



 
Hi, If you divide $f(x)$ by $x2$ then the result is a polynomial I am going to call $p(x)$ and a remainder $R.$ The the degree of the remainder $R$ is less then the degree of the divisor $x2$ so the remainder must be a constant. Hence we know that \[f(x) = 4x^415x^24 = p(x) \times (x2) + R.\] Now substitute $x = 2$ into this equation to get \[f(2) = p(2) \times (2  2) + R\] and hence $f(2) = R.$ I hope this helps,  


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