P.o.D. – Use the Quadratic Formula to solve each equation. Exact answers only in lowest, reduced form.

1.) x2−2 x+2=0

2.) 4 x2+16 x+17=0

3.) 32 x2−6 x+9=0

1.) 1±i

2.) −2± 12i

3.) 2±i√2

2.5 – Zeros of Polynomial Functions

The Fundamental Theorem of Algebra:

If f(x) is a polynomial of degree n, where n>0, then f has at least one zero in the complex number system.

Similarly stated, there will be the same number of complex solutions as the highest exponent, or degree, of the polynomial.

Linear Factorization Theorem:

If f(x) is a polynomial of degree n where n>0, then f has precisely n linear factors f ( x )=an (x−c1 ) (x−c2 )…(x−cn) where c1 , c2 ,…cn are complex numbers.

*This is very similar to the Fundamental Theorem of Algebra.

EX: Find all the zeros of each function.

a.) F(x)= x-4b.) f ( x )=x2−4 x+4

c.) f ( x )=x3+9 x

d.) f ( x )=x4−16

a.) Set the factor equal to zero and solve.

x-4=0 x=4

b.) We first need to factor the polynomial. We can do this by De-FOIL-ing

f ( x )=x2−4 x+4=(x−2)(x−2)

Now set each factor equal to zero.x−2=0→x=2

This is known as a double root, because the same answer occurs twice.

c.) Again, we must factor.f ( x )=x3+9 x=x (x2+9)

Set each factor equal to zero.x=0 x2+9=0→

x2=−9→x=√−9→x=±3 i

X = 0, 3i, and -3i

Notice that we have a degree of 3 and 3 complex solutions. This agrees with the Fundamental Theorem of Algebra.

d.) We must factor once again. This time we will factor using the Difference of Squares.

f ( x )=x4−16=¿

Next, set each factor equal to zero. This is known as the Zero Product Property.x2+4=0→x2=−4→x=√−4→x=±2 ix-2=0 X+2=0

x=2 X= -2X = 2i, -2i, 2, -2

EX: Find the rational zeros of f ( x )=x3−5x2+2 x+8.

Recall, that the word “rational” means fractional. Therefore, the only solutions which we need to find are real, fractional answers. This can easily be done graphically.

x= -1,2,4

EX: Find the rational zeros of f ( x )=x3−15x2+75 x−125

x=5

Since there is only one rational zero, we know that there must be two imaginary solutions (or a triple root) due to the Fundamental Theorem of Algebra.

EX: Find the rational zeros of f ( x )=2x4−9 x3−18 x2+71 x−30

The solutions are x= ½, -3, 2, 5.

EX: Find all the real solutions of

f ( x )=x3−7x2−11 x+14

x = -2, .85995, 8.14005

We can also solve polynomials algebraically.

EX: Solve the previous problem algebraically.

Step 1: Use any known method (typically graphing) to find at least one real, rational solution.

From the previous example, we know that x= -2 is a solution. This means that (x+2) is a factor.

Step 2: Use this factor to divide the polynomial. We can use synthetic division if we want.

1 -7 -11 14

-2 18 -14

-----------------------------------

1 -9 7 0

This means that our other factor is x2−9 x+7.

Step 3: Solve this factor set equal to zero using the quadratic formula.

x=9±√¿¿¿

-2

Thus, our three solutions are 9±√532 and -

2.

Complex (Imaginary) Zeros Occur in Conjugate Pairs:

Let f(x) be a polynomial function that has real coefficients. If a+bi is a zero, then the conjugate a-bi is also a zero of the function.

This means that if 3-4i is a solution, then 3+4i must also be a solution.

EX: Find a third degree polynomial function with integer coefficients that has 2, 7i, and -7i as zeros.

Step 1: Write all possible binomial factors.

(x-2)(x-7i)(x+7i)

Step 2: Expand (or multiply, FOIL, distribute) the factors. We will do the Difference of Squares’ factors first.( x−2 ) (x2+7xi−7 xi−49 i2 )=( x−2 ) (x2−49 (−1 ) )=( x−2 ) (x2+49 )=x3+49 x−2 x2−98=x3−2 x2+49 x−98

EX: Find all the zeros off ( x )=x3−4 x2+21x−34, given that 1+4i is a zero of f.

Step 1: Find a real, rational zero using any known method.

x=2, so (x-2) is a factor.

Step 2: Divide f by the factor we just found.

(show the synthetic division on the whiteboard).

(x3−4 x2+21x−34 )÷ ( x−2 )=x2−2 x+17

Step 4: Solve this factor set equal to zero.

x=2±√¿¿¿

The three complex zeros are 2, 1+4i, and 1-4i.

EX: Write h ( x )=x3−11 x2+41 x−51 as the product of linear factors, and list all of its zeros.

Begin by solving the function.

(show all of the work on the whiteboard)

X = 3, 4+i, 4-i

Since these are the zeros, the factors must be (x-3), (x-4-i), and (x-4+i).

Descartes’ Rule of Signs:

1. The number of positive real zeros of a function is equal to the number of sign changes of f(x) or by an even number less.

2. The number of negative real zeros of a function is equal to the number of sign changes of f(-x) or by an even number less.

EX: Describe the possible real zeros of f ( x )=−2x3+5 x2−x+8.

+1 +1 +1

There are three sign changes of f(x). This means that there are 3 or 1 positive real zero.f (−x )=−2¿+8

There are 0 sign changes in f(-x), so there are 0 possible negative real zeros.

We can make a chart explaining this situation.

Possible (+) real zeros

Possible (-) real zeros

Imaginary solutions

3 0 01 0 2Let’s examine the graph to determine which situation exists.

This graph verifies that we only have 1 positive, real zero. Now find that zero.

x = 2.8243985http://www.youtube.com/watch?v=G7T1EY9i44Y

EX: Use Descartes’ Rule of Signs to determine the possible real zeros of f ( x )=8x3−4 x2+6 x−3. Then find all of the real, rational zeros.

There are 3 sign changes in f(x), so there are 3 or 1 possible positive, real zeros.

f (−x )=8¿

There are 0 sign changes in f(-x), so there are 0 possible negative, real zeros.

x = ½

Upon completion of this lesson, you should be able to:

1. Find all zeros of a polynomial function.2. Apply Descartes’ Rule of Signs.

For more information, visit http://www.purplemath.com/modules/drofsign.htm

HW Pg. 179 6-96 6ths, 106, 113, 129-132