Problem
ALG-B1-M04-P006 Parameter and Divisibility
#6
★★☆☆☆ Level 2 of 5
Find \(a\) if \(x^3+ax^2-4x-4a\) is divisible by \(x-2\).
Substitute \(x=2\).
The value at \(x=2\) is \(8+4a-8-4a=0\). Therefore the divisibility holds for every \(a\).
Sometimes the parameter cancels completely.