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# State how many imaginary and real zeros the function has.

f(x) = x^{4} + 12x^{3} + 37x^{2} + 12x + 36

**Solution:**

Given function f(x) = x^{4} + 12x^{3} + 37x^{2} + 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 x^{4} + 12x^{3} + 37x^{2} + 12x + 36 by x = -6

⇒ (x + 6) (x^{3 }+ 6x^{2} + x +6) is the simplified form of f (x)

Again dividing x^{3} + 6x^{2} + x + 6 by x = - 6 ⇒

⇒ f(x) = (x + 6)^{2} (x^{2} + 1) = 0

⇒ (x + 6)^{2} = 0, x^{2} +1 = 0

x = -6 and x^{2} = -1 ⇒ x = +i and -i

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

## State how many imaginary and real zeros the function has.

f(x) = x^{4} + 12x^{3} + 37x^{2} + 12x + 36

**Summary: **

Imaginary and real zeros of f(x) = x^{4} + 12x^{3} + 37x^{2} + 12x + 36 are - 6, - 6, + i, -i

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