Question

find a third-degree polynomial equation with rational coefficients that has roots -4 and 2+i

Answer 1

one factor is \((x+4)\)
the harder part is finding the quadratic with zeros \(2+i\) and\(2-i\) but there are a couple quick ways to do it
do you know any?

Answer 2

no, i didn't understand this lesson when it was taught

Answer 3

ok
we can go slow

Answer 4

since \(-4\) is a zero of the polynomial, one factor of the polynomial is \(x+4\)
is that much ok?

Answer 5

yea i get that

Answer 6

ok good
now you have a complex zero \(2+i\)
is it clear that the other zero must be its conjugate \(2-i\)?

Answer 7

yes

Answer 8

ok
so what we need to do is find the quadratic (degree 2) polynomial with zeros of \(2+i\) and \(2-i\)

Answer 9

the really annoying way, that you don't want to use, is to write this in factored form as
\[(x-(2+i))(x-(2-i))\] which is not as hard as it seems but is a headache just the same
there is a much easier way
to recap, our goal is to find a quadratic \(x^2+bx+c\) whose zeros are \(2+i\) and \(2-i\)

Answer 10

best way is to work backwards
set \(x=2+i\) and see what you get
steps are
\[x=2+i\] subtract 2\[x-2=i\] square both sides (carefully) \[(x-2)^2=i^2=-1\] or \[x^2-4x+4=-1\] then add \(1\) to both sides and get
\[x^2-4x+5\]

Answer 11

this tells you \(x^2-4x+5\) is the polynomial with zeros \(2+i\) and \(2-i\)
you can check if you like by using the quadratic formula and see that that is what you get
there is also a really really snappy way that i can show you if you like

Answer 12

yes please, if you don't mind

Answer 13

it is snappy, but it requires memorizing something
the quadratic with zeros \(a+bi\) and \(a-bi\) is
\[x^2-2ax+(a^2+b^2)\]

Answer 14

in your case you have \(2+i\) so \(a=2,b=1\) please notice that there is no "\(i\)" in this calculation
so
\[x^2-2\times 2x+(2^2+1^2)\] or
\[x^2-4x+5\] is your quadratic

Answer 15

that is so much easier thank you!

Answer 16

of course your final job is to multiply out
\[(x+4)(x^2-4x+5)\] without making an algebra error
i would cheat

Answer 17

yw, but don't forget that method requires memorizing it, so if you don't you can't use it
if you practice on 2 or 3 you will probably remember it
quadratic with zeros \(3+2i\) is
\[x^2-2\times 3x+(3^2+2^2)=x^2-6x+13\]

Answer 18

ok