The Remainder Theorem states that the remainder is f(a). So, that means we have
a^3-a^2+4a = a^3+6a+10
a^2+2a+10 = 0
This has no real solutions.
I suspect a typo
As written, P is never greater than Q. See
http://www.wolframalpha.com/input/?i=plot+x%5E3-x%5E2%2B4x+,+x%5E3%2B6x%2B10
The polynomials P (x)=x^3-x^2+4x and Q(x)= x^3+6x+10 leave the same remainder when divided by x-a .find the possible values of a and solve the inequality P (x) is greater than Q (x )
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