdot \frac{- k ^{4} - 4 k ^{3} + k ^{2} \cdot 6 + 4 k}{6}

= 2 \cdot \frac{- k ^{4} - 4 k ^{3} + 6 k ^{2} + 4 k}{6}

Apply the fraction rule:

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

= \frac{2 ( - k ^{4} - 4 k ^{3} + 6 k ^{2} + 4 k )}{6}

Cancel

\frac{2 ( - k ^{4} - 4 k ^{3} + 6 k ^{2} + 4 k )}{6} : \frac{- k ^{4} - 4 k ^{3} + 6 k ^{2} + 4 k}{3}

= \frac{- k ^{4} - 4 k ^{3} + 6 k ^{2} + 4 k}{3}

(x + 6) l e f t (x - 6)

5 (x + Δ x)

(5 x - e^{- x})^{2}

\frac{1}{( s ^{2} + 4 s + 5 ) ^{2}}

(x - 3 y = 6,) - (x - 7 y = 6)

What is 2(-2k^3+2k+k^2-k^4+(-8k+5k^4+8k^3)/6) ?
The solution to 2(-2k^3+2k+k^2-k^4+(-8k+5k^4+8k^3)/6) is (-k^4-4k^3+6k^2+4k)/3