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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- rishavraj

(a - b)(a + b) = a^2 - b^2

- anonymous

that doesnt help me though because i already knew that but this equation is far more difficult and i's are involved @rishavraj

- anonymous

i need steps

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- rishavraj

see i^2 = -1

- phi

use FOIL on (5+8i)(5+8i)

- anonymous

80i-39?

- anonymous

then what

- phi

yes, but we usually put the "real" component first (tradition)
-39+80i

- phi

that is the b^2 in a^2 - b^2
a^2 is x^2
so the answer is
x^2 -( -39+80i) or
x^2 +39 -80 i

- anonymous

okay thank you so much

- rhr12

http://www.tiger-algebra.com/drill/(x-(5_8i)(x_(5_8i)/

- anonymous

i also have to multiply it by (x-4) and (x+14) or (x^2+10x-56) how would i do that? @phi

- anonymous

because when i do it i get terms with both an i and an x and i dont know how to simplify that

- phi

painfully.
you want to do
(x^2 + (39 -80 i)) (x^2+10x-56) ?

- anonymous

yeah haha i tried and it just gets too confusing after awhile

- phi

if we write it this way ,so we have "real" and imaginary terms:
\[( (x^2+39) - 80i) (x^2+10x-56) \]
and distribute the (x^2+10x-56) we get
\[ (x^2+39) (x^2+10x-56) + -80(x^2+10x-56) i \]
the first part gives a 4th order polynomial
the second part (with the "i") will be the imaginary part. (we leave the "i" on the outside)

- phi

no matter how you write it, it will be an ugly expression.

- anonymous

oh i didnt know you could do that. so should i distribute it all out?

- anonymous

i dont even know how to distribute the imaginary part

- anonymous

for the first part i got \[x ^{4}+10x ^{3}-17x ^{2}+390x-2184\]

- phi

for the imaginary part, you could distribute the -80 to get
(-80x^2-800x+4480) i
but multiplying it out really depends on what you plan to do next.
Personally I would leave things factored... unless there is a reason to multiply things out.

- anonymous

i would leave it factored also but the question was this:
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

- phi

oh. In that case we want *** real coefficients ****
and that means its roots come in complex conjugates
in other words (oh my!) we should start
5+8i and 5-8i (notice the -8i)

- phi

thus we would do
(x - (5+8i))(x - (5 -8i))

- anonymous

wow i just realized that okay thank you

- phi

or, collecting the real part
\[ ( (x-5) -8i) ( (x-5) + 8i) \]
the answer is a^2 - b^2 where a is (x-5) and b is 8i

- anonymous

so should i foil x-5 first?

- phi

yes, FOIL (x-5)(x-5)
you get \[
(x-5)^2 - (64 i^2) \\ x^2 -10x +25 + 64 \]
notice the very convenient fact, the imaginary part disappears

- anonymous

yes thank god then what about the other roots 4 and -14? do i just multiply (x-4) and (x+14) into that?

- phi

yes

- anonymous

okay thank youuuuuu ill tell you what i get

- phi

\[ ( x^2 -10x +89)(x-4)(x+14) \]

- anonymous

okay i got \[x ^{4}-67x ^{2}+1450x-4984\] sorry it took so long @phi

- phi

\[x^4-67 x^2+1450 x-4984\] looks good

Looking for something else?

Not the answer you are looking for? Search for more explanations.