looking at this proof: http://www.proofwiki...ence_of_Squares

I dont understand how it gets from (k+1)(2k2+7k+6)6

to

(k+1)(k+2)(2(k+1)+1)6

if someone can enlighten me that would be great. thanks in advance

# Induction help

Page 1 of 1## 3 Replies - 1922 Views - Last Post: 07 December 2013 - 01:47 PM

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**Replies To:** Induction help

### #2

## Re: Induction help

Posted 07 December 2013 - 11:36 AM

Well, it's been twenty years, but:

(2(k+1)+1) = 2k + 2 + 1 = 2k + 3

(k+2)(2k+3) = 2k^2 + 3k + 4k + 6 = 2k^2 + 7k + 6

(2(k+1)+1) = 2k + 2 + 1 = 2k + 3

(k+2)(2k+3) = 2k^2 + 3k + 4k + 6 = 2k^2 + 7k + 6

### #3

## Re: Induction help

Posted 07 December 2013 - 01:16 PM

yeah but where did they pull the (k+2) from? I just feel like theres steps missing.

### #4

## Re: Induction help

Posted 07 December 2013 - 01:47 PM

andrew showed you how to work backwards. We basically just reverse the process. With nice numbers, you can use a trial and error process to derive factors. You may have learned from algebra that (ax+b )(cx+d) = acx^2+(ad+bc)x+bd. So, when given an expression of the form nx^2+mx+y you need to come up with numbers for a,b,c and d such that

In your equation, n is 2, m is 7, and y is 6. The only nice numbers for ac=2 are a=1 and c=2. Now just solve the system of equations

So, bd is either (1 and 6) or (2 and 3). When we test d=3 and b=2, it satisfies both equations. So, the factored equation is (x+2)(2x+3).

ac=n ad+bc=m bd=y

In your equation, n is 2, m is 7, and y is 6. The only nice numbers for ac=2 are a=1 and c=2. Now just solve the system of equations

1*2=2 d+2b=7 bd=6

So, bd is either (1 and 6) or (2 and 3). When we test d=3 and b=2, it satisfies both equations. So, the factored equation is (x+2)(2x+3).

This post has been edited by **mojo666**: 07 December 2013 - 01:47 PM

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