Writing a polynomial as polynomial powers I want to write the polynomial
$$p(x)=x^4+3x^3+4x^2-7x+6$$ as a polynomial in powers of $x-1$.
What I am trying to do is to determine the taylor series around $x=1$.To do so, consider that $p(1)=7, p'(1)=14, p''(1)=38, p'''(1)=42, p''''(1)=24$
From this it follows that
$p(x)=7+\frac{14}{1!}(x-1)+\frac{38}{2!}(x-1)^2+\frac{42}{3!}(x-1)^3+\frac{24}{4!}(x-1)^4$
Honestly I am not convinced with my answer, since comparing with what is proposed by the book the coefficient of $(x-1)$ is $-1$ and the coefficient of $(x-1)^2$ is $4$ but I don't understand why, any suggestion?

Your expansion is correct.
Here is an unconventional way to do this.
$P(x) = x^4 + 3x^3 + 4x^2 - 7x + 6\\
P(x+1) = (x^4 + 4x^3 + 6x^2 + 4x + 1) + 3(x^3+3x^2+3x+1) + 4(x^2 + 2x+1) - 7(x+1) + 6\\
P(x+1) = x^4 + 7x^3 + 19x^2 + 14x + 7\\
P((x-1)+1) = P(x) = (x-1)^4+7(x-1)^3 + 19(x-1)^2 + 14(x-1)+7$
