POLYNOMIALS AND THEIR ZEROS
A polynomial of degree n may always be written in n n1 a standard form: f (x) an x an1x ... a1x a0 where an is the leading coefficient and (0, a0 ) is the y-intercept, or
a complete factored form: f (x) an(x xn)(x xn1)...(x x1) where is the leading coefficient and the numbers x1, …, xn are zeros of the polynomial f, which means that:
f (xn) 0, f (xn1) 0,...,f (x1) 0
In general, the zeros may be complex numbers.
This is at the heart of The Fundamental Theorem of Algebra whose consequence is that a polynomial of degree n has exactly n complex zeros, where complex numbers include real numbers.
Note: If a number z is a real zero of a function f, then a point (z, 0) is an x-intercept of the graph of f.
The non-real zeros of a function f will not be visible on a xy-graph of the function.
Examples:
Standard f (x) 3x2 3x 6 h(x) 2x4 4x3 4x 2 g(x) 2x4 4x3 4x2 4x 2 Form
Factored f (x) 3(x 1)(x 2) h(x) 2(x 1)(x 1)3 g(x) 2(x 1)2 (x i)(x i) Form
Zeros x1 1, x2 2 x1 1, x2 1, x1 1, x2 1,
x3 1, x4 1 x3 i, x4 i
FINDING ZEROS OF POLYNOMIALS
If f (x) an(x xn)(x xn1)...(x x1) then the zeros are shown explicitly ( x1,...,xn ) but if f is not given in a complete factored form then depending on the degree different techniques apply.
Examples
For a polynomial of degree 2, a quadratic function, we can always use the Quadratic Formula to find the zeros. In some cases, factoring is possible instead.
2 1. Let f (x) 3x 3x 6. Find the zeros of f, i.e. solve f(x) = 0 Factoring Quadratic Formula
3x2 3x 6 0 a 3 b 3 c 6 3(x2 x 2) 0 b b2 4ac OR x 3(x 2)( x 1) 0 2a x 2 0 x 1 0 3 (3)2 4(3)(6) x 2 x 1 x 2(3)
3 81 3 9 x 6 6 3 9 3 9 x 2 x 1 1 6 2 6
2. Let f (x) x2 8 . Find the zeros of f, i.e. solve f(x) = 0.
Factoring Quadratic Formula 2 x 8 0 a 1 b 0 c 8 (x i 8)( x i 8) 0 b b2 4ac x x i 8 i2 2 2a OR or 0 (0)2 4(1)(8) 32 4 2 x x i2 2 2(1) 2 2
x i2 2 x i2 2 1 2
FINDING ZEROS OF POLYNOMIALS For a polynomial of degree n > 2 we can try factoring techniques. If they do not apply easily or at all, there are theorems that help in narrowing down the candidates for zeros. To check if a particular number, x1, indeed is a zero of a polynomial we can divide the polynomial by the factor (x – x1). If the remainder is equal to zero than we can rewrite the polynomial in a factored form as (x x1) f1(x) where f1(x) is a polynomial of degree n 1. This process can be continued until all zeros are found.
Factoring f ( x ) 2 x 4 4 x 3 4 x 2 3 f ( x ) 2 ( x 1 )( x 1 ) 0 x1 1 x2 1 x3 1 x4 1
Division by linear factors of the form x - c Is x = 1 a zero of f? Use synthetic division to check if (x – 1) divides f without a remainder:
1 -2 -4 0 4 2 -2 -6 -6 -2 ______-2 -6 -6 -2 0 Since the remainder = 0 then the polynomial f can be rewritten as 3 2 f (x) (x 1) f1(x) (x 1)(2x 6x 6x 2)
Is x = 1 a zero of f1? 1 -2 -6 -6 -2 -2 -8 -14 ______-2 -8 -14 -16 Since the remainder is not 0 then the polynomial f has only one zero x = 1.
Is x = -1 a zero of f ? -1 -2 -6 -6 -2 2 4 2 ______-2 -4 -2 0 Since the remainder = 0 then the polynomial f can be rewritten as 2 f (x) (x 1)(x 1) f2 (x) (x 1)(x 1)(2x 4x 2) To find the remaining two zeros we can always use the Quadratic Formula.