State how many imaginary and real zeros the function has.
f(x) = x⁴ + 12x³ + 37x² + 12x + 36

Solution:

Given function f(x) = x⁴ + 12x³ + 37x² + 12x + 36

By trial and error x = -6 is to root

⇒ f (-6) = (-6)⁴ + 12 (-6)³ +37 (-6)² + 12 (-6) + 36

= 1296 - 2592 + 1332 - 72 + 36 = 0

dividing x⁴ + 12x³ + 37x² + 12x + 36 by x = -6

⇒ (x + 6) (x³ + 6x² + x +6) is the simplified form of f (x)

Again dividing x³ + 6x² + x + 6 by x = - 6 ⇒

⇒ f(x) = (x + 6)² (x² + 1) = 0

⇒ (x + 6)² = 0, x² +1 = 0

x = -6 and x² = -1 ⇒ x = +i and -i

Therefore, the zeroes of function f(x) are - 6, + i, -i

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State how many imaginary and real zeros the function has.
f(x) = x⁴ + 12x³ + 37x² + 12x + 36

Summary:

Imaginary and real zeros of f(x) = x⁴ + 12x³ + 37x² + 12x + 36 are - 6, - 6, + i, -i