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The quadratic formula
You may recall the quadratic formula for roots of quadratic polynomials ax^2 + bx + c. It says that the solutions to this polynomial are −b ±
b^2 − 4 ac 2 a
The quadratic formula
You may recall the quadratic formula for roots of quadratic polynomials ax^2 + bx + c. It says that the solutions to this polynomial are −b ±
b^2 − 4 ac 2 a
For example, when we take the polynomial f (x) = x^2 − 3 x − 4, we obtain 3 ±
which gives 4 and −1.
The quadratic formula
You may recall the quadratic formula for roots of quadratic polynomials ax^2 + bx + c. It says that the solutions to this polynomial are −b ±
b^2 − 4 ac 2 a
For example, when we take the polynomial f (x) = x^2 − 3 x − 4, we obtain 3 ±
which gives 4 and −1. Some quick terminology I (^) We say that 4 and −1 are roots of the polynomial x^2 − 3 x − 4 or solutions to the polynomial equation x^2 − 3 x − 4 = 0. I (^) We may factor x^2 − 3 x − 4 as (x − 4)(x + 1).
The quadratic formula
You may recall the quadratic formula for roots of quadratic polynomials ax^2 + bx + c. It says that the solutions to this polynomial are −b ±
b^2 − 4 ac 2 a
For example, when we take the polynomial f (x) = x^2 − 3 x − 4, we obtain 3 ±
which gives 4 and −1. Some quick terminology I (^) We say that 4 and −1 are roots of the polynomial x^2 − 3 x − 4 or solutions to the polynomial equation x^2 − 3 x − 4 = 0. I (^) We may factor x^2 − 3 x − 4 as (x − 4)(x + 1). I (^) If we denote x^2 − 3 x − 4 as f (x), we have f (4) = 0 and f (−1) = 0.
Note that in the example both roots are integers, but other times it may give numbers are not integers or even rational numbers, such as with x^2 − 5, which gives ±
5, which is a real number that is not rational.
Other times it may even give complex numbers that are not real, such as with x^2 + 1, which gives ±i.
Higher degree polynomials
If you look at a cubic polynomial a 3 x^3 + a 2 x^2 + a 1 x + a 0 or a quartic a 4 x^4 + a 3 x^3 + a 2 x^2 + a 1 x + a 0 (where the ai are all integers) there are similar (but more complicated) formulas.
Higher degree polynomials
If you look at a cubic polynomial a 3 x^3 + a 2 x^2 + a 1 x + a 0 or a quartic a 4 x^4 + a 3 x^3 + a 2 x^2 + a 1 x + a 0 (where the ai are all integers) there are similar (but more complicated) formulas. For degree 5, there are no such formulas. This is called the insolubility of the quintic and it is a famous result proved by Abel and Galois in the early 19th century. However, we will be interested in something a bit more simple to begin with: rational number solutions to polynomials with integer coefficients.
Higher degree polynomials
If you look at a cubic polynomial a 3 x^3 + a 2 x^2 + a 1 x + a 0 or a quartic a 4 x^4 + a 3 x^3 + a 2 x^2 + a 1 x + a 0 (where the ai are all integers) there are similar (but more complicated) formulas. For degree 5, there are no such formulas. This is called the insolubility of the quintic and it is a famous result proved by Abel and Galois in the early 19th century. However, we will be interested in something a bit more simple to begin with: rational number solutions to polynomials with integer coefficients. That is, we will consider polynomials of the form
f (x) = anxn^ + an− 1 xn−^1 + · · · + a 0
Rational solutions to polynomials
Note that if we have f (x) = anxn^ + an− 1 xn−^1 + · · · + a 0 and f (b/c) = 0, (where b/c is in lowest terms, i.e. b and c have no common factors) then we have
a 0 = b c
an
b c
)n− 1 +... x 1
so b must divide a 0.
Rational solutions to polynomials
Note that if we have f (x) = anxn^ + an− 1 xn−^1 + · · · + a 0 and f (b/c) = 0, (where b/c is in lowest terms, i.e. b and c have no common factors) then we have
a 0 = b c
an
b c
)n− 1 +... x 1
so b must divide a 0. Similarly, after multiplying through by (c/b)n^ we obtain
an = c b
a 0
( (^) c b
)n− 1
so c must divide an.
Polynomials in two variables
What if we look instead at polynomials in two variables? Those are polynomials like x^4 y 2 + 5xy 3 + 7x + y + 10 and y 2 − x^3 − 2 x + 1.
Polynomials in two variables
What if we look instead at polynomials in two variables? Those are polynomials like x^4 y 2 + 5xy 3 + 7x + y + 10 and y 2 − x^3 − 2 x + 1.
Example
Fermat’s last theorem (first considered by Fermat in 1637, proved by Wiles in 1994) says that for n ≥ 3, there are no positive integers A, B, and C such that
An^ + Bn^ = C n.
Even older polynomial equations in two variable
Example
Pythagorean triples A^2 + B^2 = C 2 , e.g. 3^2 + 4^2 = 5^2 , become solutions to x^2 + y 2 − 1 = 0 after dividing by C (that is, letting x = A/C and y = B/C ).
Even older polynomial equations in two variable
Example
Pythagorean triples A^2 + B^2 = C 2 , e.g. 3^2 + 4^2 = 5^2 , become solutions to x^2 + y 2 − 1 = 0 after dividing by C (that is, letting x = A/C and y = B/C ).
Example
Take the polynomial equation
y 2 = x^8 + x^4 + x^2.
Diophantus of Alexandria found that x = 1/ 4 , y = 9/16 was a solution in the third century AD.
