Question

Using the given zero, find all other zeros of f(x).

-2i is a zero of f(x) = x4 - 45x2 - 196

Answer 1

@abb0t @bridgetx516x @Christos @DragonFall @feezul_93

Answer 2

the remaining zeros are: 2i, 7, -7

Answer 3

so in total the function has four zeros:
\[2i \]
\[-2i\]
\[7\]
\[-7\]

Answer 4

however, \[\pm 2i \] are the complex/imaginary zeros.
\[\pm 7 \] are the real solutions.

Answer 5

If \(x=-2i\) is a root (aka zero), then we know that its conjugate \(x=2i\) is also a root because we have only real coefficients in the equation, and the only way to get the complex numbers to vanish is to have pairs of them.
This means that (x-2i) and (x+2i) are both factors of the polynomial, and if we multiply them together and divide the original polynomial f(x) by the product, we'll have a polynomial with the remaining roots as zeros.

\[(x-2i)(x+2i) = x^2+2ix - 2ix -4i^2 = x^2-4i^2\]but \(i^2=-1\) by definition, so that simplifies to \[x^2-4i^2=x^2-4(-1)=x^2+4\]

Now if we do long division of our original polynomial by x^2 + 4, what do we get?

x^2 - 49
____________________________
x^2 + 4 | x^4 - 45x^2 - 196
- ( x^4 + 4x^2)
-------------
-49x^2 - 196
- ( -49x^2 - 196 )
-------------
0

So our remaining polynomial is \(x^2-49 = 0\) which is the difference of two squares and easily factored to yield the remaining 2 roots.

We know by the fundamental theorem of algebra that because the highest power of x in our polynomial is 4 that we have 4 roots.