Question

18-Zeroes of p(z) = z² – 36 are ______ and ______ . *

Answer 1

Answer:

+6 , -6

Explanation:

p(z) = z²-36

p(z) = 0

z²- 36 = 0

z² = 36

z = root 36

z = + 6 , -6

Answer 2

Answer:

6 and -6

Explanation:

p(z)= z^2-36

we can write 36 as 6^2

so,

= z^2- 6^2

by using identity a^2 - b^3 = (a+b)(a-b)

we get,

= (z+6)(z-6)

z = - 6 or 6