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

$\dfrac{x^4-ax^3+bx+c}{x-2}=x^3+2x^2+(4-a)x+8+b-2a+\dfrac{16+c+2b-4a}{x-2}$

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

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

$\dfrac{x^4-ax^3+bx+c}{(x-2)^2}=x^2+4x+12-a+\dfrac{32+b-4a}{(x-2)^2}$

$\implies32+b-4a=0$

and again:

$\dfrac{x^4-ax^3+bx+c}{(x-2)^3}=x+6+\dfrac{24-a}{(x-2)^3}$

$\implies24-a=0$

Then

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

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

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