InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The polynomial `x^6+4x^5+3x 64+2x^3+x+1`is divisible by_______ where `w`is the cube root of units`x+omega`b. `x+omega^2`c. `(x+omega)(x+omega^2)`d. `(x-omega)(x-omega^2)`where `omega`is one of the imaginary cuberoots of unity.A. `x + omega `B. ` x + omega^(2)`C. `( x+ omega) (x + omega^(2))`D. `(x + omega)(x - omgea^(2))` |
|
Answer» Correct Answer - D Let `f(x) = x^(6) + 4x^(5) + 3x^(4) + 2x^(3) + x+ 1`. Hence, `f(omega) = omega^(6) + 4omega^(5) + 3omega^(4) + 2omega^(3) + omega + 1` `= 1+ 4 omega ^(2)+ 3omega^(4) + 3omega + 2 + omega +1` `= 4(omega^(2) + omega + 1)` `= 0` Hence, `f(x)` is divisible by `x-omega` Then `f(x)` is also divisible by `x - omega^(2)` (as complex roots occur in conjugate pairs). `f(-omega) = (-omega)^(6) + 4(-omeaga)^(5) + 3(-omega)^(4) + 2(-omega)^(3) + (-omega) +1` `= omega^(6) - 4omega^(5) + 3omega^(4) - 2omega^(3) - omega + 1` ` = 1 - 4omega^(2) + 3omega- 2 - omega + 1` ` ne 0`. |
|