Question

NEED HELP
Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

1 - 2i is a zero of f(x) = x4 - 2x3 + 6x2 - 2x + 5.

Answer 1

if one zero is \(1-2i\) then the other is the conjugate \(1+2i\)

Answer 2

and therefore as we just learned, the quadratic polynomial is comes from is
\[x^2-2x+5\]

Answer 3

that is \(a+bi=1+2i\) so the quadratic is
\[x^2-2\times 1x+(1^2+2^2)=x^2-2x+5\]

Answer 4

then factor as
\[ x^4 - 2x^3 + 6x^2 - 2x + 5=(x^2-2x+5)(\text{something})\]

Answer 5

Knowing that x^2-2x+5 is a factor (or two) of f(x), you could now divide f(x) by x^2-2x+5 and have a much simpler quadratic to analyze for the other 2 roots.

Answer 6

what @whpalmer4 said
i would divide by thinking

Answer 7

\[x^4 - 2x^3 + 6x^2 - 2x + 5=(x^2-2x+5)(\text{something})\] and as for the "something" it is pretty clear that the first term must be \(x^2\) to give an answer of \(x^4\) and also pretty clear that the constant must be \(1\) because \(5\times 1=5\)

Answer 8

so \(\text{something}\) is \(x^2+bx+1\) and you can figure out what \(b\) is in a couple of ways

Answer 9

\[(x^2+bx+1)(x^2-2x+5)=x^4 - 2x^3 + 6x^2 - 2x + 5\] you see that \(-2x+5bx=-2x\implies b=0\)

Answer 10

or for that matter \(-2x^3+bx^3=-2x^3\implies b=0\)

Answer 11

or you can figure as @whpalmer4 said if the roots are rational they must be 1 and -1 and so finally you get
\[(x^2-2x+5)(x^2-1)\]

Answer 12

that is a bit dangerous though, because they don't actually have to be rational

Answer 13

right, I realized that and yanked the post!

personally, I would have taken the first divide step of multiplying \(x^2\) by \((x^2-2x+5)\) and observed that what was left after subtracting was just \(x^2-2x+5\)., then spent a few seconds verifying that I wanted a "+1" not a "-1" after the \(x^2\). RRT plus Horner's method would be a quick check.

it is \((x^2 -2x+5)(x^2+1)\) whether you go around the tree to the left or to the right...giving roots of \(x = \{\pm i, 1\pm 2i\}\)