Question

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

-2i is a zero of f(x) = x4 - 45x2 - 196 (1 point)

Answer 1

9x(x^2-5x+4)=0

Answer 2

The first root will be 9x=0 -> x=0. Then factor out x^2-5x+4 and set it equal to 0.

Answer 3

Well, 0 has to be a solution, as you see by looking at the equation. Can you see why? Then divide both sides by x and then it's reduced to a quadratic equation you can solve.

Answer 4

So the right answer would be o, 1, and 4

Answer 5

Sorry to have to object, but I don't believe zero is a root. Could you explain why you think it is one? What about that 196 in the definition of f(x)? If x=0, the function value is 196 (not zero and thus zero is not a root / solution.)

Answer 6

@damandan: you opened this post. Please become involved if you want ongoing help.

Answer 7

sorry I had dinner but I am back now

Answer 8

The possible answers are
a 2i, 14i, -14i
b. 2i, 7, -7
c. 2i, 14, -14
d. 2i, 7i, -7i

Answer 9

The complex conjugate of a root is also a root so 2i is also a root. For the other two roots you do it the easy way which is to just insert the different answers into the polynomial and get the other two roots to be 7 and -7.

If you didn't have the answer though you could do it like this:

Answer 10

complex roots always occur in pairs.

Answer 11

To find out that it was 7 and negative seven did you synthetically divide?

Answer 12

You know you can factor the polynomial like this where r are is the roots
\[(x-r_1)(x-r_2)(x-r_3)(x-r_4)\]
Since you know the first two roots you can write
\[(x+2i)(x-2i)(x-r_3)(x-r_4)=x^4 - 45 x^2 - 196\]\]
Then you can divide and expand
\[(x-r_3)(x-r_4)=\frac{ x^4 - 45 x^2 - 196 }{ x^2+4 } \]
Doing the polynomial long division you get
\[(x-r_3)(x-r_4)=x^2-49=(x-7)(x+7)\]
And there you get the two last root 7 and -7

Answer 13

\[\left( x+2 \iota \right)~and~\left( x-2 \iota \right)are~ factors~ of~f(x)\]
\[(x+2 \iota)(x-2 \iota )=x^2-(2 \iota)^2=x^2-4 \iota^2=x^2+4\]
divide f(x) byx^2+4 and proceed

Answer 14

Thank you!