# Re: equation: question, Mr G

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Posted by Denis Borris on October 25, 2002 at 13:46:45:

In Reply to: Re: equation posted by T.Gracken on October 25, 2002 at 12:57:27:

: : When ax^3-x^2-x+b is divided by x-1, the remainder is 6.
: : When it is divided by x+2, the remainder is 9.
: : What are a and b?
: : Doesn't seem possible to solve. Help please.
: certainly it does! but you have to use the remainder thereom for it to make sense.
: that is, if f is a polynomial function divided by (x-k), then f(k) is the remainder.
: so if the remainder is 6 when dividing by (x-1), then replace the x's in ax^3-x^2-x+b with 1 and it must equal 6
: that is... a-1-1+b=6
: or a + b = 8.
: do the same for the other given information; that is if divide by (x+2) the remainder is 9. note that x+2= x-(-2), so replace x's with -2 and we get
: -8a-4+2+b=9
: or -8a + b = 11
: now you have two equations with two unknowns, solve for a and b.

I follow that; a = -1/3 and b = 8 1/3
BUT:
if x,a and b are restricted to integer range 1 to 99, there are 1806 solutions:
(x,a,b in increasing x order)
1: 8,1,49
2: 8,2,27
3: 8,2,97
...(did you think I was gonna type 'em all?!)
1805: 98,66,39
1806: 99,33,73

Where, oh where have I screwed up, praytell.

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