Remainder Theorem

The term containing the highest power of $x$ in the polynomial $f(x)$ is $2x^4$. Two of the roots of the equation $f(x)=0$ are -1 and 2. Given that $x^2-3x+1$ is a quadratic factor of $f(x)$, find the remainder when $f(x)$ is divided by $2x-1.$ Since degree of $f(x)$ is $4$ and $x^2-3x+1$ is a factor of $f(x)$, it can be written as product of two quadratics Therefore, $f(x) = ( x^2-3x+1 ) (ax^2+bx+c )$. Again since, $2x^4$ is the leading term $a$, must be equal to $2$ So, $f(x) = ( x^2-3x+1 ) (2x^2+bx+c )$. Given $-1$ and $2$ are the roots of $f(x)$ $\Rightarrow$ $f(-1)=0$ & $f(2)=0$ Note that $x^2-3x+1$ does not vanishes at $x = -1 , 2$ $\Rightarrow$ $2x^2+bx+c$ must vanishes at this two points. $\Rightarrow 2-b+c = 0 , 8+2b+c=0$ Solving the equations we get $b=-2, c=-4$ Thus $f(x) = ( x^2-3x+1 ) (2x^2-2x-4 )$. Required remaninder is $f(\frac{1}{2}) = \frac{9}{8}$