DESCRIBING POLYNOMIALS

A polynomial is one term or the sum or difference of two or more terms. A term is a number or the product or quotient of a number and a variable. A polynomial has no variables in a denominator. For a term that has only one variable, the degree of a term is the exponent of the variable.

Example:

has degree 3, has degree 1, has degree 2, and degree 0. Notice that the degree of a constant is 0.

The degree of a polynomial is the same as the degree of the term with the highest degree. You can name a polynomial by its degree or by the number of its terms. When refering to degree 1 we use the name linear. When refering to degree 2 we use the name quadratic. When refering to degree 3 we use the name cubic. When talking about terms of polynomials we refer to 1 term as a monomial, 2 terms as a binomial, and 3 terms as a trinomial.

Example:

is degree 1, linear, and is binomial because it has 2 terms.

is degree 2, quadratic, and is trinomial because it has 3 terms.

is degree 3, cubic, and is monomial because it has 1 term.

5 is degree 0, constant, and is monomial because it has 1 term.

COEFFICIENTS,DEGREE, and OPERATIONS OF POLYNOMIALS

A coefficient is an integer in front of the variable. As stated before, degree refers to the type of polynomial. All the operations can also be performed on polynomials.

> poly:=5*x^3;

> degree(poly,x);

> coeff(poly,x^3);

Two examples of operations on polynomials is that of addition and multiplication.

> poly:=(2*x^2-3*x+4)+(3*x^2+2*x-3);

> poly:=(2*x^2-3*x+4)*(3*x^2+2*x-3);

When multipliing two binomials we use the foil method. Foil is just a name for first, outer, inner, and last which helps you to remember how to use the distributive property. An example:

> poly:(3*x+1)*(3*x+1);

In this, we multiplied the first two terms , then the outer two terms , then the inner two terms , and finally the last two terms 1*1.

COMMUTING POLYNOMIALS

We can also do composition with polynomials. A composition means that a polynomial is obtained by substituting one polynomial into another polynomial for each occurence of x in the polynomial.

Example:

> f:=2*x^2+1;

> g:=2*x^2+1;

> fg:=subs(x=g,f);

When the composition of f and g, and the composition of g and f are 0, then we say that the two polynomials commute with each other.

Example:

Show that if p and q are constant polynomials which commute, then p=q.

> p:=c;

> q:=c1;

> pq:=subs(x=q,p);

> qp:=subs(x=p,q);

> dif:=collect(qp-pq,x);

> c2:=coeff(dif,x);

since , then , and therefore , so solution is complete.

There is something fishy about your argument. A good exam question would be to identify the fish.

SEMIGROUPS OF POLYNOMIALS

A semigroup of polynomials is a set of polynomials where the composition of two polynomials in the set is also in the set.

Example:

Show that the linear polynomials of slope 1 form a semigroup of polynomials. Is it commutative?

S,a set of linear polynomials with slope 1 is of the form y = x +a.

> restart;

> y1:= x + a1;

> y2:= x + a2;

> y1y2:=subs(x=y2,y1);

> y2y1:=subs(x=y1,y2);

> collect(y1y2,x);

> collect(y2y1,x);

> dif:=collect(y1y2-y2y1,x);

y1(y2) and y2(y1) are both of the form y = x + a so they form a semigroup. It is commutative because y1(y2) - y2(y1) =0.

THE BINOMIAL THEOREM and APPLICATIONS OF THE BINOMIAL THEOREM

.

As stated before, a binomial has two terms, but what happens when these two terms are raised to a certain power.

Example:

Here is where the binomial theorem can be useful. Basically, the binomial theorem is a formula for the expansion of (a+b)^n for n any positive integer. In sequence form, the binomial theorem looks like this:

(x+a)^n= n choose 0 * x^n + n choose 1 * a*x^n-1 + .... + n choose n * a^n.

The n choose j, where j above is 0,1,2,..,n, are the numerical coefficients that appear in the expansion of .

The n choose j is called the binomial coefficient.

The binomial theorem is useful because it allows us to find the coefficients of terms in the expansion of a binomial. Also, the theorems applications extend into probability, as well as into combinations and permutations.

Example:

Suppose we have a device with a .5 success rate and a .5 failure rate, and out of 7 devices we want to know the probability of more than 2 devices being defected. The binomial theorem will do this for us, as I will show in your example section.

When talking about n choose j, remember this is . Ouch! How are they supposed to know this?

EXAMPLE PROBLEMS OF POLYNOMIALS

Here are some example problems on polynomials.

1. Find the degree of this polynomial. .

> poly:=5*x^7+1;

> degree(poly,x);

2. Find the coefficient in front of in the polynomial .

> restart;

> poly:=5*x^7+1;

> coeff(poly,x^7);

3. Find the composition of .

> restart;

> f:=5*x+7;

> g:3*x+9;

> fg:=subs(x=g,f);

There is something wrong here.

4. Find all the linear polynomials that commute with .

> restart;

> f:=a*x+n;

> g:=2*x+1;

> fg:=subs(x=g,f);

> gf:=subs(x=f,g);

> dif:=collect(gf-fg,x);

all linear polynomials in the form of will commute with when , so this completes the solution.

5.Show that if p and q are linear polynomials which commute with , then p and q commute.

> restart;

> p:=a*x+n;

> q:=b*x+m;

> r:=2*x+1;

> pr:=subs(x=r,p);

> rp:=subs(x=p,r);

> dif:=collect(rp-pr,x);

> qr:=subs(x=r,q);

> rq:=subs(x=q,r);

> dif:=collect(rq-qr,x);

here we see that commutes with when and commutes with when .

> pq:=subs(x=q,p);

> qp:=subs(x=p,q);

> dif:=collect(qp-pq,x);

> solve(dif,a);

I have shown that both and commute with , and now when , p and q commute with each other, and this completes the solution.

6.Show that the linear polynomials with y-intercept 0 form a group of polymonials. Is it commutative?

S is a set of linear polynomials with y-int = 0 of the form y = x.

> restart;

> y1:= x;

> y2:= x;

> y1y2:=subs(x=y2,y1);

> y2y1:=subs(x=y1,y2);

> collect(y1y2,x);

> collect(y2y1,x);

> dif:=collect(y1y2-y2y1,x);

y1(y2) and y2(y1) are both of the form y = x so they form a semigroup. It is commutative because y1(y2) - y2(y1) =0

7.Because of a mistake in packaging, a case of 3 bottles of red wine had no labels. All the bottles have a .5 probability of tasting good or a .5 probability of not. What is the probability that all 3 taste good?

> answer:=(3!/(3!*0!))*(.5)^3*(.5)^0;

8.What is the coefficient of in the expansion of ?

> poly:=(2*y+3)^10;

> coeff(poly,y^8);

>

ASSIGNMENT FOR CLASS

Here are some problems for you guys to work on.

Don't ever class a bunch of high school student 'you guys'

1.Find the coefficient of in the expansion of .

2.Find the composition of the two polynomials and Do they commute with each other?

3.What is ? What is the degree of your answer? What is the name of your degree?(meaning is it linear, quadratic, etc.)

4.Find a quadratic polynomial that commutes with . If one exists.

5.Is there a cubic polynomial that commutes with ?

6.Show that is a semigroup of polynomials. If it is at all.

7.Does the two sets and form a semigroup? Why or why not?

8.Challenge problem.Prove the binomial theorem.

9.How could sets of semigroups be connected to the real world?(hint:think of computers)

10.Since we must know what a combination is for the use of the binomial theorem, find C(3,1),C(6,3),and C(n,n).

WHAT I HOPE THAT YOU LEARN FROM THIS WORKSHEET

I hope that this worksheet has taught you something about polynomials. After completing this worksheet you should be able to describe polynomials from degree to name, do the operations on polynomials, define composition and commuting of polynomials, tell me what a semigroup is, explain the binomial theorem and tell some of its applications, and most importantly be able to tell me how polynomials relate to real world issues, such as working in a store, finding probability of success and failure in a factory setting, and engineering issues.