What is the Manipulation of Algebraic Expressions?

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The discussion revolves around manipulating algebraic expressions to simplify and prove equality between two sides of an equation. Participants explore how to transform the right-hand side expression into a form that matches the left-hand side, specifically focusing on the expression involving (k+1). There is a debate about whether it's necessary to show each step of manipulation and the validity of using substitutions to simplify complex expressions. Ultimately, the conversation highlights the importance of correct factorization and understanding the relationships between algebraic terms. The resolution comes when a participant realizes the correct factorization leads to the desired equality.
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Homework Statement



I need some help with this Algebra problem. In the following I don't know how they manipulated the RHS to get to the LHS:

http://img402.imageshack.us/img402/7176/solns.gif

Homework Equations



The Attempt at a Solution



\frac{1}{4}(k+1)^2 .k^2 + (k+1)^3

\frac{1}{4}(k+1)^2 .k^2 + k^3 + 3k^2 +3k +1

I don't know how to manipulate k^2 + k^3 + 3k^2 +3k +1 into (k^2+4k+4)
 
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Are you required to carry steps to change the right-side to be equal to the left-side? You should be allowed to carry steps on both sides so that you can show the right and left sides are equal to a third expression. The idea is, that if a=b, and if b=c, then a=c.
 
another opinion of mine personally is to resolve 'complex' question initially is to model them by substitution.
Like letting another alegrabic representation (eg) Let a = K+1 and then it will look simpler.

Hope it helps.
 
symbolipoint said:
Are you required to carry steps to change the right-side to be equal to the left-side? You should be allowed to carry steps on both sides so that you can show the right and left sides are equal to a third expression. The idea is, that if a=b, and if b=c, then a=c.

Well I'm not sure what they've done there. I mean how they simplified [\frac{1}{2}k(k+1)]^2 + (k+1)^3 into \frac{1}{4} (k+1)^2 (k^2+4k+4)

Anyway, here's the rest (it's from a proof by induction problem):

http://img200.imageshack.us/img200/6978/62815040.gif
 
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[\frac{1}{2}k(k+1)]^2 + (k+1)^3=(\frac{1}{2})^2k^2(k+1)^2+(k+1)(k+1)^2

Now just factor (k+1)2 and you are done. :approve:
 
njama said:
[\frac{1}{2}k(k+1)]^2 + (k+1)^3=(\frac{1}{2})^2k^2(k+1)^2+(k+1)(k+1)^2

Now just factor (k+1)2 and you are done. :approve:

But that doesn't work! If I factor out the (k+1)2 I will have:

\frac{1}{4}(k+1)^2 (k^2+k+1)

Which is not the same as:

\frac{1}{4} (k+1)^2 (k^2+4k+4)

:rolleyes:
 
roam said:
But that doesn't work! If I factor out the (k+1)2 I will have:

\frac{1}{4}(k+1)^2 (k^2+k+1)

Which is not the same as:

\frac{1}{4} (k+1)^2 (k^2+4k+4)

:rolleyes:

You are wrong.

Do the factorization again.

=(k+1)^2(\frac{1}{4}k^2+k+1)

Now factor 1/4 and see what will you come up with.
 
Awww! Gee! I see what you mean now! Thanks a lot for the help.
 
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