Question

for the polynomial, one zero is given. find all others.
p(x)=x^3+2x^2-6x+8; 1+i

Answer 1

ooh kay
now we have a question
one zero is \(1+i\) the other is its conjugate \(1-i\)
lets find the quadratic with those zeros
there are three ways
one is annoying
one is easy
one is really really easy

Answer 2

I would prefer the very easy way :)

Answer 3

lets do the easy way
work backwards
\[x=1+i\] subtract 1\[x-1=i\] square both sides (carefully)
\[(x-1)^2=i^2\\
x^2-2x+1=-1\] add 1
\[x^2-2x+2\] is your quadratic

Answer 4

ok very very easy way is very very easy but it requires memorizing something
if the zeros are \(a+bi\) then the quadratic is
\[x^2-2ax+a^2+b^2\]

Answer 5

in your case \(1+i\) you have \(a=1,b=1\) quadratic is
\[x^2-2\times 1+1^2+1^2=x^2-2x+2\]

Answer 6

this means
\[p(x)=x^3+2x^2-6x+8=(x^2-2x+2)(something)\] and the "something" should be real easy to find

Answer 7

@satellite73 thanks for sharing the "very very easy way" don't remember if I learnt that or proves that it's not easy if you don't remember

Answer 8

yw
it is usually the case that the more you do something the easier it becomes, but memorizing something always risks memorizing it incorrectly
the very hard way is usually what is taught, multiplying
\[(x-(a+bi))(x-(a-bi))\] which is a pain

Answer 9

oh i know how to do it just don't recall the formula you provided

Answer 10

i learnt the hard way and yes it comes easy with practice

Answer 11

@Alexy
did you find the part after \[x^3+2x^2-6x+8=(x^2-2x+2)(something)\]?