Question

Find a polynomial with the lowest degree, with real coefficients whose zeros include, 4,-10, 3+4i
I've gone through the problem and i dont have an answer that matches the choices im not sure if i did it wrong or if the paper is incorrect so checking here!

Answer 1

hmm so...hmmm
say... if the polynomials real zeros were say 1 and 2
how would you find it then?

Answer 2

What i was trying in the book was that you basically multiply all the 0's together in a (x-c) format in which I did but I got an incorrect answer.

Answer 3

gimme a sec

Answer 4

so your roots are
4,-10, 3+4i

there's a complex one
complex roots do not travel all by their lonesome
they come along with their "conjugate" buddy

so that means, there is a 4th root
that is \(\Large 3-4i\)

Answer 5

http://openstudy.com/study#/updates/50fc3fdde4b010aceb3332a3

Answer 6

so gimme a few secs

Answer 7

hold the mayo

Answer 8

\(\bf x\to
\begin{cases}
4\\-10\\ 3+4i\\ 3-4i
\end{cases}\implies (x-4)(x+10)(x-3-4i)(x-3+4i)
\\ \quad \\
(x-4)(x+10)({\color{brown}{ [x-3] }} -4i)({\color{brown}{ [x-3] }}+4i)
\\ \quad \\
recall\implies \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\qquad thus
\\ \quad \\
(x-4)(x+10)({\color{brown}{ [x-3] }}^2 -[4i]^2)
\\ \quad \\
\textit{keep in mind that }i^2=-1\qquad thus
\\ \quad \\
(x-4)(x+10)([x-3]^2 -[4^2i^2])
\\ \quad \\
(x-4)(x+10)([x-3]^2 -[16{\color{brown}{ (-1)}}])
\\ \quad \\
(x-4)(x+10)([x-3]^2+16)\\ \quad \\ (x-4)(x+10)([x^2-6x+9]+16)
\\ \quad \\
(x-4)(x+10)(x^2-6x+25)\)


and from there is just a straight multiplication

Answer 9

Oh wow! that is the right answer, apparently some how I was getting x^2-6x-7 and not + 25 and that screwed me over lol and I did the problem twice honestly im not sure where I went wrong but I'll lo0k over your answer

Answer 10

Realized where I went wrong, I did not know i^2 was -1 I assumed it wuld just cancel out into 1.

Answer 11

Thank you!