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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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(a - b)(a + b) = a^2 - b^2
that doesnt help me though because i already knew that but this equation is far more difficult and i's are involved @rishavraj
i need steps

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Other answers:

see i^2 = -1
  • phi
use FOIL on (5+8i)(5+8i)
80i-39?
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
okay thank you so much
http://www.tiger-algebra.com/drill/(x-(5_8i)(x_(5_8i)/
i also have to multiply it by (x-4) and (x+14) or (x^2+10x-56) how would i do that? @phi
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) ?
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.
oh i didnt know you could do that. so should i distribute it all out?
i dont even know how to distribute the imaginary part
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.
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))
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
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
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
okay thank youuuuuu ill tell you what i get
  • phi
\[ ( x^2 -10x +89)(x-4)(x+14) \]
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

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