The question is "Show that z-2 is a factor of P(z)= z^3+mz^2+(3m-2)z-10m-4 for all values of m. For what values of m is (z-2)^2 a factor of P(z)?"

Question

I've proven the first part, using synthetic division, for all values of m. But the second part, I don't understand why the divisor must be divide again by 2, in order to show the values of m. If it's trying to find the values of P(z), shouldn't be P(z) that should be divided?

Answer 1

after your first division by z-2 using symthetic division you should have had
z^2 + (m+4)z + 5m+2

divide it again by z-2, you should now have a remainder of 7m+10

but if the division is exact that remainder of 7m+10 has to be zero
so
7m+10 = 0
m = -10/7

(dividing by (z-2)^2 is the same as dividing by z-2 and then dividing that result again by z-2.
Just like dividing 20 by 4,
divide 20 by 2 to get 10, then divide by 2 again to get 5)

Answer 2

Thats the results I got, thank you =D

Answer 3

Arithmetic progression