How do you long divide # (t^5 - 3t^2 - 20) (t-2)^-1#?
1 Answer
Mar 1, 2016
t^4 + 2 t^3 + 4 t^2 + 5 t + 10
Explanation:
The answer is 'a biquadratic in t + k / ( t
Use this to 'add and subtract suitable multiples of powers of t in the numerator to get factor (t-2 ), in successive pairs'.
Numerator
= ( t^5
= ( t.
So, t