Consider P(x)=3x^3-2x+1 e Q(x)=4x-3. Calculate the quotient and the remainder of the division P(x)/Q(x).
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You need to use reminder theorem such that:
`P(x) = Q(x)*Q'(x) + R(x)`
`Q'(x)` represents the quotient
`R(x)` represents the reminder
You need to notice that P`(x) ` is of order 3 and `Q(x)` is of first order, hence `Q'(x)` needs to be of order 2 and reminder `R(x)` is constant.
Replacing `3x^3-2x+1` for `P(x)` , `4x-3` for `Q(x)` , `(ax^2 + bx + c)` for `Q'(x)` and d for `R(x)` yields:
`3x^3 - 2x + 1 = (4x - 3)(ax^2 + bx + c) + d`
Open the brackets such that:
3`x^3 - 2x + 1 = 4ax^3 + 4bx^2 + 4cx - 3ax^2 - 3bx - 3c + d`
`3x^3 - 2x + 1 = 4ax^3 + x^2(4b - 3a) + x(4c - 3b) - 3c + d`
Equating the coefficients of like powers yields:
`4a = 3 => a = 3/4`
`4b - 3a = 0` (the term that contain `x^2` , to the left side, is missing, hence, its coefficient is 0)
`4b - 3*(3/4) = 0 => 4b = 9/4 => b = 9/16`
`4c - 3b = -2 => 4c = 3b - 2 => 4c = 27/16 - 2 => c = -5/64`
`d = 49/64`
Hence, evaluating the quotient `Q'(x)` and reminder `R(x)` yields `Q'(x) = 3/4 x^2 + 9/16 x - 5/64` and `R(x) = 49/64` .
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