ernestbafful1332 ernestbafful1332
  • 09-01-2018
  • Mathematics
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What are the zeros of p(m)=(m^2-4)(m^2+1) jmap?

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Аноним Аноним
  • 09-01-2018
(m^2 - 4)(m^2 + 1) = 0

(m - 2)(m + 2)(m^2 + 1) = 0

(m - 2)= 0  gives m = 2
(m + 2)= 0 gives m = -2
m^2 + 1 = 0
m^2 = -1
 m = i, -i

The zeros are    -2, 2 , i, -i

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