# Homework Help: Factorizing & Manipulation

1. Sep 3, 2009

### roam

1. The problem statement, all variables and given/known data

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 [Broken]

2. Relevant equations

3. 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 $$(k^2+4k+4)$$

Last edited by a moderator: May 4, 2017
2. Sep 3, 2009

### symbolipoint

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.

3. Sep 3, 2009

### NJunJie

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.

4. Sep 3, 2009

### roam

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 [Broken]

Last edited by a moderator: May 4, 2017
5. Sep 3, 2009

### njama

$$[\frac{1}{2}k(k+1)]^2 + (k+1)^3=(\frac{1}{2})^2k^2(k+1)^2+(k+1)(k+1)^2$$

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

6. Sep 3, 2009

### roam

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:

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

:uhh:

7. Sep 3, 2009

### njama

You are wrong. :yuck:

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.

8. Sep 3, 2009

### roam

Awww! Gee! I see what you mean now! Thanks a lot for the help.