Question 154518
Let {{{f(x)=-x^3+3x^2-3x+1}}}, and {{{g(x)}}} be {{{f(x)}}} divided by {{{1-x}}}; solve for {{{g(x)}}} if {{{1-x}}} is a factor of {{{f(x)}}}.
a. {{{g(x)=x^4-4x^3+6x^2-4x+1}}}
b. {{{g(x)=x^3-3x^2+3x-1}}}
c. {{{g(x)=-x^2+2x-1}}}
d. {{{g(x)=x^2-2x+1}}}
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Let's first factor {{{f(x)}}}: 

{{{f(x)=-x^3+3x^2-3x+1}}} 

Regroup the terms:

{{{f(x)=1-x^3-3x+3x^2}}},

Factor the first two terms as the difference of cubes:
Factor out {{{-3x}}} out of the last two terms:

{{{f(x)=(1-x)(1+x+x^2)-3x(1-x)}}}

Factor out {{{(1-x)}}}

{{{f(x)=(1-x)((1+x+x^2)-3x)}}} 

Remove the inner parentheses:

{{{f(x)=(1-x)(1+x+x^2-3x)}}}

Combine the two like terms:

{{{f(x)=(1-x)(1-2x+x^2)}}}

Factor the second parentheses:

{{{f(x)=(1-x)(1-x)(1-x)}}}

Write as a cube:

{{{f(x)=(1-x)^3}}}

Now since

g(x) is f(x) divided by 1-x

{{{g(x)=f(x)/(1-x)=(1-x)^3/(1-x)=(1-x)^2=1-2x+x^2=x^2-2x+1}}}

So the correct choice is (d)

Edwin</pre>