Express the polynomial $x^3-4x-4$ as a linear combination of $x-2$, $(x-2)^2$ and $(x-2)^3$
I've been looking everywhere but I still don't quite understand the question. I know that a linear combination is like a matrix consisting of a specific combination of vectors multiplied by a coefficient. In the form…
$$a_1v_1+a_2v_2 +a_3v_3 ,\text{for }a_1 \,to \, a_n \, real \, numbers$$
So to express it as the question asks i think i have to find the coefficients off…
$$x^3-4x-4 = a(x-2)+b(x-2)^2 +c(x-2)^3$$
But i'm not too sure about it or where to go. I thought I'd had to involve vectors and matrices somehow. Please help, I really want to understand this content well, I've been having trouble picking up content from this new class.
We've also been talking about basis and span. I think a vector forms a basis for a system. If for a $R^n$ system if for 3 row (have 3 pivots showing 1 = 0, after row reducing) then the vectors/columns corresponding to those rows make a basis for the system and the system hence contains/span all of "R^3".
EDIT: You guys are right the question given to me was inconsistent. I asked the teacher who then admitted there was a typo and re-wrote the question. I think I've got plenty to work on anyway with this already.
He actually meant to ask
Express the polynomial $x^3-2x-4$ as a linear combination of $x-2$, $(x-2)^2$ and $(x-2)^3$
To wich i got C=1, B=6, a=10
Best Answer
It cannot be done. For any linear combination of $x-2$, its square, and its cube has $2$ as a root, but our given polynomial does not.
We can do it if in addition to the given polynomials we use $(x-2)^0$, that is, $1$.