Linear Algebra II - Linear Transformation Assume that $T$ is a linear transformation. If $T: P_2 \to P_2$ where $P_2$ is the polynom vector space with polynomials of degree $2$, and $T(x+1) = x$, $T(x-1) =1$, $T(x^2)=0$ find $T(2+3x-x^2)$
I was just wondering if it is possible to do $$T(2+3x-x^2)  $$ as ..
$$(a,b,c) = a(x+1) + b(x-1) + c(x^2)
= ax+a+bx-b+cx^2$$
Then let $a=2, b=-x, c= 2/x$
to get 
$$T (2+3X-x^2) = 2T(x+1) -xT(x-1) +2/xT(x^2)$$
to eventually get $(-x^2+3x+1)$

$a(x+1)+b(x-1)+cx^2=ax+a+bx-b+cx^2$ is a relevant quantity here, but remember that the scalars should be numbers, not things like $-x$ or $\frac2x$. Can you find numbers $a,b,c$ so that $a(x+1)+b(x-1)+cx^2=2+3x-x^2$? Once you find those numbers, then you can get your answer using the linearity of $T$.
