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April 26th, 2014, 10:03 PM   #1
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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?


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April 27th, 2014, 05:36 AM   #2
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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 $p+b$.
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April 27th, 2014, 08:05 AM   #3
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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.
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