Pre-Calc Help: Complex Division with Imaginary Numbers

[ddono9090]

Given -4i is a root, determine all other zeros of f(x)=3x³(x-4)(3x+9)⁴

 

[DrPhil]

Given -4i is a root, determine all other zeros of f(x)=3x³(x-4)(3x+9)⁴

Fortunately, if the function has real coefficients, then you never have to divide by a complex number. The reason for that is that complex roots always come in complex-conjugate pairs. Thus if -4i is a root, so is +4i, and
......(x - 4i)(x + 4i) = (x^2 + 16)
is a factor of the function, which you can divide out and reduce the order of the function by 2.

The given f(x) is completely factored and has all real roots .. must be a mix-up in the question.

 

[HallsofIvy]

Given -4i is a root, determine all other zeros of f(x)=3x³(x-4)(3x+9)⁴

There is really no need to be given that -4i is root- you could find that yourself. All "zeros" of \(\displaystyle f(x)= 3x^3(x- 4)(3x+ 9)^4\) satisfy \(\displaystyle f(x)= 3x^3(x- 4)(3x+ 9)^4= 0\). The only way a product of numbers can be 0 is if at least one of the factors is 0. So we must have x= 0 or x- 4= 0 or 3x+ 9= 0. Hmm, none of those has -4i as a root so the "Given -4i is a root" is not true!