TITLE: Express the polynomial $x^3-4x-4$ as a linear combination of $x-2$, $(x-2)^2$ and $(x-2)^3$
QUESTION [8 upvotes]: 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

REPLY [6 votes]: 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, $1$.

REPLY [5 votes]: As written, it is impossible. Make the substitution $x=2.$ The right-hand side becomes $0,$ but the left-hand side does not.
Unfortunately, the three terms you're given do not comprise a basis for the space of polynomials of degree no greater than $3.$ A fourth element is needed--any non-zero constant will do.

REPLY [4 votes]: Notice, $$x^3-4x-4=a(x-2)+b(x-2)^2+c(x-2)^3$$ $$x^3-4x-4=cx^3+(b-6c)x^2+(a-4b+12c)x-2(a-2b+4c)$$  Now, comparing the corresponding coefficients, we get $$c=1$$ $$b-6c=0\iff b=6\times 1=6$$ $$a-4b+12c=-4\iff a=-4+4(6)-12(1)=8$$
But, these values have to satisfy $$-2(a-2b+4c)=-4$$$$-2(8-2(6)+4(1))=-4$$
$$ 0\neq-4$$
Hence, the given polynomial $x^3-4x-4$ can't be expressed in the linear form of $(x-2)$

REPLY [3 votes]: You also need a constant term, else when $x=2$ we have $LHS=p(2)=-4$ but for any linear combination we have $RHS=0$.
However, this is not then a linear combination of the given powers of $(x-2)$.

If you allow the constant term, then:
So $$p(x)=x^3-4x-4=a_0+a_1(x-2)+a_2(x-2)^2+a_3(x-2)^3 \tag{1}$$
Evaluate at $x=2: 8-8-4=a_0 \implies \boxed{a_0=-4}$
Taking derivative of (1):
$p'(x) = 3x^2-4 = a_1 + 2a_2(x-2) + 3a_3(x-2)^2 \tag{2}$
Evaluate (2) at $x=2$:
$3(4)-4 = a_1 \implies \boxed{a_1 = 8}$
Taking further derivatives will enable you to find $a_2,a_3$.

REPLY [3 votes]: Let $f(x)=x^3-4x-4$, then expanding we get
$$f(y+2)={y}^{3}+6{y}^{2}+8y-4$$
Let $y=x-2$ to get
$$f(x)=(x-2)^{3}+6(x-2)^{2}+8(x-2)-4$$

REPLY [2 votes]: An alternate route. For all polynomials of degre 3, by the Taylor expansion, you have

$$
P(x)=P(a)+P'(a)(x-a)+\frac{P^{(2)}(a)}{2!}(x-a)^2+\frac{P^{(2)}(a)}{3!}(x-a)^3
$$

You may apply it to $P(x)=x^3-4x-4$ and $a=2$. Observe that here $P(2)\neq0$.