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March 1, 2015

March 1, 2015

Posted by **Ann** on Monday, September 10, 2012 at 1:41pm.

- maths -
**Steve**, Monday, September 10, 2012 at 2:58pmI assume you want to find a and b so it divides evenly?

x^3 + 2(a-2b)x^2 - b(8a-3b)x + 6ab^2

-------------------------------------

x+2a

= x^2 - 4bx + 3b^2 remainder 6ab(1-b)

so, to divide evenly, either a=0 or b=1

check:

a=0: x^3 - 4bx^2 + 3b^2x + 0 = x(x^2-4bx+3b^2) for any b

b=1: x^3 + 2(a-2)x^2 - (8a+3)x + 6a

= (x+2a)(x^2 - 4x - 3) for any a

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