95th Problem 2016

$METHOD 1\rightarrow Using Remainder Theorem:$ For $(x-\frac {5}{2})$ to be a factor of $f(x)=2x^3-3x^2+x-15$ the expression $f(\frac {5}{2})$ must be equal to 0. Now we will check it by substituting $\frac {5}{2}$ in $f(x)$ ,we evaluate the expression $2(\frac {5}{2})^3-3 (\frac {5}{2})^2+\frac {5}{2}-15$ which evaluates to 0 proving that $(x-\frac {5}{2})$ is a factor.

$METHOD 2 \rightarrow Using Direct Divison:$ Divide $2x^3-3x^2+x-15$ by $(x-\frac {5}{2})$ if it divides $f(x)$ completely i.e. without leaving remainder(which it does) then $(x-\frac {5}{2})$ is a factor of $f(x)$

$f(x) = (x - \frac{5}{2}) \cdot p(x) + C$ where $p(x)$ is a polynomial and $C$ a constant ; $f(\frac{5}{2}) = 0 = p(\frac{5}{2}) (\frac{5}{2} -\frac{5}{2}) + C = C \Rightarrow C = 0 \Rightarrow (x - \frac{5}{2}) | f(x)$

Or use the theorem:

$a$ is a root of a polynomial $f(x)$ ,i.e, $f(a) = 0 \iff (x - a) | f(x)$