UnkleRhaukus
  • UnkleRhaukus
Simplify : \[=\frac{1}{4(b-a)^2}\left(b^3-3a^2b+6a^3\right)\] NB: has a nice solution
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
should i try; partial fractions, polynomial long division, some other complicated method, .../?
UnkleRhaukus
  • UnkleRhaukus
i am not confident in these methods
KingGeorge
  • KingGeorge
First guess is to use polynomial long division.

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UnkleRhaukus
  • UnkleRhaukus
bother, polynomial long division is not my strong suite.
KingGeorge
  • KingGeorge
If I use long division I get\[{1 \over 4}\cdot\left({{4a^3 \over {(b-a)}^2}+2a+b}\right)\]Which is a little bit nicer.
UnkleRhaukus
  • UnkleRhaukus
the solution in the back of my book says \[=\frac{2a+b}{4}\]
KingGeorge
  • KingGeorge
Well, \[{a^3 \over {(b-a)}^2} \neq 0\]so it looks like they left out a term to me.
UnkleRhaukus
  • UnkleRhaukus
yeah the solution could be incorrect,
UnkleRhaukus
  • UnkleRhaukus
\[=\frac{1}{4(b-a)^2}\left(b^3-3a^2b+6a^3\right)\] \[=\frac{1}{4} \times (b-a)^2 \overline {\bigg)b^3-3a^2b+6a^3}\] \[=\cdots \]
UnkleRhaukus
  • UnkleRhaukus
can you tell me the steps i always forget
KingGeorge
  • KingGeorge
First FOIL \((b-a)^2=b^2-2ab+a^2\). Then let's divide with respect to \(b\) first. so we need to find what we multiply \(b^2\) by to get \(b^3\). This is \(b\). So that's the first thing we put on top of the division symbol thing.
UnkleRhaukus
  • UnkleRhaukus
\[\quad\qquad\qquad\qquad b\]\[=\frac{1}{4} \times (b-a)^2 \overline {\bigg)b^3-3a^2b+6a^3}\]\[\quad\qquad\qquad\qquad b^3-2b^2a+a^2b\]
UnkleRhaukus
  • UnkleRhaukus
\[\quad\qquad\qquad\qquad b\]\[=\frac{1}{4} \times (b-a)^2 \overline {\bigg)b^3-3a^2b+6a^3}\]\[\quad\qquad\qquad\qquad b^3-2b^2a+a^2\]\[\quad\qquad\qquad\qquad \quad-3a^2b+2b^2a+5a^2\]
KingGeorge
  • KingGeorge
So now you have (with some steps not included) \[={1 \over 4}\left(b+ {2ab^2-4a^2b+6a^3 \over b^2-2ab+a^2}\right)\]Since \[b^3-3a^2b+6a^3 - (b^3-2b^2a+a^2b)=2ab^2-4a^2b+6a^3\]Remember you're subtracting an \(a^2b\) not just an \(a^2\)
UnkleRhaukus
  • UnkleRhaukus
\[\quad\qquad\qquad\qquad b-\frac{3}{2} a \]\[=\frac{1}{4} \times (b-a)^2 \overline {\bigg)b^3-3a^2b+6a^3}\]\[\quad\qquad\qquad\qquad b^3-2b^2a+a^2\]\[\quad\qquad\qquad\qquad \quad-3a^2b+2b^2a+5a^2\]\[\quad\qquad\qquad\qquad \quad \frac{3}{2}ab^2-3a^2b+\frac{3}{2}a^3\]
KingGeorge
  • KingGeorge
Now that we have this new equation, do long division one more time with respect to a this time. If I've done everything correct, it should be what we're looking for.
KingGeorge
  • KingGeorge
|dw:1333344745070:dw|
KingGeorge
  • KingGeorge
If we do the same process again except dividing with respect to a instead, we find that we have \[={1 \over 4}\left( b+6a+{8a^2b-4ab^2 \over b^2-2ab+a^2} \right)\]
KingGeorge
  • KingGeorge
Now we finally do one more long division step where we divide with respect to \(2ab\) going into \(8a^2b\). This results in the equation\[{1\over4}\left( b+6a-4a+{4a^3 \over (b-a)^2} \right)={a^3 \over (b-a)^2}+{2a+b \over 4}\]Which is what we wanted.
UnkleRhaukus
  • UnkleRhaukus
i am sorry the solution is not as nice suggested
KingGeorge
  • KingGeorge
It's still fairly nice. Instead of dividing a large polynomial by another polynomial, we're only dividing a monomial by a polynomial now.
UnkleRhaukus
  • UnkleRhaukus
OK so i gave you the wrong question but i have managed to solve the problem i was having by studying your working. I should have been asking \[=\frac{1}{4} \times (b-a)^2 \overline {\big)b^3-3a^2b+2a^3}\]
UnkleRhaukus
  • UnkleRhaukus
but using polynomial long division i have solved it \[\quad\qquad\qquad\qquad\qquad\qquad b\]\[=\frac{1}{4} \times (b^2-2ab+a^2) \overline {\bigg)b^3-3a^2b+2a^3}\]\[\qquad\qquad\qquad\qquad\qquad b^3-2ab^2+a^2b\]\[\quad\qquad\qquad\qquad\qquad b+2a\]\[=\frac{1}{4} \times (b^2-2ab+a^2) \overline {\bigg)b^3-3a^2b+2a^3\quad}\]\[\qquad\qquad\qquad\qquad\qquad \underline{b^3-2ab^2+a^2b\quad}\]\[\qquad\qquad\qquad\qquad\qquad 2ab^2-4a^2b+2a^3\]\[\qquad\qquad\qquad\qquad\qquad 2ab^2-4a^2b+\underline{2a^3}\]\[\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad 0\]\[=\frac{b+2a}{4}\]
UnkleRhaukus
  • UnkleRhaukus
I thank you for your efforts. this problem is just a step in a larger question, this was the final step
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KingGeorge
  • KingGeorge
Oh wow that looks intense.
UnkleRhaukus
  • UnkleRhaukus
yeah i have been on this question for weeks, but now i can finally move on, ill be more confident in my polynomial long division for next time

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