AND THEIR ZEROS

A of degree n may always be written in n n1  a standard form: f (x)  an x  an1x ... 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  xn1)...(x  x1) where is the leading coefficient and the x1, …, xn are zeros of the polynomial f, which means that:

f (xn)  0, f (xn1)  0,...,f (x1)  0

In general, the zeros may be complex numbers.

This is at the heart of The Fundamental Theorem of whose consequence is that a polynomial of degree n has exactly n complex zeros, where complex numbers include real numbers.

Note:  If a z is a real zero of a 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  xn1)...(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 , we can always use the 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.