Acommittee of 5 people is to be chosen from a group of 8 women and 10 men. how many diffferent committees are possible? how many are possible if the committee must feature both men and women? how many are possible if the committee must feature 3 women and 2 men? how many are possible if the committee must have more women than men?

A) 5 to be chosen among a Total : 10 Men + 8 Women

¹⁸C₅ = (18!)/(5!)(13!) = 8,568 groups of five

b) A must to have men and women. If so we have to deduct all groups of 5 that are all men and all group of 5 that are all women
Groups of 5 with only men: ¹⁰C₅ = 252
Groups of 5 with only women: ⁸C₅ = 56

So number of committees of 5 men and women mixed =
8568 - 252 - 56 = 8,260 committees 

c) 3 Women and 2 Men:

⁸C₃ x ¹⁰C₂ = 2,520 groups of 3 W and 2 M

d) More women than men, it means:
  3 W + 2 M OR (we have found it in c) = 2,520)
  4 W + 1 M OR   ⁸C₄ x ¹⁰C₁   →→→→ =    700
  5 W + 0 M OR   ⁸C₅ x ¹⁰C₀   →→→→  =     56

Total where W>M = 3,276 groups of 5 where women are at least 3

Answer to your question: a. the fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. the roots are ​ x=1±i7√2 .number of roots=degree of polynomialhere, degree of polynomial is 2 so number of roots will be 2.hope it *_*