Question

answer the following
●17.1,
●17.2,
●21.1,
●22.2......​

Answer 1

17.1- The polynomial might be p(x)=4, where 4 is constant.

It also can be written as,

$p(x) = 4 {x}^{0}$

because ,

${x}^{0} = 1$

17.2- Let R be a commutative, unital ring. According to wikipedia, a monomial order is a total order ≤ on the space of monomials of a given polynomial ring of R such that

1. If u≤v and w is some monomial, then uw≤vw;

2. 1≤u for all u.

It is stated that monomial orders are well-orderings. This is not clear at all to me. I googled it but I couldn't find anything proving this. However, there are several places (lecture notes, mostly) where a monomial order is defined as a well-ordering in monomials respecting multiplication.

21.1-

${x}^{2} + 2x + 1 = 0$

The above equation have only one zero which is x=1

22.2- A “double zero” means that two of the factors of the equation have zero as a solution when you solve for that factor equal to zero. A factor that has a particular solution is (X - that solution), or in your case, (X-4) which has X=4 as a zero, because if X = 4, then (X-4) = 0.

That gives us two of your factors, (X-4)(X-4) or (X-4)²

A “simple zero” means that one factor of the equation has zero as a solution when you solve (that factor = 0). Again, that factor would be (X - solution number), or in this case, (X - (-3) ), or (X+3). If X =-3 as given in your problem, then (X+3) = (-3 + 3) = 0.

So, is there one answer to your question? No. There are an infinite number of polynomials that have a double zero of 4 and a single zero of -3. Here are a few answers, just multiply them out to get your answer:

7 (x-4)² (x+3) = 0

2 (x-4)² (x+3) (x - 3) = 0

(x-4)² (x+3) (x + 7) (x - 9) = 0

All of these equations have your required answers, but some have additional zeroes.