anonymous
  • anonymous
1.Factorise using the remainder theorem 2x^3 - 9x^2 - 11x + 30. 2.Factorise : (2y+x)^2(y-2x)+(2x+y)^2(2x-y)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
plzz answer..>!!:((
anonymous
  • anonymous
@jdoe0001 plzz cn u answer any 1.?? plzz
jdoe0001
  • jdoe0001
hmmm, do you know what the remainder theorem is? because if the exercises asks for it,... it's assume you're supposed to know it already :S

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anonymous
  • anonymous
ik wat is remainder theoram..bt i dk hw to do dis 1.. HELP:( ;9
jdoe0001
  • jdoe0001
well,... I looked and looked, and I don't think the remainder ALONE will do it the way I've do it is using the "rational root test" and the rational root test uses the remainder theorem but just a bit
anonymous
  • anonymous
k..cn u shw hw??
anonymous
  • anonymous
@jdoe0001 cn u help me??
jdoe0001
  • jdoe0001
http://www.youtube.com/watch?v=GTx-rXTQg1o <---
jdoe0001
  • jdoe0001
check that, let me know for anything confusing :)
anonymous
  • anonymous
do u noe hw to do d second question?
jdoe0001
  • jdoe0001
well, the second is just factoring you have 2 polynomials being + ADDED, so the way to factor them is to get "common factor" on both
anonymous
  • anonymous
cn u shw hw??plzz step by step
jdoe0001
  • jdoe0001
ok, lemme give you.... a quick example ab + bc what would be the common factor there? :)
anonymous
  • anonymous
b
jdoe0001
  • jdoe0001
yeah, so lemme show you the one you have there now \(\bf \color{blue}{(2y+x)^2}(y-2x)+\color{blue}{(2x+y)^2}(2x-y)\) see any common factors?
anonymous
  • anonymous
2
jdoe0001
  • jdoe0001
so, you take out the common factors out
anonymous
  • anonymous
k den
jdoe0001
  • jdoe0001
then, that's about it on factoring, you end up with several "factors" or item multiplying each other
anonymous
  • anonymous
wat do u gt?
anonymous
  • anonymous
idk hw to do it..cn u shw fully hw to do dis equation??
jdoe0001
  • jdoe0001
in the previous example ab + bc # common factor is "b" so if we take out "b", that is, we take away "b" from "ab" and leave only "a" by its lonesome and take away "b" from "bc" and leave "c" all by its lonesome too we're left with \(\bf b(a+c)\)
jdoe0001
  • jdoe0001
so you'd do the same with any polynomial to get a common factor
jdoe0001
  • jdoe0001
some are just longer than just 2 terms, it could be any amount of terms
anonymous
  • anonymous
ya...i gt dat..bt hw do u do ni dis equation is it 2[(y+x)(y-x)+(x+y)(x-y)] =2(y+x+y) =2y^2x
anonymous
  • anonymous
u cn do cancellation..rie??
jdoe0001
  • jdoe0001
\(\bf (2y+x)^2(y-2x)+(2x+y)^2(2x-y)\\ \textit{let's call }(2y+x)^2 = a\\ a(y-2x)+a(2x-y) \implies a[(y-2x)+(2x-y) ]\\ \implies (2y+x)^2[(y-2x)+(2x-y) ]\)
anonymous
  • anonymous
ohhk
anonymous
  • anonymous
so d final anser is dat??
jdoe0001
  • jdoe0001
on another note, if you were to expand say -(2x -y) what would it look like?
anonymous
  • anonymous
i hv another questin
anonymous
  • anonymous
|dw:1378328224060:dw|
jdoe0001
  • jdoe0001
can't say right off, might be better to post anew so we all can see and help and revise each other
anonymous
  • anonymous
hmmk

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