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A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
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
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
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