Introduction to polynomials
Definition, vocabulary, domain and range
Polynomial
A polynomial is a mathematical expression which is the sum of monomial terms.
The following are polynomials:
- x^7+3x^3-17
- x^{101}
- -3x+4x^2
Degree of a polynomial
The degree of a polynomial is the value of the largest exponent of a monomial term in the polynomial with a nonzero coefficient.
Consider the following polynomial:
x^3+7x^2
Its degree is 3.
Consider the following polynomial:
x^5-7x^2+20x^{11}
Its degree is 11.
A polynomial of degree 2 is a quadratic.
Leading coefficient
The leading coefficient of a polynomial is the coefficient of the monomial which determines the degree of the polynomial.
Consider the following:
7x^3+11x^2
Its leading coefficient is 7.
Consider the following:
4x^2+11x^100 + x^{2018}
Its leading coefficient is 1.
The domain of a polynomial function is all real numbers.
- If a polynomial function has an odd degree, then its range is all real numbers.
- If a polynomial function has an even degree, then it is more difficult to find its range. The only even degree polynomial we can always find the range of are quadratic functions.
Consider the following function:
f(x)=x^7+3x^2-1.
The domain of f(x) is (-\infty, \infty) and the range of f(x) is (-\infty,\infty) because f(x) is an odd degree polynomial.
Basic operations with polynomials
Addition and subtraction of polynomials
If ax^n and bx^n are monomials of the same degree then:
ax^n+bx^n=\left(a+b\right)x^n
This rule for adding monomials can be extended to the addition of polynomials.
If f\left(x\right) and g\left(x\right) are polynomials then f\left(x\right)+g\left(x\right) is the polynomial obtained by adding the monomial terms of f\left(x\right) with the monomial terms of the same degree of g\left(x\right).
Consider the following functions:
- f\left(x\right)=3x^2+2x-7
- g\left(x\right)=-7x^2+1
The sum f\left(x\right)+g\left(x\right) is computed as follows:
f\left(x\right)+g\left(x\right)=3x^2+2x-7-7x^2+1\\=\left(3x^2-7x^2\right)+2x+\left(-7+1\right)\\=-4x^2+2x-6
Subtracting two polynomial expressions is performed in a similar manner.
Consider the following functions:
- f\left(x\right)=3x^2+2x-7
- g\left(x\right)=-7x^2+1
The difference f\left(x\right)-g\left(x\right) is computed as follows:
f\left(x\right)-g\left(x\right)=3x^2+2x-7-\left(-7x^2+1\right)\\=3x^2+2x-7+7x^2-1\\=10x^2+2x-8
Multiplication of polynomials
If ax^n and bx^m are monomials, then the multiplication of these two monomials are computed as follows:
ax^n \cdot bx^m=abx^{n+m}
The multiplication of a monomial and a polynomial extends this notion by distributing the monomial term.
Multiplcation of a monomial and polynomial
Consider the monomial ax^n and the polynomial b_mx^m+b_{m-1}+\cdots +b_1x+b_0. The multiplication of the monomial and the polynomial is computed as follows:
ax^n\left(b_mx^m+b_{m-1}x^{m-1}+\cdots +b_1x +b_0\right)=ab_mx^{n+m}+ab_{m-1}x^{n+m-1}+\cdots +ab_1x^{n+1}+ab_0x^n
Consider the following monominal and polynomial:
- 2x^7
- 3x^2+7x-1
The multiplication of the two is found as follows:
2x^7\left(3x^2+7x-1\right)=6x^9+14x^8-2x^7
Multiplication of two polynomials is computed by distributing one polynomial to all monomial terms of the other polynomial and then performing multiplication as above.
Consider the following functions:
- f\left(x\right)=x^2-2x
- g\left(x\right)=x^6+7
f\left(x\right)g\left(x\right) is computed as follows:
f\left(x\right)g\left(x\right)=\left(x^2-2x\right)\left(x^6+7\right)\\=x^2\left(x^6+7\right)-2x\left(x^6+7\right)\\=x^8+7x^2-2x^7-14x
Consider the following functions:
- f\left(x\right)=x^3-3
- g\left(x\right)=x^3-1
f\left(x\right)g\left(x\right) is computed as follows:
f\left(x\right)g\left(x\right)=\left(x^3-3\right)\left(x^3-1\right)\\=x^3\left(x^3-1\right)-3\left(x^3-1\right)\\=x^6-x^3-3x^3+3\\=x^6-4x^3+3
Division of polynomials
Long division
If a polynomial can be factored, then it is relatively straightforward to divide the polynomial by one of its factors.
The polynomial x^2-3x+2 can be factored as x^2-3x+2=\left(x-2\right)\left(x-1\right). Therefore:
\dfrac{x^2-3x+2}{x-1}=\dfrac{\left(x-1\right)\left(x-2\right)}{x-1}=x-2.
Consequence of Polynomial long division
Suppose f\left(x\right) and g\left(x\right) are polynomial functions. Then there exists polynomial functions q\left(x\right) and r\left(x\right) such that the following hold:
- f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)
- The degree of r\left(x\right) is strictly less than that of g\left(x\right).
Consider the polynomial functions f\left(x\right)=2x^3+3x+1 and g\left(x\right)=x^2-4x. We have:
- q\left(x\right)=2x+8
- r\left(x\right)=35x+1
Then:
f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)
More specifically:
2x^3+3x+1=\left(x^2-4x\right)\left(2x+8\right)+\left(35x+1\right)
Observe that the degree of g\left(x\right)=x^2-4x is 2, whereas the degree of r\left(x\right)=35x+1 is 1, which is strictly less than 2.
The algorithm of polynomial long division reduces dividing polynomials by dividing monomials.
Dividing 3x^3 and 12x^5 is done as follows:
\dfrac{12x^5}{3x^3}=\dfrac{12}{3}x^{5-3}=4x^2
We simply divided the coefficients and subtracted the exponents. The process of polynomial long division will require us to do so repeatedly.
Suppose f\left(x\right) and g\left(x\right) are polynomials with the degree of g\left(x\right) being no more than the degree of f\left(x\right). For simplicity, write the terms of f\left(x\right) and g\left(x\right) in descending order.
We can follow 4 steps to produce the polynomials q\left(x\right) and r\left(x\right) satisfying the following equation:
f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)
- Divide the leading monomial term of g\left(x\right) into the leading monomial term of f\left(x\right).
- Multiply g\left(x\right) by the additive inverse of the monomial term obtained in Step 1.
- Add the polynomial f\left(x\right) and the polynomial obtained in Step 2 to obtain a new polynomial whose degree is strictly less than that of f\left(x\right).
- If the polynomial obtained in Step 3 has degree less than that of g\left(x\right), then stop. If not, repeat steps 1-3 with the polynomials g\left(x\right) and the one obtained in Step 3.
Divide the polynomial x^2-3x+2 by the polynomial x-1.
Divide the leading monomial term of g\left(x\right) into the leading monomial term of f\left(x\right).
Dividing x^2 by x gives:
\dfrac{x^2}{x}=x
Multiply g\left(x\right) by the additive inverse of the monomial term obtained in Step 1.
Multiply x by x-1 and take its additive inverse:
-x\left(x-1\right)=-x^2+x
Add the polynomial f\left(x\right) and the polynomial obtained in Step 2 to obtain a new polynomial whose degree is strictly less than that of f\left(x\right).
Add the polynomials x^2-3x+2 and -x^2+x.
If the polynomial obtained in Step 3 has a lesser degree than that of g\left(x\right), then stop. If not, repeat Steps 1-3 with the polynomials g\left(x\right) and the one obtained in Step 3.
Divide -2x by x :
\dfrac{-2x}{x}=-2
Multiply x-1 by -2 and take its additive inverse:
-\left(-2\right)\left(x-1\right)=2x-2
Add the polynomials -2x+2 and 2x-2 :
-2x+2+2x-2=0
The algorithm stops here because the polynomial 0 has a lesser degree than the original polynomial x-1. It follows by the polynomial long division algorithm that:
x^2-2x+2=\left(x-1\right)\left(x-2\right)+0=\left(x-1\right)\left(x-2\right)
Synthetic division
The process of synthetic division allows us to determine if a polynomial f\left(x\right) is divisible by a polynomial of the form \left(x-a\right) where a is some number. In fact, synthetic division is the same as polynomial long division in disguise.
To perform synthetic division between a polynomial g\left(x\right) and a linear polynomial \left(x-a\right), begin by writing the terms of g\left(x\right) in descending order and include the 0 terms.
Consider the polynomial g\left(x\right)=x^3+2x-x^5. To perform synthetic division on g\left(x\right), rewrite g\left(x\right) as:
g\left(x\right)=-x^5+0x^4+x^3+2x+0
Division box
Suppose g\left(x\right) a polynomial written in descending order, 0 's included, and x-a is a linear polynomial. The division box of g\left(x\right) and x-a is obtained by writing the terms of g\left(x\right) in order and the coefficient a to the side:
The division box of x^3-0x^2+2x+1 and x-7 is the following:
To perform synthetic division between a polynomial g\left(x\right) and a linear polynomial x-a, begin by writing the terms of g\left(x\right) in descending order and include the 0 terms.
- Step 1: Write the division box corresponding to g\left(x\right) and x-a.
- Step 2: Bring the leading coefficient straight down.
- Step 3: Multiply the number brought down with a and put the resulting number below the first row in the next column.
- Step 4: Add the two numbers in the column where the number obtained in Step 3 was placed and put this result directly below the two numbers.
- Step 5: Repeat steps 3 and 4 with the numbers obtained in Step 4 until there are no more columns left.
The numbers placed at the bottom row of the division box (except for the last number) are the coefficients in descending order of the polynomial q\left(x\right) such that:
f\left(x\right)=\left(x-a\right)q\left(x\right)+r
Where r is the last number in the bottom row of the division box.
Apply the process of synthetic division to the polynomials f\left(x\right)=x^3+2x+1 and x-7.
Set up the division box
Bring down the 1
Multiply 7 and 1
Add 0 and 7
Repeat steps 3 and 4
Multiply 7 and 7.
Add 2 and 49.
Multiply 7 and 51.
Add 1 and 357.
The process of synthetic division stops here since there are no more columns. The process of synthetic division informs us that:
x^3+2x+1=\left(x-7\right)\left(x^2+7x+51\right)+358
Roots of polynomials and factorization
General properties
Root of a polynomial
Suppose f\left(x\right)=a_nx^n+\cdots +a_1x+a_0 is a polynomial function. Then a number c is called a root of f\left(x\right) if:
f\left(c\right)=0
Consider the following polynomial function:
f\left(x\right)=x^3-2x^2+1
The real number 1 is a root of f\left(x\right) because:
f\left(1\right)=1-2+1=0
Consider the following polynomial function:
f\left(x\right)=x^2+1
The complex number i=\sqrt{-1} is a root of f\left(x\right) because:
f\left(i\right)=i^2+1=-1+1=0
Fundamental Theorem of Algebra
Let f\left(x\right) be a polynomial function with a degree of at least 1. Then there exists a number c, which could be a complex number, such that:
f\left(c\right)=0
Consider the following polynomial function:
f\left(x\right)=x^2+1
f\left(x\right) has two roots, namely i and -i. Therefore, if only real roots of polynomial functions are allowed, then the polynomial x^2+1 would have no roots. In particular, the fundamental theorem of algebra is not valid if we only allowed for real roots.
Since every non-constant polynomial function has a root, it is natural to find methods of finding roots of polynomial functions. This can be difficult and often impossible for polynomials of high degree. However, we can find the roots of some polynomial functions.
Finding roots of polynomials
Roots of a linear function
Consider a linear function of the following form:
f\left(x\right)=ax+b
f\left(x\right) has only one root that corresponds with the x -intercept of f\left(x\right) :
\dfrac{-b}{a}
Consider the following linear function:
f\left(x\right)=2x-3
It has only one root:
c=\dfrac{3}{2}
Roots of a quadratic and its discriminant
Consider the followingquadratic function:
f\left(x\right)=ax^2+bx+c
Let D=b^2-4ac be its discriminant. Then we have the following:
- If D \gt 0, then f\left(x\right) has two real roots.
- If D=0, then f\left(x\right) has only one root that is real.
- If D \lt 0, then f\left(x\right) has two complex roots which are not real.
Consider the following quadratic function:
f\left(x\right)=2x^2-3x+4
The discriminant of f\left(x\right) is -23. Therefore, f\left(x\right) has two complex roots which are not real.
Quadratic formula
Consider the following quadratic function:
f\left(x\right)=ax^2+bx+c
Let D=b^2-4ac be its discriminant. Then the roots of f\left(x\right) are:
- \dfrac{-b+\sqrt{D}}{2a}
- \dfrac{-b-\sqrt{D}}{2a}
Consider the following quadratic function:
f\left(x\right)=x^2-3x-4
The discriminant of f\left(x\right) is 9+16=25. Therefore, f\left(x\right) has two real roots:
- \dfrac{3+\sqrt{25}}{2}=4
- \dfrac{3-\sqrt{25}}{2}=\dfrac{-1}{4}
Descartes' rule of signs
Suppose f\left(x\right)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0 is a polynomial function. Then the number of positive real roots of f\left(x\right) is no more than the number of sign changes from the list of numbers \{a_n,a_{n-1},\ldots,a_1,a_0\}.
Consider the following polynomial function
f\left(x\right)=3x^5-17x^3-x^2+x+9
The list of coefficients is:
\{3,-17,-1{,}1{,}9\}
In the above list of numbers, there are two places where the sign switches twice (from positive to negative or negative to positive). Therefore, there are at most two positive real roots of the polynomial function f\left(x\right).
Roots and factorizations
Suppose c is a real number and consider the following polynomial function:
f\left(x\right)=x-c
Observe that c is a root of f\left(x\right). The following theorem says that if c is a root of a polynomial function, then that polynomial is divisible by x-c.
Roots and factorization
Let f\left(x\right) be a polynomial function. Then a real number c is a root of f\left(x\right) if and only if x-c divides f\left(x\right). In other words, there is a polynomial function g\left(x\right) such that:
f\left(x\right)=\left(x-c\right)g\left(x\right)
Consider the following polynomial function:
f\left(x\right)=x^2-3x+2
1 is a root of f\left(x\right) and:
\dfrac{x^2-3x+2}{x-1}=x-2
With a remainder of 0.
In particular:
x^2-3x+2=\left(x-1\right)\left(x-2\right)
Consider the following polynomial function:
f\left(x\right)=x^4+x^3+x^2+2x-5
1 is a root of f\left(x\right) and:
\dfrac{x^4+x^3+x^2+2x-5}{x-1}= x^3+2x^2+3x+5
With a remainder of 0.
In particular:
x^4+x^3+x^2+2x-5=\left(x-1\right)\left(x^3+2x^2+3x+5\right)
Rational roots theorem
Suppose f\left(x\right)=a_nx^n+\cdots +a_1 x+a_0 is a polynomial function. The numbers a_n,\ldots, a_1, a_0 are integers and both a_n, a_0 are not zero. Let \dfrac{p}{q} be a rational number in reduced form. If \dfrac{p}{q} is a root of f\left(x\right), then q divides a_n and p divides a_0.
Consider the following polynomial function:
f\left(x\right)=3x^3+2x+1
If f\left(x\right) were to have a rational root, then it must be one of the following rational numbers:
- 1
- -1
- \dfrac{1}{3}
- \dfrac{-1}{3}
But:
- f\left(1\right)=6
- f\left(-1\right)=2
- f\left(\dfrac{1}{3}\right)=\dfrac{16}{9}
- f\left(\dfrac{-1}{3}\right)=\dfrac{2}{9}
Therefore, the polynomial f\left(x\right) does not have any rational roots.
Pascal's triangle and the binomial theorem
Pascal's Triangle
Pascal's triangle is the infinite triangular array of numbers satisfying the following two properties:
- The sides of the triangular array of numbers consists entirely of 1 's.
- Every other number in the triangle is the sum of the two numbers directly above it.
The following graphic contains the first seven rows of Pascal's triangle.
The triangular array of numbers satisfies the following two properties:
- The sides of the triangular array of numbers consists entirely of 1 's.
- Every other number in the triangle is the sum of the two numbers directly above it.
For example, the number in blue is the sum of the two red numbers directly above it.
The rows and the positions of the numbers are both counted by starting from 0.
For example, in the following image of Pascal's Triangle:
- The 0 th number in the 5 th row is in blue.
- The 3rd number in the 4th row is in green.
Binomial coefficient
Suppose n\geq 0 is an integer and i is an integer satisfying 0\leq i\leq n. Then the i th number in the n th row of Pascal's triangle is denoted:
n \choose i
The 0 th number in the 5 th row of Pascal's triangle is 1:
{5\choose 0}=1
The 3 rd number in the 4 th row of Pascal's triangle is 4:
{4\choose 3}=4
Factorial
Suppose n\geq 1 is an integer. Then the factorial of n is:
n!=n\left(n-1\right)\cdots 3\left(2\right)\left(1\right)
Furthermore, we have:
0!=1
The following are factorial computations:
- 3!=3\left(2\right)\left(1\right)=6
- 5!=5\left(4\right)\left(3\right)\left(2\right)\left(1\right)=120
Computing binomial coefficients
Let n\geq 0 be an integer and i be an integer satisfying 0\leq i\leq n. Then:
{n\choose i}=\dfrac{n!}{i!\left(n-i\right)!}
The theorem allows us to compute {5\choose 3} as follows:
{5\choose 3}=\dfrac{5!}{3!2!}=\dfrac{5\left(4\right)\left(3\right)\left(2\right)\left(1\right)}{3\left(2\right)\left(1\right)\left(2\right)\left(1\right)}=\dfrac{5\left(4\right)}{2}=10
The theorem allows us to compute {7\choose 2} as follows:
{7\choose 2}=\dfrac{7!}{2!5!}=\dfrac{7\left(6\right)\left(5\right)\left(4\right)\left(3\right)\left(2\right)\left(1\right)}{2\left(1\right)\left(5\right)\left(4\right)\left(3\right)\left(2\right)\left(1\right)}=\dfrac{7\left(6\right)}{2}=21
Binomial Theorem
Suppose a and b are numbers and n\geq 1 is an integer. Then:
\left(a+b\right)^n=\sum_{i=1}^n { n \choose i}a^ib^{n-i}
If we substitute a=b=1 into the equation of the binomial theorem, we see that:
\left(1+1\right)^n=2^n=\sum_{i=1}^n{n\choose i}1^i1^{n-i}=\sum_{i=1}^n{n\choose i}
Applying the binomial theorem when n=2 produces the following known formula:
\left(a+b\right)^2=a^2+2ab+b^2
The binomial theorem presents itself in probability theory. If a binomial experiment is being run and the probability of a success in a single trial is p, then the probability that there will be k successes in n trials is given by the following expression:
{n\choose k}p^k\left(1-p\right)^{n-k}
Suppose a fair coin is flipped 5 times. What is the probability that four of the flips are heads?
To answer the question, we consider the binomial experiment of flipping a coin. Consider landing on heads a success, which has a probability of .5 of happening on each experiment. Thus the probability that there are four successes in five experiments is:
{5\choose 4}\left(.5\right)^4\left(1-.5\right)^{5-4}=0.15\ 625
Hence there is approximately a 16\% chance of flipping four heads out of five flips of a fair coin.