Question 210907
 If P(x)=3x^3+2x^2+x+20 and we write 

P(x) in the form P(x)=A+B(x-1)+C(x-1) (x-2)+D(x-1)(x-2)(x-3) A=? and B=?

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Set both expressions for P(x) equal to each other:


{{{3x^3+2x^2+x+20 = A+B(x-1)+C(x-1)(x-2)+D(x-1)(x-2)(x-3)}}}

Choose values for x that will cause some of those expresions
in parentheses to become 0.


Substitute 1 for x in the above:

{{{3(1)^3+2(1)^2+(1)+20 = A+B(1-1)+C(1-1)(1-2)+D(1-1)(1-2)(1-3)}}}

{{{3+2(1)+(1)+20 = A+B(1-1)+C(1-1)(1-2)+D(1-1)(1-2)(1-3)}}}

{{{3+2+1+20 = A+B(0)+C(0)(-1)+D(0)(-1)(-2)}}}

{{{26 =A}}}

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{{{3x^3+2x^2+x+20 = A+B(x-1)+C(x-1)(x-2)+D(x-1)(x-2)(x-3)}}}

Now substitute 2 for x in the above:

{{{3(2)^3+2(2)^2+(2)+20 = A+B(2-1)+C(2-1)(2-2)+D(2-1)(2-2)(2-3)}}}

{{{3(8)+2(4)+(2)+20 = A+B(1)+C(1)(0)+D(1)(0)(-1)}}}

{{{24+8+2+20 = A+B}}}

{{{54=A+B}}}

And since A=26, substituting:

{{{54=A+B}}}

{{{54=26+B}}}

{{{28=B}}}

So {{{A=2}}} and {{{B=28}}}

Edwin</pre>