Question 166514
For {{{x^2(a-b)+a^2(b-x)+b^2(x-a)}}} show that {{{(x-a)}}} is a linear factor.
<pre><font size = 4 color = "indigo"><b>
We use the fact that if a number is substituted for x 
in a polynomial, the result is the same as the remainder 
left when that polynomial is divided by x minus that number.

So we substitute {{{x=a}}} in the polynomial to see if the
remainder is 0, and if it is then we'll know that if we 
had divided the polynomial by {{{x-a}}}, the reaminder
would be 0, making {{{x-a}}} a factor:

So we substitute {{{x=a}}} into the polynomial:

{{{x^2(a-b)+a^2(b-x)+b^2(x-a)}}}
{{{a^2(a-b)+a^2(b-a)+b^2(a-a)}}}
{{{a^3-a^2b+a^2b-a^3+b^2(0)}}}
{{{a^3-a^2b+a^2b-a^3+0}}}

and everything cancels out and we have

0

so 0 would be the remainder if 

{{{x^2(a-b)+a^2(b-x)+b^2(x-a)}}}

were divided by {{{x-a}}}, so 

{{{x-a}}} is a factor of  {{{x^2(a-b)+a^2(b-x)+b^2(x-a)}}}.

And of course {{{x-a}}} is linear because the highest power
of x is 1.

Edwin</pre>