Let, $ f(x+h) \; = \; 7 + (x+2) + 3(x+2)^3 + (x+2)^4 - (x+2)^5 \\ \; \\ Here \; h \; = \; 2 \\ \; \\ \therefore f(h) \; = \; 7+h+3h^3+h^4-h^5 \\ \; \\ \therefore f(2) \; = \; 7+2+3(8)+16-32 \\ \; \\ \therefore f(2) \; = \; 17 \\ \; \\ \; \\ f'(h) = 1+9h^2+4h^3-5h^4 \\ \; \\ \therefore f'(2) = 1+36+32-80 \; = \; -11 \\ \; \\ \; \\ f''(h) = 18h+12h^2-20h^3 \\ \; \\ \therefore f''(2) = 36+48-160= \; -67 \\ \; \\ \; \\ f'''(h) = 18+24h-60h^2 \\ \; \\ \therefore f'''(2)=18+48-240 \; = \; -174 \\ \; \\ \; \\ f''''(h)=24-120h \\ \; \\ \therefore f''''(2)=24-240 \; = \; -216 \\ \; \\ \; \\ f'''''(h) \; = \; -120 \\ \; \\ \; \\ By \; taylor's \; theorem, \\ \; \\ f(x+h) \; = \; f(h)+xf'(h)+\dfrac{x^2}{2!} f''(h) + \dfrac{x^3}{3!} f'''(h) + \ldots \\ \; \\ \therefore f(x+2) \; = \; f(2)+xf'(2)+\dfrac{x^2}{2!} f''(2) + \dfrac{x^3}{3!} f'''(2) + \ldots \\ \; \\ \therefore f(x+2) \; = \; 17+x(-11)+\dfrac{x^2}{2} (-76) + \dfrac{x^3}{6} (-174) + \dfrac{x^4}{24} (-216) + \dfrac{x^4}{120} (-120) \\ \; \\ \therefore f(x+2) = 17-11x-38x^2-29x^3-9x^4-x^5 \\ \; \\ \; \\ \therefore 7 + (x+2) + 3(x+2)^3 + (x+2)^4 - (x+2)^5 \; = \; 17-11x-38x^2-29x^3-9x^4-x^5 $