When the Polynomial p(x)=ax^3+5x^2+bx-2 where a and b are constants is divided by 2x^2+3x-2, the remainder is 9x-6.
(I) Find the values of a and b
(II) Using these values of a and b, factorize p(x) completely
(III) Find the remainder when p(x) is divided by (3x+1).

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  1. After half a page of a messy long division of ax^3+5x^2+bx-2 by 2x^2+3x-2
    I had a remainder of (b+a - (30-9a)/4)x -2+(10a-3a)/2
    matching this with 9x - 6
    we get b+a - (30-9a)/4 = 9 and -2+(10a-3a)/2 = -6

    I will leave it up to you to solve for a and b
    Hint: it comes out to "nice" integers

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  2. Let p(x) = (2x^2+3x-2)(mx+c)+(9x-6)
    If you expand that out, you get
    2mx^3 + (3m+2c)x^2 + (3c-2m+9)x - (2c+6) = ax^2 + 5x^2 + bx - 2
    That means you just have to solve the equations
    2m = a
    3m+2c = 5
    3c-2m+9 = b
    2c+6 = 2
    Looks daunting, but if you start at the bottom, you get a value for c right away, and the rest falls out easily.

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