Graded Warm Up

Complete the graded warm up on your desk by yourself. There should be no talking.

2.3 Real Zeros of Polynomial Functions

Back to elementary school…

Let’s say you need to find the prime factors of a rather large number. We’ll use 126 as an example. How do you start?

You pick a number that you know is a factor and divide them.

Then keep going with the answer you get from the division.

We know 2 is a factor because it’s even, so we divide 126 by 2 and get 63.

Now we can find the factors of 63, which are 9 and 7.

Now we find the factors of 9, which are 3 and 3.

We use the same idea with polynomials!

Long Division Reminder

Long Division of Polynomials

f(x) = 6x3 – 19x2 + 16x – 4

Where are the zeros?

We can use the zero we know to find the others.

If x=2 is a zero of the polynomial, than (x-2) is a factor.

We will divide 6x3 – 19x2 + 16x – 4 by (x-2) and then factor whatever is left to get the rest of the factors.

f(x)=6x^3-19x^2 + 16x-4

1 2 3 4

-3

-2

-1

1

x

y

Back to our starting problem: finding the zeros of f(x) = 6x3 – 19x2 + 16x – 4

We know one of the zeros is x=2, so we can divide 6x3 – 19x2 + 16x – 4 by x-2

Now we’re left with 6x2-7x+2, which we can factor into ( )( )

So 6x3 – 19x2 + 16x – 4 = (x-2)(2x-1)(3x-2)

Can you find the zeros now?

4161962 23 xxxx

Practice!

Divide 6x5 + 5x4 – 16x3 +20x2 -7x by 2x2 - x

Practice!Divide x2 + 3x + 5 by x + 1

We get a remainder this time, which means that (x+1) is NOT a factor and that x=-1 is NOT a zero.

It also means that x2 + 3x + 5 = (x+1)(x+2) + 3

The Division Algorithm If f(x) and d(x) are polynomials such that d(x) ≠ 0, and the

degree of d(x) is less than or equal to the degree of f(x), there exist unique polynomials q(x) and r(x) such that

f(x) = d(x)q(x) + r(x)

where r(x) = 0 or the degree of r(x) is less than the degree of d(x).

Dividend

Quotient

Divisor

Remainder

Synthetic Division Only can be used for linear divisors: (x

– k)

Divide by the zero, not the factor!!!

1) Divide x4 – 10x2 -2x + 4 by x + 3

1. Write the zero to the top left and put it in a box. Write out the coefficients of each term, remembering to insert a zero for any missing terms.

2. Bring down the first coefficient.

3. Multiply that number by the divisor.

4. Write the answer underneath the second coefficient.

5. Add and write the answer below the line.

6. Multiply this answer by the divisor

7. Lather, rinse, repeat.

Synthetic Practice

1. Divide -2x3 + 3x2 + 5x – 1 by x + 2

2. Divide x4 - 4x2 + 6 by x – 4

The Remainder Theorem

If a polynomial f(x) is divided by x – k, the remainder is r = f(k)

Use the remainder theorem to evaluate the following function at x = -2

f(x) = 3x3 + 8x2 + 5x – 7

Factor Theorem

A polynomial f(x) has a factor (x – k) if and only if f(k) = 0

You can use synthetic division to test whether a polynomial has a factor (x – k) If the last number in your answer line is a zero, then (x – k) is a

factor of f(x)

Repeated Division

Show that (x – 2) and (x + 3) are factors of

f(x) = 2x4 + 7x3 – 4x2 – 27x – 18

Then find the remaining factors of f(x)

Using the Remainder in Synthetic Division In summary, the remainder r, obtained in the synthetic division of f(x)

by (x – k), provides the following information.

1) The remainder r gives the value of f at x = k. That is, r = f(k)

2) If r = 0, (x – k) is a factor of f(x)

3) If r = 0, (k, 0) is an x-intercept of the graph f

Example

(a) verify the given factors of the function

(b) find the remaining factor(s) of f

(c) use your results to write the complete factorization of f

(d) list all real zeros of f

(e) confirm your results by using a graphing utility to graph the function.

f(x) = 2x3 + x2 – 5x + 2 (x + 2) and (x – 1)

Rational Zero Test

If the polynomial

f(x) = anxn + an-1xn-1 + …a2x2 + a1x + a0

Has integer coefficients, every rational zero of f has the form

Rational Zero =

Where p and q have not common factor other than 1, p is a factor of the constant term a0, and q is a factor of the leading coefficient an

Finding the possible real zeros

1) f(x) = x3 + 3x2 – 8x + 3

2) f(x) = 2x4 – 5x2 + 3x – 8

Descartes’ Rule of Signs

Let f(x) = anxn + an-1xn-1 + …a2x2 + a1x + a0 be a polynomial with real coefficients and a0 ≠ 0

1) The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by an even integer

2) The number of negative real zeros of f is either equal to the number of variations in sign of f(-x) or less than that number by an even integer.

Using Descartes’ Rule of Signs

Describe the possible real zeros of

a) f(x) = 3x3 – 5x2 + 6x – 4

b) f(x)=4x4+3x3-2x2+7x-9

Finding the Zeros of a Polynomial Function

f(x) = 6x3 – 4x2 + 3x - 2

Finding the Zeros of a Polynomial Function

f(x) = 6x3 – 4x2 + 3x - 2

HomeworkPg 123-125 #11-17 odd, 25, 27, 32, 39-43

odd