Question 696139
Given the identity Ax^3+x^2-2x-3=(3x+B)(x-1)(x+2)+C(x-1)+D , find the values of the constants A,B,C and D. Hence or otherwise, find the remainder when Ax^3+x^2-2x-3 is divided by x–1.


{{{(3x + B)(x - 1)(x + 2) + C(x - 1) + D = Ax^3 + x^2 - 2x - 3}}}
{{{(3x + B)(x^2 + x - 2) + C(x - 1) + D = Ax^3 + x^2 - 2x - 3}}}
{{{3x^3 + 3x^2 - 6x + Bx^2 + Bx - 2B + Cx - C + D = Ax^3 + x^2 - 2x - 3}}}
{{{3x^3 + 3x^2 + Bx^2 - 6x + Bx + Cx - 2B - C + D = Ax^3 + x^2 - 2x - 3}}} ----- Rearranging terms
{{{(3x^3) + (3x^2 + Bx^2) + (- 6x + Bx + Cx) + (- 2B - C + D) = (Ax^3) + (x^2) +  (- 2x) + ( - 3)}}}


From the above, we can see that:
{{{3x^3 = Ax^3}}} -------- {{{highlight_green(3 = A)}}}


{{{3x^2 + Bx^2 = x^2}}} ------ {{{Bx^2 = x^2 - 3x^2}}} ------ {{{Bx^2 = - 2x^2}}} ----- {{{highlight_green(B = - 2)}}}


{{{- 6x + Bx + Cx = - 2x}}} ----- {{{x(- 6 + B + C) = x(- 2)}}} ---- {{{- 6 + B + C = - 2}}} ----- {{{B + C = 4}}}
{{{- 2 + C = 4}}} ----- Substituting – 2 for B
C = 4 + 2, or {{{highlight_green(C = 6)}}}


{{{- 2B - C + D = - 3}}} 
{{{- 2(- 2) - 6 + D = - 3}}} ----- Substituting – 2 for B, and 6 for C
{{{4 - 6 + D = - 3}}}
D = - 3 + 2, or {{{highlight_green(D = - 1)}}}



With {{{A = 3}}}, {{{Ax^3 + x^2 - 2x - 3}}} becomes: {{{3x^3 + x^2 - 2x - 3}}}, and when divided by x – 1, we have a remainder, or f(1) = {{{3(1)^3 + (1)^2 - 2(1) - 3}}}, or {{{3 + 1 - 2 - 3}}}, or {{{highlight_green(- 1)}}}


You can do the check!!


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