Question

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

3, -13, and 5 + 4i

Answer 1

you know it is going to look like
\[(x-3)(x+13)(something)\] where the "something" is a quadratic (degree two) polynomial with zeros \(5+4i\) and \(5-4i\)
there are a couple easy ways to find that quadratic
one is really really easy

Answer 2

is the first part at least clear?

Answer 3

remember that if you have only real coefficients, any complex zeros must come in conjugate pairs \(a\pm bi\)

Answer 4

I'm eagerly awaiting the "really really easy" way :-)

Answer 5

what @whpalmer4 said is true, but the method i use doesn't even bother with that fact
ok i will show it second

Answer 6

I was thinking that I would do (3-x)(-13-x)((5+4i)-x)((5-4i)-x)

Answer 7

method one
set \[x=5+4i\] and work backwards
subtract \(5\) and get
\[x-5=4i\] then square both sides (carefully) and get
\[(x-5)^2+(4i)^2=-16\] or
\[x^2-10x+25=-16\] then add \(16\) to get
\[x^2-10x+41\]

Answer 8

so your quadratic with zeros \(5+4i\) and \(5-4i\) is \(x^2-10x+41\)
that way requires some algebra but nothing more
the real real easy way requires remembering something
the quadratic with zeros \(a+bi\) and \(a-bi\) and leading coefficient 1 is
\[x^2-2ax+(a^2+b^2)\]

Answer 9

so if i see \(5+4i\) i know it is
\[x^2-2\times 5x+(5^2+4^2)=x^2-10x+41\] easy but you have to memorize it, so it is good to know the other method as well

Answer 10

you could write what you had
\[ (3-x)(-13-x)((5+4i)-x)((5-4i)-x) \] or better yet
\[(x-3)(x+13)(x-(5+4i))(x-(5-4i))\] but then you would have the unenviable task of multiplying out that the last two terms

Answer 11

@whpalmer4 real real easy right?

Answer 12

Oh okay thanks sooo much I'm taking an online class and youtube nor the textbook was helping but your help sure did!

Answer 13

it is, except for the pesky remembering it part :-)

Answer 14

I've seen you use that before, and I ought to remember by now, but...I don't :-)

Answer 15

@nylearns you still have to multiply that mess out without making an algebra mistake
i would cheat
and yes, the remembering part is the pesky part, but it is not nearly as hard to remember as say the quadratic formula
you do one or two and you know it

Answer 16

Yea but for me the simpler it is the better! @satellite73

Answer 17

if it was me, and i was taking an on line class i would do this
http://www.wolframalpha.com/input/?i=%28x-3%29%28x%2B13%29%28x^2-10x%2B41%29
and be sure to get it right
you can check that the zeros are the ones you want

Answer 18

So could I solve this :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) = x^4 - 2x^3 + 6x^2 - 2x + 5.
On there? @satellite73

Answer 19

Oh, for that you can use the part I mentioned: you've got a polynomial with only real coefficients, and you have a complex zero, so it has a conjugate pair: \(a\pm bi\). You've got the \(a-bi\) case, so the other zero would be \(a+bi\).