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April 26th, 2014, 10:03 PM  #1 
Newbie Joined: Apr 2014 From: United States Posts: 2 Thanks: 0  Division/Factoring of Polynomials Ok so I am familiar with these problems, but on mymathlab the examples are not that detailed. It shows what they factor into, but not how or why.... Could I get some help breaking this problem down? Thank You 
April 27th, 2014, 05:36 AM  #2 
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 
Since this equals \[ \frac{(p^2+pr+pb+rb)(p+b)}{(pb)(p+r)} \] it is natural to try to divide $p^2+pr+pb+rb$ by $p+r$. If \[ p^2+pr+pb+rb=(x+y)(p+r) \] then \[ p^2+pr+pb+rb=xp+xr+py+ry \] We see that this is possible if $x=p$ and $y=b$. Well, I am not sure I would recommend doing it exactly like this, but it is clear that if $p^2+pr+pb+rb$ is divisible by $p+r$, the other factor must have $p$ (because of $p^2$) and $b$ (because of $rb$).Therefore, it is natural to try $p+b$. 
April 27th, 2014, 08:05 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
Hope you can follow this "stepbystepper": [(p^2+pr+bp+br) / (pb)] / [(p+r) / (p+b)] [(p^2+bp+pr+br) / (pb)] * [(p+b) / (p+r)] [(p(p+b) + r(p+b)) / (pb)] * [(p+b) / (p+r)] Let k = p+b: [(pk + rk) / (pb)] * [k / (p+r)] (pk^2 + rk^2) / [(pb)(p+r)] k^2(p+r) / [(pb)(p+r)] k^2 / (pb) ; substitute back in: (p+b)^2 / (pb) You can check these yourself by assigning values to the variables; try p=2,r=3,b=4 Substitute in original expression, and in simplified expression; in this case, you'll get both = 18 And of course (p+b)^2 / (pb) needs to be footnoted with p<>b. 

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division or factoring, polynomials 
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