*Of the constant ''α''. I'm quite stuck at this problem literally. I'm at the point that I compared the coefficients (separately), but I can't think of a connection. I've attempted letting R(1) = 1/2*R(2); then calculating R(1) by putting the compared coefficients into it, but I didn't get the needed value; and, to make worse, ''c'' troubled me. Any help would be appreciated.
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Verified answer
let f (x) = x³ - 2x² + αx + 5
f (3)
= (3)³ - 2(3)² + 3(a) + 5
= 27 - 18 + 3α + 5
= 14 + 3α
f (-1)
= (-1)³ - 2(-1)² + α(-1) + 5
= -1 - 2 - α + 5
= -α + 2
f (3) = 2 f (-1)
14 + 3α = 2(-α + 2)
14 + 3α = -2α + 4
5α = -10
α = -2
x is all numbers
Change x to 1
(x^3-2x^2+αx+5)/(x-3) = 2 (x^3-2x^2+ax+5)/(x+1)
(1 - 2 + a + 5) / -2 = 2 (1 - 2 + a + 5)/2
(4 + a) / -2 = 2 (4 + a) / 2
(4 + a) / -2 = 4 + a
4 + a = -2 (4 + a)
4 + a = -8 + -2a
4 + 8 = -2a - a
12 = -3a
a = -4
This will help
http://www.youtube.com/watch?v=ztZgL87jSgI