Question

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

8, -14, and 3 + 9i

Answer 1

first part is easy right?
\[(x-8)(x+14)\] for the zeros of \(8\) and \(-14\)
second part is annoying

Answer 2

there is however a more or less snappy way to find the quadratic with the zeros \(3+9i\) and \(3-9i\)

Answer 3

I know that you have to multiply the zeros together, but I don't know how to multiply (3+9i) and (3-9i) with the other zeros. could you help explain how to do that?

Answer 4

one way is to work backwards
start with
\[x=3+9i\] then subtract 3 to get
\[x-3=9i\] square both sides and get
\[(x-3)^2=-81\] or
\[x^2-6x+9=-81\] then add \(81\) and get
\[x^2-6x+90\]

Answer 5

@suzanna you do not multiply the zeros together

Answer 6

you have to multiply all the factors together
i.e. you have to multiply
\[(x-8)(x+14)(x-(3+9i))(x-(3-9i))\] so as you can see it is kind of a drag

Answer 7

to recap \((x-8)(x+14)\) is a routine multiplication right?

Answer 8

yes I can see, and that is where I get confused as to how to multiply them together and yes that is easy!

Answer 9

i will show you a really snappy way
but again lets make it clear \((x-8)(x+14)\) is the easy part
\((x-(3+9i)(x-(3-9i)\) is the hard part

Answer 10

yes i got x^2+6x-112 for (x-8)(x=14)

Answer 11

that is right , yes

Answer 12

okay, now how would i do the rest? I just get confused after that.

Answer 13

ok lets do it two ways
you want the real fast way first? comes with no explanation, just mechanics

Answer 14

yes please!

Answer 15

suppose you want a quadratic polynomial with leading coefficient 1 and zero \(a+bi\) which of course means it also has the zero \(a-bi\)
the polynomial is
\[x^2-2ax+(a^2+b^2)\]

Answer 16

for example, if you want a zero of \(4+3i\) then the quadratic is
\[x^2-2\times 4x+(4^2+3^2)=x^2-8x+25\]

Answer 17

in your case it will be
\[x^2-2\times 3x+3^2+9^2=x^2-6x+90\]

Answer 18

fast right?

Answer 19

yes it is! Is that all you have to do?

Answer 20

however, it comes at the disadvantage of having to memorize it and no explanation

Answer 21

so let me repeat what i had in the first post
suppose you want a zero of \(3+9i\) you can put \(x=3+9i\) and work backwards

Answer 22

i.e. put
\[x=3+9i\] subtract 3 and get
\[x-3=9i\] square both sides to get
\[(x-3)^2=-81\] or
\[x^2-6x+9=-81\] then add \(81\) to both sides and get
\[x^2-6x+90=0\] same thing obviously

Answer 23

yes i see!

Answer 24

of course if you have the stomach for it, you can always multiply \[(x-(3+9i)(x-(3-9i)\]which is not as bad as it looks
first is \(x^2\)
last is \((3+9i)(3-9i)=3^2+9^2=90\)
and "outer and inner" the \(9i\) part will add to zero and you will get \(-3x-3x=-6x\) for a final answer of
\[x^2-6x+90\]

Answer 25

but if you can memorize
\[x^2-2ax+a^2+b^2\] as the quadratic with zeros \(a+bi\) and \(a-bi\) then while your fellow students are struggling in their exam you can be done in like 30 seconds

Answer 26

okay, now would you multiply (x^2-6x+90) and (x^2+6x-112) together?

Answer 27

yeah exactly
this is donkey work
i would cheat

Answer 28

the command "expand" will do it for this one
http://www.wolframalpha.com/input/?i=expand+%28x-8%29%28x%2B14%29%28x^2-6x%2B90%29

Answer 29

wow is this really your first question here?

Answer 30

yes it is! and thank you for helping me!

Answer 31

yw
summer math, hope it is over soon for you
good luck !