Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.
4, -14, and 5 + 8i

Answer 1

use the provided roots and multiply them to each other
keep in mind that you have a complex there, make sure to include its conjugate

Answer 2

I understand that I need to change them like (x-4)(x+14)(5+8i) I just dont know how to multipy with that stupid i

Answer 3

hehe, thus you use the conjugate

Answer 4

i = sqrt(-1) so just treat is as another variable. the same way you treat x

a term can look this: 8ix^2

Answer 5

it's really adding the complex conjugate => \(\bf (x-4)(x+14)(5+8i)(5-8i)\)

to multiply the complex paiir, keep in mind that \(\bf (a-b)(a+b) = (a^2-b^2)\)

Answer 6

so when i multiply the conjugates i get 25 - 16i^2?

Answer 7

\(\bf (5+8i)(5−8i) \implies (5^2-(8i)^) \implies (25 -64(i^2))\\
i^2 = (\sqrt{-1})^2 \implies \sqrt{(-1)^2} \implies -1\\
(25 -64(i^2)) \implies (25 +64)\)

Answer 8

\(\bf (5+8i)(5−8i) \implies (5^2-(8i)^2)\) that is, hehe

Answer 9

wow im dumn 8 x 8 is 64 haha but thank you this really helped Ill do the rest on my own! :)

Answer 10

well, I kinda skipped the whole root itself
keep in mind that that 5 + 8i really means

Answer 11

x = 5 + 8i, x = 5 - 8i
(x -5-8i) (x-5+8i)

Answer 12

okay i tried doing it and i didnt get anything close to the answer..... these are my answer choices
f(x) = x4 - 362.5x2 + 1450x - 4984

f(x) = x4 - 9x3 + 32x2 - 725x + 4984

f(x) = x4 - 67x2 + 1450x - 4984

f(x) = x4 - 9x3 - 32x2 + 725x - 4984

Answer 13

lemme correct a few mistakes there

Answer 14

okay!

Answer 15

\(\bf (x -5-8i) (x-5+8i) \implies ((x -5)-(8i)) ((x-5)+(8i))\\
\color{blue}{\textit{now keep in mind that } (a-b)(a+b) = (a^2-b^2)}\\
((x -5)-(8i)) ((x-5)+(8i)) \implies ((x -5)^2-(8i)^2)\\
\left[(x^2-2(x)(5)+5^2)\right] - \left[64i^2\right]\\
\color{blue}{i^2 = (\sqrt{-1})^2 \implies \sqrt{(-1)^2} \implies -1 \textit{ therefore } 64i^2 \implies -64}\\
x^2-10x+25-(-64) \implies x^2-10x+25 +64 \implies x^2-10x+89\)

Answer 16

so you'd have to multiply \(\bf (x-4)(x+4)(x^2-10x+89)\)

Answer 17

darn, yet another typo

Answer 18

so you'd have to multiply \(\bf (x-4)(x+14)(x^2-10x+89)\)

Answer 19

Thank you!

Answer 20

yw

Answer 21

Whenever you work with conjugates, it helps to know that:
(a+bi)(a-bi)=a^2+b^2
So for example, in
(x-5+2i)(x-5-2i)
take a=x-5, b=2, so
(x-5+2i)(x-5-2i)
=(x-5)^2+4 which expands readily to
=x^2-10x+25+4
=x^2-10x+29