Question

\[\lim_{x \rightarrow -1} \frac{ x^3+ax^2+bx+c }{ (x+1)^3 }\] is equal to a finite non zero number find the value of 2A+3B+4C ?

Answer 1

@hartnn @zepdrix @bahrom7893 @Zarkon @phi

Answer 2

Well, you have (x+1)^3 as a possibility foe the polynomial.

Answer 3

when x^3+ax^2+bx+c=(x+1)^3

Answer 4

as x--> -1 , the denominator -->0

to get a finite limit, we will want the top =0, so we can use L'Hopital's rule
(3 times)

assuming the top does evaluate to 0 at x=-1 you can
take the derivative of top and bottom

3x^2 +2ax + b and 3(x+1)^2
again assuming the top =0, apply L'Hopital again
6x +2a 6(x+1) and again
6 6 and we get a limit of 6/6 =1

To use L'Hopital, each derivative must evaluate to 0 at x=-1

so we find at x=- 1
6*-1 +2a =0 , a=3
3x^2 +2ax + b =0 or 3 +6*-1 + b=0, b=3

finally
x^3 +ax^2 +bx + c =0
-1 +3*1 + 3* -1 + c =0, c =1


2A+3B+4C becomes 2*3 + 3*3+4*1= 6+9+4= 19