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Apologies in advance if this problem is a duplicate. I'm not even sure how to properly research it.

Studying algorithms (actually still in the prerequisite step of data structures) and an explanation of the Principal of Induction demonstrates why:

starting equation

I was able to deduct that (j)(j + 1) + 2(j + 1) = j^2 + j + 2j + 2, because you can separately multiply the elements in the parentheses, so (j)(j + 1) can be simplified to j x j or j^2 plus j x 1 or j, and similarly with 2(j + 1).

But going from

middle equation

to

end equation

I can't quite figure out why (j + 1)(j + 2) = j^2 + 3j + 2.

I can fill in j with a number and see that the value is the same, but why?

Wait - I think I'm figuring it out. At least if I do it backwards. Same principal.

Take the j in the first paragraph and multiply it by the one in the second: j^2, then multiply it by the 2, so we've got 2j, which leaves us with 1 x J or j and 1 x 2 or 2. So that's a total of 3j, j^2 and 2.

Going forwards from the first equation to the second one, we can take j^2 and make it (j)(j), but how do 3j and 2 get incorporated into it?

Wait wait - to get the 3j we can add 2 to the second (j) (making (j + 2)), so that the first (j) is multiplied by 2 to give us 2j. Then we add a 1 to the first (j) (making (j + 1)), which is multiplied by each item in (j + 2) resulting in 1j or (j); the 3rd j in 3j, and the + 2 is the result of the 1 x 2.

Is there a simple rule or formula, or combination of them to get from j^2 + 3j + 2 to (j + 1)(j + 2)?

MikeiLL
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    I am having some trouble following your question, but I think what you are asking is: How do I factor a quadratic polynomial? There are a number of basic tutorials on this online, e.g. http://www.purplemath.com/modules/factquad.htm. – mweiss Sep 05 '14 at 15:07
  • Yes. I think I need to add the following three words to my vocabulary (once I figure out what they mean) factor, quadratic and polynomial. – MikeiLL Sep 05 '14 at 15:27

1 Answers1

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You have deduced that $$j(j+1)+2(j+1)=j^2+j+2j+2$$Now you can change the order of the middle terms and reverse the procedure to get $$=j^2+2j+j+2=j(j+2)+(j+2)=(j+1)(j+2)$$

That depends on noticing something special about your particular case. A similar thing can be done in many other cases, but that depends on knowing what you are looking for.

Mark Bennet
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  • Yes. I think I need to research the phrase "middle term" and the words factor, quadratic and polynomial. – MikeiLL Sep 05 '14 at 15:29
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    @MikeiLL "Middle term" is rather informal. To factor the quadratic $x^2+(a+b)x+ab$ you can rewrite as $x^2+ax+bx+ab=x(x+a)+b(x+a)=(x+b)(x+a)$ or exchange the role of $a$ and $b$. Notice how the coefficient $x$ is the sum $a+b$ and the constant term is the product $ab$. – Mark Bennet Sep 05 '14 at 15:33
  • I don't even remember what coefficient means, man. Maybe I looked at that word in the 1980's in high school. Need to spend a little more intimate time with PurpleMath to refresh on all this. Didn't realize website development would end up involving so much math. This is algebra, right? – MikeiLL Sep 05 '14 at 17:06