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

Latest revision as of 17:48, 1 July 2011

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:

$\boxed{(ab+1+a)(ab+1+b)}$

@@ Line 2: / Line 2: @@
 {{:AoPSWiki:Problem of the Day/June 29, 2011}}
 ==Solution==
-{{potd_solution}}
+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>
-First we have the question: <math>(ab+1)(a+1)(b+1)+ab<math>
+This makes the equation <math>(ab+1)(ab+1+a+b)+ab</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>
 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
@@ Line 19: / Line 16: @@
 Thus we get a factored form of:
-<math>(ab+1+a)(ab+1+b)</math>
+<math>\boxed{(ab+1+a)(ab+1+b)}</math>
-That is the solution