I used Descartes’ rule of signs and from there determined that there were no true positive zeros. So I discovered that there could be 4, 2 or 0 true negative zeros. How do you know if there are 4 reals and 0 imaginary, 2 reals and 2 imaginary, or 0 reals and 4 imaginary zeros?
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Graph the function on a calculator. The xI intercepts are the true zeros. I would have 4 x-int so there are 0 imaginary zeros. Or if there are 2 real, then 2 imaginary and so on
Source(s): algebra student 2
Hello,
ƒ( ) = ⁴ + 12³ + 37² + 12 + 36
= ⁴ + 12³ + 36² + ² + 12 + 36
= ²(² + 12 + 36) + (² + 12 + 36)
= (² + 1)(² + 12 + 36)
= (² + 1)(² + 2× × 6 + 6²)
= (² + 1)( + 6)² ←←← Since a²+2ab+b²=(a+b)²
= (² – i²)( + 6)² ←←← Since i²=-1
= ( – i)( + i)( + 6)² ←←← As a²–b²=(a–b)(a+b)
And using the zero product property, you conclude that:
ƒ( ) = 0
has four zeros, two of which are real and equal (multiplicity of 2):
₀ = -6
and two of them imaginary and distinct:
₁ = -i
₂ = +I
compliments,
Dragon.Jade
((x + 6)^2) * ((x^2) + 1) as +/- i as 1 real and 2 imaginary
x^4+12x^3+36x^2+x^2+12x+36
x^2(x^2+12x+36)+(x^2+12x+36)
(x^2+1)(x+6)^2
x^2+1=0
x=±i (each with multiplicity 1)
x+6=0
x=-6 (multiplicity 2)
the total multiplicities must be equal to the degree of the function.
imaginary roots always come in pairs (conjugates)
the even multiplicity of the real root means that it does not cross the x-axis at this point. reflect from there