Determine to the 24 +22+8=0 Need (f(2)| > | g (2) on 121= 2 19 (2)| = 22 +8 ≤ 2/2/+8 = 2·2+8 12 If(z) | = |21ª = 16 | f(z) | > |g(z) | So 16 >12 T So of za f zeroes and = and ftg in 121<2 +22+8 4 Zeroes number with 2 4 on of solutions 12/<2 4 zeroes 121=2 counting multipitciy (use Rourke' on g(2)= 22 +8 have the same number f(2)=24

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