anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
If 5 + 8i is a zero, then 5 - 8i must also be a zero. Now combine the zeros and multiply \[f(x)=(x-4)(x+14)(x-5+8i)(x-5-8i)\]
anonymous
  • anonymous
f(x)=(x−4)(x+14)(x−5+8i)(x−5−8i)

Looking for something else?

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

More answers

anonymous
  • anonymous
x^4-9x^3
anonymous
  • anonymous
how would i do the rest
anonymous
  • anonymous
id either get 725x or 1450x
anonymous
  • anonymous
I'm not sure how you got x^4-9x^3. You need to multipliy these using the distributive property. Start with (x−5+8i)(x−5−8i) to eliminate the imaginary parts
anonymous
  • anonymous
x^2-5x-8ix-5x+25+40i+8ix-40i
anonymous
  • anonymous
you're missing -64i², which simplifies to 64. Now combine like terms. All the parts with i will cancel out
anonymous
  • anonymous
x^2-10x+25-64i^2+81x
anonymous
  • anonymous
there's no 81x. Just \(x^2-10x+25-64i^2\) and -64i² = 64. So once you combine it's \[x^2-10x+89\]
anonymous
  • anonymous
Set that off to the side and multiply (x-4)(x+14)
anonymous
  • anonymous
x^2+14x-4x-56 x^2+10x-56
anonymous
  • anonymous
Right :) so now all that's left is to multiply the two quadratics \[(x^2-10x+89)(x^2+10x-56)\]
anonymous
  • anonymous
(x2−10x+89)(x2+10x−56) x^4+10x^3-56x^2-10x^2+560+89X^2+890x-4989
anonymous
  • anonymous
x^4+10x^3+23x+890x-4429
anonymous
  • anonymous
you missed a term (-10x)(10x) and 89*56 = 4984. Altogether it should be \[x^4+10x^3-56x^2-10x^3-100x^2+560x+89x^2+890x-4984\]
anonymous
  • anonymous
x^4+67x^2+1450x-4984
anonymous
  • anonymous
C thank you!
anonymous
  • anonymous
\[x^4-67x^2+1450x-4984\] yes C

Looking for something else?

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