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 #-#2).
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 #-# 2 t^4) + (2 t^4 #-#4 t^3) + (4 t^3 #-#8 t^2 ) + (5 t^2 #-#10 t ) + (10t -20)
= ( t.#-#2 )( t^4 + 2 t^3 + 4 t^2 + 5 t + 10
So, t #-#2 is a divisor. k = 0.