Question 1040787
By the remainder theorem, if x-k is factor of p(x), then p(k) = 0.


In this case, x-k = x-1, so k = 1. Therefore, if we find a function p(x) such that p(k) = p(1) = 0 then we have found the answer.


Essentially we plug in x = 1 into each answer choice. Then we see which results in 0. 


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Let's go through the answer choices one by one. Starting with choice A


{{{p(x) = x^3+x^2-2x+1}}}
{{{p(1) = 1^3+1^2-2(1)+1}}} Replace every x with 1
{{{p(1) = 1+1-2(1)+1}}}
{{{p(1) = 1+1-2+1}}}
{{{p(1) = 1}}}


Since the result is NOT equal to 0, this means that x-1 is NOT a factor of p(x).


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Now onto choice B. Like before, plug in x = 1 to see if we get a result of 0.


{{{q(x) = 2x^3-x^2+x-1}}}
{{{q(1) = 2(1)^3-(1)^2+1-1}}} Replace every 'x' with 1
{{{q(1) = 2(1)-1+1-1}}}
{{{q(1) = 2-1+1-1}}}
{{{q(1) = 1}}}


Again we don't get 0. So choice B is not the answer either. 
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Part C)


{{{r(x) = 3x^3-x-2}}}
{{{r(1) = 3(1)^3-1-2}}} Plug in x = 1
{{{r(1) = 3(1)-1-2}}}
{{{r(1) = 3-1-2}}}
{{{r(1) = 0}}}


We got a result of 0, so x-1 is a factor of r(x). We have our answer. Let's check part D just to complete the problem.
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Part D)


{{{s(x) = -3x^3+3x+1}}}
{{{s(1) = -3(1)^3+3(1)+1}}} Plug in x = 1
{{{s(1) = -3(1)+3(1)+1}}}
{{{s(1) = -3+3+1}}}
{{{s(1) = 1}}}


The nonzero result means x-1 is not a factor of s(x)
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To summarize, the final answer is <font color=red>Choice C</font>