A3 3.4 Zeros of Polynomial Functions

Homework: p. 387-388 1-31 eoo, 39-51 odd

Rational Zeros TheoremReal zeros of polynomial functions are either rationalzeros or irrational zeros. Examples:

24 9f x x 2 3 2 3x x The function has rational zeros –3/2 and 3/2

2 2f x x 2 2x x The function has irrational zeros – 2 and 2

Rational Zeros TheoremSuppose f is a polynomial function of degree n > 1 of the form

11 0

n nn nf x a x a x a

with every coefficient an integer and . If x = p /q isa rational zero of f, where p and q have no common integerfactors other than 1, then

0 0a

• p is an integer factor of the constant coefficient , and0a

• q is an integer factor of the leading coefficient .na

RZT – Examples:Find the rational zeros of 3 23 1f x x x The leading and constant coefficients are both 1!!!

The only possible rational zeros are 1 and –1…check them out:

3 21 1 3 1 1 3 0f 3 21 1 3 1 1 1 0f

So f has no rational zeros!!!(verify graphically?)

RZT – Examples:Find the rational zeros of 3 23 4 5 2f x x x x

Potential Rational Zeros:

1, 2 Factors of –2

Factors of 3 1, 3 1 2

1, 2, ,3 3

Graph the function to narrow the search…

Good candidates: 1, – 2, possibly –1/3 or –2/3

Begin checking these zeros, using synthetic division…

RZT – Examples:Find the rational zeros of 3 23 4 5 2f x x x x

1 3 4 –5 –2

3 7 2

3 7 2 0

Because the remainder is zero,x – 1 is a factor of f(x)!!!

23 7 2 1f x x x x Now, factor the remaining quadratic…

3 1 2 1f x x x x

The rational zeros are 1, –1/3, and –2

RZT – Examples:Find the polynomial function with leading coefficient 2 that hasdegree 3, with –1, 3, and –5 as zeros.

First, write the polynomial in factored form:

2 1 3 5f x x x x

Then expand into standard form:

22 2 3 5f x x x x

3 2 22 5 2 10 3 15x x x x x 3 22 6 26 30x x x

RZT – Examples:Using only algebraic methods, find the cubic function with thegiven table of values. Check with a calculator graph.

x

2 1 5f x k x x x

–2 –1 1 5

f(x) 0 24 0 0

(x + 2), (x – 1), and (x – 5)must be factors…

But we also have : 1 24f 1 2 1 1 1 5 24k 2k

2 2 1 5f x x x x

22 2 5x x x 3 22 8 14 20x x x

Properties of Roots of Polynomial Equations

1. A polynomial equation of degree n has n roots, counting repeated roots separately

2. If a+bi is a root to the polynomial equation with real coefficients ( ), then the imaginary number a-bi is also a root. Imaginary roots, if the occur, always occur in conjugate pairs

0b

Theorem: Linear Factorization Theorem

If f (x) is a polynomial function of degree n > 0, thenf (x) has precisely n linear factors and

1 2 nf x a x z x z x z where a is the leading coefficient of f (x), and are the complex zeros of f (x).the are not necessarily distinct numbers;some may be repeated.

1 2, , , nz z ziz

Fundamental Polynomial Connections in the Complex Case

The following statements about a polynomial function f areequivalent even if k is a nonreal complex number:

1. x = k is a solution (or root) of the equation f (x) = 0.

2. k is a zero of the function f.

3. x – k is a factor of f (x).

One “connection” is lost if k is a complex number…

k is not an x-intercept of the graph of f !!!

and now for some cool little theorems…Fundamental Theorem of Algebra: if f(x) is a polynomial of degree n, then the equation f(x)=0 has at least one complex root.

Linear Factorization Theorem: giventhen the linear factorization is:

Notice: the a’s are the same, and the linear factors are the zeros…

11 0( ) ...n n

n nf x a x a x a

1 2 3( ) ( )( )( )...( )n nf x a x c x c x c x c

some examples…

1. Find a 4th degree polynomial function with real coefficients that has zeros of -2, 2, and i such that f(3)=-150. Write the equation in factored form, and in general form.

2. Find an n-th degree polynomial function with real coefficients. Write the complete linear factorization and the polynomial in general form. n=3 (degree), x=6 and – 5+2i, f(2)=- 636