Question

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

you know how to find a quadratic polynomial with zeros \(1-2i\) and \(1+2i\) ?

Answer 2

no

Answer 3

@satellite73

Answer 4

ok there are a few ways
one is easy, the other is really really easy
lets do the easy way first

Answer 5

working backwards from
\[x=1+2i\] subtract 1 and get
\[x-1=2i\] then square (carefully)

\[(x-1)^2=(2i)^2\\
x^2-2x+1=-4\] then add \(4\) and get
\[x^2-2x+5\]

Answer 6

the real real easy way is to memorize that if \(a+bi\) is a zero of a quadratic, then it is
\[x^2-2ax+(a^2+b^2)\] so in your case, since the zero is \(1+2i\) it is
\[x^2-2\times 1+(1^2+2^2)=x^2-2x+5\]

Answer 7

wow thankyou

Answer 8

last job is to factor
\[x^4 - 2x^3 + 6x^2 - 2x + 5=(x^2-2x+5)(something)\]

Answer 9

i didnt look at it like that

Answer 10

factoring should be pretty easy now right?

Answer 11

yes
!

Answer 12

k good
you should get
\[(x^2-2x+5)(x^2+1)\] so other zeros are \(\pm i\)