Mathematics
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OpenStudy (anonymous):
PLS pls pls help
The polynomial P(x) is divided by (x-1)(x+2). Find the remainder if, given that P(1)=2 and P(-2)= 5
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OpenStudy (anonymous):
@satellite73 help pls pls pls. I been stuck on this one for days...
OpenStudy (amistre64):
do you recall the remainder thrm?
OpenStudy (anonymous):
yea
P(x) = (x-1)(x+2) Q(x) + R
OpenStudy (amistre64):
Since the P(1) and P(-2) are not zero, we know that the given is not a divisor right?
OpenStudy (anonymous):
ye
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OpenStudy (amistre64):
what ideas have you run thru so far?
OpenStudy (anonymous):
i tried to sub P(1) and P(-2) into (x-1)(x+2) Q(x) + R
OpenStudy (amistre64):
yeah, all that really tells us is that R = 1 or -2 :)
OpenStudy (amistre64):
we are given 2 points, the least poly to fit between 2 points is a line, right?
OpenStudy (anonymous):
oh yea...
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OpenStudy (amistre64):
and since the given is an x^2, the top would have to be the remainder ... of at most x
OpenStudy (amistre64):
you get where this is going?
OpenStudy (anonymous):
kinda...
OpenStudy (amistre64):
\[\frac{P(x)}{(x-1)(x+2)}=Q(x)+\frac{R(x)}{(x-1)(x+2)}\]
OpenStudy (amistre64):
R(x) is a poly of at most, degree 1; and since its value changes, it cant be a constant and must be a line
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OpenStudy (amistre64):
the line between 1,2 and -2,5
OpenStudy (anonymous):
oh....
OpenStudy (amistre64):
im assuming you know how to define that line ...
OpenStudy (anonymous):
with gradient -1 right?
OpenStudy (amistre64):
maybe, i hadnt quite gotten to the line :)
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OpenStudy (anonymous):
then i got line 3-x
OpenStudy (amistre64):
thats a great line
OpenStudy (anonymous):
would that really be the answer?
OpenStudy (amistre64):
and any multiple of it would work as well
OpenStudy (amistre64):
\[R(x)=k(3-x)\]
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OpenStudy (amistre64):
the k migh tbe a little much tho; 3-x is sufficient
OpenStudy (anonymous):
yea thanks so much!!!
OpenStudy (amistre64):
youre welcome
OpenStudy (anonymous):
you're* LOLOLOL
OpenStudy (amistre64):
.... they dont put that key in an opportune spot ;)