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October 30, 2014

October 30, 2014

Posted by **KnowsNothing** on Friday, October 12, 2012 at 7:39pm.

- math -
**Count Iblis**, Friday, October 12, 2012 at 8:08pmIf p(x) is the polynomial, then you have:

p(x) = (x+2)q1(x) - 19

for some poynomial q1(x). You see that the remainder of -19 is the value of

p(x) at x = -2. We also have:

p(x) = (x-1)q2(x) + 2

Therefore p(1) = 2.

Then if you divide p(x) by (x-1) (x+2), the remainder will be a first degree polynomial, so we have:

p(x) = (x-1)(x+2)q3(x) + r(x)

Then if you put x = 1 in here and use that p(1) = 2, you find:

r(1) = 2

Putting x = -2 and using that

p(-2) = -19 yields:

r(-2) = -19

These two values of r(x) fix r(x) as

r(x) is of first degree. We have:

r(x) = (-19)/(-3) (x-1) + 2/3 (x+2) =

7 x - 5

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