Question

MEDALS AWARDED!!!!!

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

4, -8, and 2 + 3i

Answer 1

Each root c corresponds to a factor, (x-c)

Also, for any complex root a + bi, the complex conjugate a - bi must also be a root.

So that means the 3 roots you're given REALLY tell you FOUR roots.... so line up those 4 corresponding factors:

(x-4)(x+8)[x-(2 + 3i)][x-(2 - 3i)]

And multiply it out.

Answer 2

yeah, you can multiply out, but there are two easier ways to find the quadratic with zeros \(2+3i\) and \(2-3i\)

Answer 3

one way is to work backwards
set
\[x=2+3i\] subtract \(2\) and get
\(x-2=3i\) square to get
\[(x-2)^2=-9\] or '
\[x^2-4x+4=-9\] then add 9 to get
\[x^2-4x+13\] as your quadratic

Answer 4

the real real easy way, although it requires memorization, is to know that if the quadratic has zeros \(a+bi\) and \(a-bi\) then it is
\[x^2-2ab+(a^2+b^2)\] but you have to memorize that
it is not hard though and would save you lots of time on a quiz or test

Answer 5

oops that was wrong
it is
\[x^2-2ax+(a^2+b^2)\]

Answer 6

so in your example since \(a+bi=2+3i\) you get
\[x^2-2\times 2x+(2^2+3^2)=x^2-4x+13\]

Answer 7

Nice, that's slick! Great tip.

I know that you can make the complex factors easier to multiply by rewriting as a product of sum and difference:

[x-(2 + 3i)][x-(2 - 3i)]=
[(x-2) - 3i)][(x-2) + 3i)]

Then you get:
\((x-2)^2 - (3i)^2=x^2-4x+4-(-9)=x^2-4+13\)

which is kind of a similar idea, but I think I might like the "backwards" way that @satellite73 showed. :)

Answer 8

.... even better, I mean. lol

Answer 9

pretty much the exact same computation right?

Answer 10

i.e. you end up squaring \(x-2\) and then adding \(9\)

Answer 11

thank you i got it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!