Question

If $-3,1,8$ are the roots of $p{x}^3+q{x}^2+rx+s=0$, then the roots of $p(x-3{)}^3+q(x-3{)}^2+r(x-3)+s=0$ are

• A. $0,4,11$
• B. $-6,-2,5$
• C. $-2,2,9$
• D. $-1,3,10$

Answer 1

Answer: (A)

$0,4,11$

The equation whose roots are $\alpha ,\beta$ and $\gamma$ is $f\left(x\right)=0$ (say)
The equation whose roots are $\alpha +3,\beta +3$ and $\gamma +3$ is $f(x-3)=0$
Thus, the roots of above equation would be $0,4,11$