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

Question

asked Nov 27, 2020 33.3k views

1 Answer

Lakmal · Answer 1 · 2020-12-01T07:41:00+0000

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

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

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

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