By taylor series,

$f(x)= f(a)+ (x-a)f'(a)+ \dfrac {(x-a)^2}{2!}f''(a) + \dfrac {(x-a^2)}{3!} f'''(a)+ \cdots $

$Here, \ a=0$

$f(x)=7+(x+2)+ 3 (x+2)^3 +(x+2)^4 - (x+2)^5 $

$f(0)=7+2+3 \times 2^3 + 2^4 -2^5 =17$

$ f'(x)= 0 +1+9 (x+2)^2 +4(x+2)^3 -5 (x+2)^4 $

$f'(0)=1+9\times 2^2 +4 \times 2^3 -5\times 2^4 =-11 $

$f''(x)=18 (x+2)+ 12 (x+2)^2 -20 (x+2)^3 $

$f''(0) = 18 \times 2 +12 \times 2^2 -20 \times 2^3 = -76 $

$f'''(x) = 18 +24 (x+2)-60 (x+2)^2 $

$f'''(0)=18 +24 \times 2 -60 \times 2^2 = -174 $

$f^{iv}(x)= 24-120 (x+2) $

$f^{iv}(0)= 24 -120 \times 2 = -216 $

$f^v (x)= -120 $

$ f^v (0) = -120 $

$f(x)= f(0)+ (x)f'(0)+ \dfrac {(x)^2}{2!}f''(0)+ \dfrac {(x)^3}{3!} f'''(0)+ \dfrac {(x)^4}{4!}f^{iv}(0)+ \dfrac {(x)^5}{5!} f^v(0)$

$f(x)= 17 -11 x - \dfrac {76x^2}{2}- \dfrac {174x^3}{6}- \dfrac {216x^4}{24}- \dfrac {120 x^5}{120} $

$f(x)=17-11x-38x^2-29x^3-9x^4-x^5$