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小 發表於 2020-12-25 08:36 PM (第 1190 天)
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(a) By the remainder theorem, we know that x^3+4x^2+x+m=0 when x=-n. Hence,
(-n)^3+4(-n)^2+(-n)+m=0
-n^3+4n^2-n+m=0
n^3-4n^2+n-m=0 ---(*)
(b) (i) putting n=-1 in (*), we have
(-1)^3-4(-1)^2+(-1)-m=0
-1-4-1=m
m=-6
(ii) since x^3+4x^2+x-6 is divisible by x-1, x-1 is a factor of x^3+4x^2+x-6.
x^3+4x^2+x-6=x^2(x-1)+5x^2+x-6=x^2(x-1)+5x(x-1)+6x-6=x^2(x-1)+5x(x-1)+6(x-1)
[you can achieve the above result using long division]
=(x-1)(x^2+5x+6)
=(x-1)(x+2)(x+3)