Question 152145
Without actual division prove that {{{2x^4-6x^3+3x^2+3x-2}}} is exactly divisible by {{{x^2-3x+2}}}. How can we prove that without actual division? I tried to understand but it's too tough for me. Thanks

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First we factor {{{x^2-3x+2}}} as {{{(x-1)(x-2)}}}

So {{{2x^4-6x^3+3x^2+3x-2}}} will be divisible by {{{x^2-3x+2}}}
if and only if both {{{x-1}}} and {{{x-2)}}} are both factors
of {{{2x^4-6x^3+3x^2+3x-2}}}.

Now we know by the remainder theorem that if 
{{{2x^4-6x^3+3x^2+3x-2}}} were to be divided by {{{x-1}}}, 
the remainder would have the same value as 
{{{2x^4-6x^3+3x^2+3x-2}}} with x replaced by +1. 

Therefore the remainder of the division would be

{{{2x^4-6x^3+3x^2+3x-2}}}
{{{2(1)^4-6(1)^3+3(1)^2+3(1)-2}}}
{{{2(1)-6(1)+3(1)^2+3(1)-2}}}
{{{2-6+3+3-2}}}
{{{0}}}.

And since the remainder is 0, {{{2x^4-6x^3+3x^2+3x-2}}} is
divisible by {{{x-1}}}.  Now we do the same with the other
factor {{{x-2}}}:

As before we know by the remainder theorem that if
{{{2x^4-6x^3+3x^2+3x-2}}} were to be divided by {{{x-2}}}, the 
remainder would have the same value as {{{2x^4-6x^3+3x^2+3x-2}}}
with x replaced by +2. 

Therefore the remainder of the division would be

{{{2x^4-6x^3+3x^2+3x-2}}}
{{{2(2)^4-6(2)^3+3(2)^2+3(2)-2}}}
{{{2(16)-6(8)+3(4)+3(2)-2}}}
{{{32-48+12+6-2}}}
{{{0}}}.

And since the remainder is 0, {{{2x^4-6x^3+3x^2+3x-2}}} is
also divisible by {{{x-2}}}.

And since {{{2x^4-6x^3+3x^2+3x-2}}} is divisible both
by {{{x-1}}} and {{{x-2}}} it is divisible by their
product {{{(x-1)(x-2)}}} or {{{x^2-3x+2}}}, and we didn't
do any division!

Edwin</pre>