Simplifying (k(k+1)(2k+1))/6 + (k+1)^2

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
ScripterKitty
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Simplifying (k(k+1)(2k+1))/6 + (k+1)^2

Postby ScripterKitty » Tue Oct 09, 2012 9:42 pm

All,

Looking for some help in factoring the following equation:

(k(k+1)(2k+1))/6 + (k+1)^2

I have gotten this far:

(k(k+1)(2k+1)+6(k+1)^2)/6

I was checking the exponent rules and got the following for 6(k+1)^2:

6k^2 but I ran the same equation through the mathway solver and got 6k^2 + 12k + 6. I checked purple math tutorial and it said the exponent should apply to the sum of the numbers inside the parenthesis but in this case it seems it is applying to the numbers separately and the one on the outside of the parenthesis. I don't understand why or if this is even the next step I should be taking. Any help is appreciated.

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stapel_eliz
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Postby stapel_eliz » Tue Oct 09, 2012 11:16 pm

ScripterKitty wrote:I was checking the exponent rules and got the following for 6(k+1)^2:

6k^2 but I ran the same equation through the mathway solver and got 6k^2 + 12k + 6.

To learn how to multiply polynomials, please try this lesson. It will explain why 6(k + 1)(k + 1) does not equal 6(k)(k), and how to get the Mathway result. :wink:


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