Question 265051
given that the remainder when f(x) is divided by (x-1) is equal to the remainder when f(x) is divided by (2x+1). f(x)= px^3 + 6x^2 + 12x + q
find the value of p 
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The remainder theorem states that the remainder when a polynomial {{{f(x)}}} 
is divided by {{{k(x - a)}}} is {{{f(a)}}}.

Therefore the remainder when {{{f(x)}}} is divided by {{{x-1}}} is {{{f(1)}}}, which is

{{{"f(1)" = p*(1)^3 + 6*(1)^2 + 12*(1) + q}}} 
{{{"f(1)" = p + 6 + 12 + q}}}
{{{"f(1)" = p + 18 + q}}}

Also the remainder when {{{f(x)}}} is divided by {{{2x+1}}}, or {{{2(x+1/2)}}}
is {{{f(-1/2)}}}, which is

{{{f(-1/2) = p*(-1/2)^3 + 6*(-1/2)^2 + 12*(-1/2) + q}}} 
{{{f(-1/2) = p*(-1/8) + 6*(1/4) - 6 + q}}}
{{{f(-1/2) = -p/8 + 3/2 - 6 + q}}}

We are told that these are equal:

{{{p + 18 + q = -p/8 + 3/2 - 6 + q}}}

{{{8p + 144 + 8q = -p + 12 - 48 + 8q}}}

{{{8p + 144 + 8q = -p = -36 + 8q}}}

{{{9p=-180}}}

{{{p = -20}}}

Edwin</pre>