Question

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

Answer 1

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

Answer 2

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

Answer 3

i need steps

Answer 4

see i^2 = -1

Answer 5

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

Answer 6

80i-39?

Answer 7

then what

Answer 8

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

Answer 9

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

Answer 10

okay thank you so much

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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)

Answer 17

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

Answer 18

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

Answer 19

i dont even know how to distribute the imaginary part

Answer 20

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

Answer 21

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.

Answer 22

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

Answer 23

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)

Answer 24

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

Answer 25

wow i just realized that okay thank you

Answer 26

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

Answer 27

so should i foil x-5 first?

Answer 28

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

Answer 29

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

Answer 30

yes

Answer 31

okay thank youuuuuu ill tell you what i get

Answer 32

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

Answer 33

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

Answer 34

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