Question

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

7, -11, and 2 + 6i

Answer 1

ok here we go

Answer 2

you know it is going to look like
\[(x-7)(x+11)(x-(2+6i))(x-(2-6i))\] when you multiply out
multiplying \((x-7)(x+11)\) is routine
the question is, what do you get when you multiply
\[(x-(2+6i))(x-(2-6i))\] and this is not nearly as hard as it looks

Answer 3

you have a choice
you can actually multiply all that mess out, which is really not that bad
however it is easier to work backwards
\[x=2+6i\\x-2=6i\\(x-2)^2=-36\\x^2-4x+4=-36\\x^2-4x+40=0\] so the original quadratic with the zeros \(2+6i\) and \(2-6i\) is \(x^2-4x+40\)

Answer 4

what is actually easiest is to know that if \(a+bi\) is a zero of a quadratic with leading coefficient one, then the quadratic is
\[x^2-2ax+(a^2+b^2)\]

Answer 5

great so i just multiple x^2 - 4x + 40 to x+11 and x-7

Answer 6

in your case \(a=2,b=6\) to the polynomial was
\[x^2-4x+40\]

Answer 7

yes, final answer will be whatever you get when you multiply
\[(x^2-4x+40)(x-7)(x+11)\]

Answer 8

Yay thank you!

Answer 9

if i were you, i would cheat!

Answer 10

cheat ha? its a simple problem! Just take some time :)

Answer 11

yeah, i would probably make an algebra error
here is the answer, and you can see from the zeros it is right
\[x^4-53 x^2+468 x-3080\]
http://www.wolframalpha.com/input/?i=%28x^2-4x%2B40%29%28x-7%29%28x%2B11%29