Algebra Pre-Algebra and Basic Algebra Math Forum

 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 $\frac{(p^2+pr+pb+rb)(p+b)}{(p-b)(p+r)}$ 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 "step-by-stepper": [(p^2+pr+bp+br) / (p-b)] / [(p+r) / (p+b)] [(p^2+bp+pr+br) / (p-b)] * [(p+b) / (p+r)] [(p(p+b) + r(p+b)) / (p-b)] * [(p+b) / (p+r)] Let k = p+b: [(pk + rk) / (p-b)] * [k / (p+r)] (pk^2 + rk^2) / [(p-b)(p+r)] k^2(p+r) / [(p-b)(p+r)] k^2 / (p-b) ; substitute back in: (p+b)^2 / (p-b) 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 / (p-b) needs to be footnoted with p<>b. Tags division or factoring, polynomials Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post J.Xavier Algebra 1 May 5th, 2013 05:29 PM alexmath Algebra 2 April 18th, 2013 06:53 AM cafegurl Algebra 2 September 26th, 2010 02:55 PM watkd Algebra 13 August 20th, 2010 11:24 PM Voltman Algebra 2 April 9th, 2009 07:14 AM

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