polynomial long division: (a^4 + a^2 - 20) / (a - 2)

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santasad13

Junior Member
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Jan 26, 2006
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more help please-
divide a^4 + a^2 - 20 by a - 2

a. use long division

i did that and ended up with a^3 + 2a^2 + 5a + 10

b. factor a^4 + a^2 - 20 first, than divide by a - 2

i cant figure out how to factor it and get the answer it says below

c. show how you got the same answer.
 
You posted late at night and waited a whole hour. Show a little patience, will you! If you want instant repsonse, you must pay for a service that offers instant response. Try to keep that in mind.

Your long division looks fine. Were you doubting?

Can you factor \(\displaystyle (x^{2}\;+\;x\;-\;20)\)?
 
oh, so it would be (a^2 - 4) (a^2 + 5)=( a+2)( a -2)(a^2 + 5)=( a+2)(a^2 + 5)=a^3 + 2a^2 + 5a + 10

when you put it that way, it make sense. thanks!
 
That's the spirit. Learn to recognize a quadratic form.

(three dogs and a cow)^2 - (three dogs and a cow) - 20 = 0

You should be able to factor that.
 
(three dogs and a cow)^2 - (three dogs and a cow) - 20 = 0

(three dogs and a cow - 5)(three dogs and a cow +4)=0
 
Very nice. How about this one...

\(\displaystyle \L\;9v^{\frac{1}{4}}\;-\;v^{\frac{1}{2}}\;-\;20\;=\;0\)

1) It's a trick. Consider it carefully before proceeding.
2) If you get it on the first try, I'll give you a gold star and dub you "The Quadratic Factor Master".
 
what i do know is that the first term is a perfect square, but the other two are not, which rules out perfest square trinomial and difference of perfect squares. i cant seem to figure out any combination that would work when you multiply it back out. there must be something im missing..............
although im not giving up
 
You fell for it.

Hint:
\(\displaystyle v^{1/2}*v^{1/2}\;=\;v\)

\(\displaystyle v^{1/4}*v^{1/4}\;=\;v^{1/2}\)
 
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