Question

1.Given that x=2 is a triple root of f(x)=x*4-ax^3+bx+c=0, find the values of a,b and c

Answer 1

`f(x)=x^4-ax^3+bx+c`

By the polynomial remainder theorem, dividing
`f(x)` by
`x-2` three times leaves a remainder of 0; we have

`(x^4-ax^3+bx+c)/(x-2)=x^3+2x^2+(4-a)x+8+b-2a+(16+c+2b-4a)/(x-2)`

`\implies16+c+2b-4a=0`

Dividing the quotient by
`x-2` again leaves a remainder of 0:

`(x^4-ax^3+bx+c)/((x-2)^2)=x^2+4x+12-a+(32+b-4a)/((x-2)^2)`

`\implies32+b-4a=0`

and again:

`(x^4-ax^3+bx+c)/((x-2)^3)=x+6+(24-a)/((x-2)^3)`

`\implies24-a=0`

Then

`24-6a=0\implies a=24`

`32+b-4a=0\implies b=64`

`16+c+2b-4a=0\implies c=-48`