Question

Given that $b^3+m{b}^2+nb+c$ is divisible by (b-s) if $s^3+m{s}^2+ns+c=0.$ Also given that $d^3+d^2+ed+1$ is divisible by (d-1) and $d^3-4d^2+fd-3$ is divisible by (d-3). What is the value of e+f ?

• A. -1
• B. 1
• C. 2

Answer 1

The correct option is B 1

$d^3+d^2+ed+1$ is divisible by (d-1). As 1 is a root, Put d=1 in the equation and equate it to 0, to get the value of e

1+1+e+1 = 0 $\Rightarrow$ e = -3

Similarly put d=3 in $d^3-4d^2+fd-3$ to get the value of f

27 - 36 + 3f - 3 = 0 $\Rightarrow$3f = 12 $\Rightarrow$ f = 4

e+f = -3+4=1