Question

How do you long divide `(t^5 - 3t^2 - 20) (t-2)^-1`?

Answer 1

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.