Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 29, 2011"

(Solution)
(Solution)
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First we have the question: <math>(ab+1)(a+1)(b+1)+ab<math>
+
First we have the question: <math>(ab+1)(a+1)(b+1)+ab</math>
  
We multiply <math>(a+1)(b+1)<math> to get <math>(ab+1+a+b)<math>
+
We multiply <math>(a+1)(b+1)</math> to get <math>(ab+1+a+b)</math>
  
This makes the equation <math>(ab+1)(ab+1+a+b)+ab<math>
+
This makes the equation <math>(ab+1)(ab+1+a+b)+ab</math>
  
 
Now we seperate the equation to  <math>(ab+1)(ab+1)+(ab+1)(a+b)+ab</math>
 
Now we seperate the equation to  <math>(ab+1)(ab+1)+(ab+1)(a+b)+ab</math>
  
We get <math>(ab+1)^2 +(a+b)(ab+1)+ab<math>
+
We get <math>(ab+1)^2 +(a+b)(ab+1)+ab</math>
  
 
Now this is just a quadratic equation
 
Now this is just a quadratic equation

Revision as of 23:10, 28 June 2011

Problem

AoPSWiki:Problem of the Day/June 29, 2011

Solution

This Problem of the Day needs a solution. If you have a solution for it, please help us out by adding it.


First we have the question: $(ab+1)(a+1)(b+1)+ab$

We multiply $(a+1)(b+1)$ to get $(ab+1+a+b)$

This makes the equation $(ab+1)(ab+1+a+b)+ab$

Now we seperate the equation to $(ab+1)(ab+1)+(ab+1)(a+b)+ab$

We get $(ab+1)^2 +(a+b)(ab+1)+ab$

Now this is just a quadratic equation

Thus we get a factored form of:

$(ab+1+a)(ab+1+b)$

That is the solution