¿Como puedo expresar el polinomio x^4-5x^3-3x^2+7x+6 en potencias de x-2??

A

F(X)=x^4-5x^3-3x^2+7x+6

formula de Taylor ,em torno do ponto x= "a"

F(X)= F(a)+ F´(a)(x-a) + F´´(a)(x-a)^2/2!

+F´´´(a)(x-a)^3/3!+ F´´´´(a)(x-a)^4/4! + ......+0+0

F(X)= F(2)+ F´(2)(x-2) + F´´(2)(x-2)^2/2!

+F´´´(2)(x-2)^3/3!+ F´´´´(2)(x-2)^4 /4!+ ......+0+

**********************************************

F(X)=x^4-5x^3-3x^2+7x+6

F´(X)=4X^3-15X^2-6X+7

F´´(X)=12X^2-30X-6

F´´´(X)=24X-30

F´´´´(X)=24

F´´´´(X)=0

F(2)=-16

F´(2)=-33

F´´(2)=-18

F´´´(2)=18

F´´´´(2)=24

entonces

F(X) =-16+ -33(X-2)-1(8/2)(X-2)^2

+(18/6)(X-2)^3+(24/24)(X-2)^4

1)F(X) = -16 -33(X-2)-9(x-2)^2+3(x-2)^3+(x-2)^4

2) F(X)=x^4-5x^3-3x^2+7x+6

check

em (1) para X=7

F(7)=594

em 2) para X=7

F(7) =594~~ok

entoces

x^4-5x^3-3x^2+7x+6 =

-16 -33(X-2)-9(x-2)^2+3(x-2)^3+(x-2)^4