How do you simplify $\frac{40 k^{4}}{4 k} - \frac{44 k^{3}}{4 k} - \frac{60 k^{2}}{4 k} - \frac{4 k}{4 k} - \frac{21}{4 k}$ $\frac { 40k ^ { 4} } { 4k } - \frac { 44k ^ { 3} } { 4k } - \frac { 60k ^ { 2} } { 4k } - \frac { 4k } { 4k } - \frac { 21} { 4k }$ ?

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How do you simplify $\frac{40 k^{4}}{4 k} - \frac{44 k^{3}}{4 k} - \frac{60 k^{2}}{4 k} - \frac{4 k}{4 k} - \frac{21}{4 k}$ $\frac { 40k ^ { 4} } { 4k } - \frac { 44k ^ { 3} } { 4k } - \frac { 60k ^ { 2} } { 4k } - \frac { 4k } { 4k } - \frac { 21} { 4k }$ ?

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Answer 1 · 2018-04-24T22:52:53

$\frac{40 k^{4} - 44 k^{3} - 60 k^{2} - 4 k - 21}{4 k}$ $(40k^4-44k^3-60k^2-4k-21)/(4k)$ or

$10 k^{3} - 11 k^{2} - 15 k - 1 - (\frac{21}{4} k)$ $10k^3-11k^2-15k-1-(21/4k)$

Explanation:

The denominator is the same for every term, so this will just be equal to

$\frac{40 k^{4} - 44 k^{3} - 60 k^{2} - 4 k - 21}{4 k}$ $(40k^4-44k^3-60k^2-4k-21)/(4k)$

The numerator is unfactorable, but we can divide all terms by $4 k$ $4k$ . We get:

$10 k^{3} - 11 k^{2} - 15 k - 1 - (\frac{21}{4} k)$ $10k^3-11k^2-15k-1-(21/4k)$

Answer 2 · 2018-04-24T22:57:17

$10 k^{3} - 11 k^{2} - 15 k - 1 - \frac{21}{4 k}$ $10k^3 - 11k^2 - 15k - 1 - 21/(4k)$

Explanation:

Notice that the variable, k, cancels out as:
$\frac{k^{a}}{k^{b}} = k^{a - b}$ $k^a/k^b = k^(a-b)$

This goes for any variable in division. The coefficients, or the constants, simplify as normal division. Such as 40/4 is 10.

$\frac{21}{4}$ $21/4$ does not cancel or simply, therefore it can be left alone. We could take it to $\frac{1}{4} \cdot (\frac{21}{k})$ $1/4*(21/k)$ if we wanted, but it's more simple as is.

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How do you simplify $\frac{40 k^{4}}{4 k} - \frac{44 k^{3}}{4 k} - \frac{60 k^{2}}{4 k} - \frac{4 k}{4 k} - \frac{21}{4 k}$ $\frac { 40k ^ { 4} } { 4k } - \frac { 44k ^ { 3} } { 4k } - \frac { 60k ^ { 2} } { 4k } - \frac { 4k } { 4k } - \frac { 21} { 4k }$ ?

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Explanation:

Explanation:

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Explanation:

Explanation:

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How do you simplify $\frac{40 k^{4}}{4 k} - \frac{44 k^{3}}{4 k} - \frac{60 k^{2}}{4 k} - \frac{4 k}{4 k} - \frac{21}{4 k}$ $\frac { 40k ^ { 4} } { 4k } - \frac { 44k ^ { 3} } { 4k } - \frac { 60k ^ { 2} } { 4k } - \frac { 4k } { 4k } - \frac { 21} { 4k }$ ?