Question

Expand 2 𝒙³ + 7 𝒙² + 𝒙 – 6 in power of (𝒙 – 2) by using Taylors Theorem.

Answer 1

By Taylor’s series,

`f(x)=f(a)+(x+a)f'(a)+(x+a)^2/(2!)f''(a)+(x+a)^2/(3!)f'''(a)+`.........

Here,
f(𝑥) = 2 𝑥³ + 7 𝑥² + 𝑥 – 6
f(2) = 2 (2)³ + 7 (2)² + 2 – 6 = 40
f’(𝑥) =6x²+ 14x + 1
f’(2) = 6 (2)² + 14 (2) + 1 = 53
f’’(𝑥) = 12x + 14
f’’(2) = 12(2) + 14 = 38
f’’’(𝑥) = 12
f’’’(2) = 12
f’’’’(𝑥) = 0 .

`f(x)=f(2)+(x-2)f'(2)+(x-2)^2/(2!)f''(a)+(x-2)^3/(3!)f'''(2)+0`

2 𝒙³ + 7 𝒙² + 𝒙 – 6 = 40 + (x - 2) (53)+`(x-2)^2/(2!)(38)+(x-2)^3/(3!)(12)`

2 𝒙³ + 7 𝒙² + 𝒙 – 6 = `2(x - 2)^3+19(x - 2)^2+53(x-2)+40`