Odpowiedzi

2009-04-14T23:29:11+02:00
W(x) = x³ - (a+b)x² - (a-b)x+3, r1=1, r2=3
W(1) =0--->1-(a+b)-(a-b)+3=0
W(3) =0--->27-9(a+b)-3(a-b)+3=0 rozwiazuje ten uklad aby obliczyc a ib
1-a-b-a+b+3=0----->2a=4--->a=2
-9a-9b-3a+3b=-30
-12a-6b=-30
-24-6b=-30------>6b=6------->b=1
W(x) = x³ - 3x² - x+3
x³ - 3x² - x+3=0
x²(x-3)-(x-3)=0
(x-3)(x²-1)=0
x²-1=0 -----> x3=-1