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