anonymous
  • anonymous
Given that 5x^3 + 6x^2 -x + 2 = (x+1) (x-2) Q(x) + ax + b find the value of a and b. Question on Identities :)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Q(x) is an unknown polynomial?
anonymous
  • anonymous
yea
anonymous
  • anonymous
then... a = -1, b = 2?...dont think it should be this simple

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anonymous
  • anonymous
that isnt the answer lol
anonymous
  • anonymous
answer is a = 20 b = 24
anonymous
  • anonymous
but i cant get it :(
anonymous
  • anonymous
if Q(x) = 5x-9 ?
anonymous
  • anonymous
o..o maybe you could work out the full solution and see if it matches my answer :P
anonymous
  • anonymous
Thank you.
anonymous
  • anonymous
\[ (x+​1)*​(x+​2)*​(5*​x-​9) = 5x^3 + 6x^2 -17x - 18 + 16x +20=5x^3 + 6x^2 -x + 2\]
anonymous
  • anonymous
anonymous
  • anonymous
oo then how should i carry on
anonymous
  • anonymous
hmm a is 16 and b is 20 for me
anonymous
  • anonymous
check for any careless mistake? :O
anonymous
  • anonymous
no i don't think so
anonymous
  • anonymous
can you upload the picture of your working?
anonymous
  • anonymous
i didn't do it on my notebook, in head.
anonymous
  • anonymous
\[(x+​1)(x+​2)(5x−​9) +ax + b = 5x^3+6x^2−17x−18+ ax+b\]
cwrw238
  • cwrw238
i did it also and got same result as ishaan
anonymous
  • anonymous
i need the working :(
anonymous
  • anonymous
or rather simple explanation towards the steps
cwrw238
  • cwrw238
equating coefficients of x gives -1 = 10 - 27 + a a = 27-11 = 16
anonymous
  • anonymous
use q(x) as px +q, suggested by you earlier if you really need to show working
anonymous
  • anonymous
can i let Q(x) = px + q will i be able to get the answer?
anonymous
  • anonymous
oh but i cant seem to get the answer somehow :(
anonymous
  • anonymous
i get decimal numbers :(
anonymous
  • anonymous
show me your working i will try to guide you
anonymous
  • anonymous
kk i try again
anonymous
  • anonymous
well you can get multiple answers, maybe.
anonymous
  • anonymous
only a and b
anonymous
  • anonymous
(x+1) (x-2) *********
anonymous
  • anonymous
Ishaan i edited the question
anonymous
  • anonymous
sry about it
cwrw238
  • cwrw238
i can see how ishaan got p = 5 because of the 5x^3 but the q = -9 is not obvious = 5x^3 + 15x^2 + 10x + qx^2 + 3qx + 2q + ax + b = 5x^3 + (15 + q)x^2 + (10 + 3q + a)x + 2q + b compare coefficients with 5x^3 + 6x^2 - x + 2 of x^2: q + 15 = 6 so q = -9 of x: 10 + a + 3q = 10 + a - 27 = a - 17 and a -17 = -1 so a = 16 of x^0 : 2q + b = 2 so -18 + b = 2 so b = 20 a=16, b = 20
anonymous
  • anonymous
Thanks :)
cwrw238
  • cwrw238
yw

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