Creating Polynomials Given the Zeros.. What do we already know about polynomial functions? They are...

Creating Polynomials Given the Zeros.

What do we already know about polynomial functions?

They are either ODD

functions

They are either EVEN

functions

Lineary = 4x - 5

Cubicy = 4x3 - 5

Fifth Powery = 4x5 –x + 5

Quadraticsy = 4x2 - 5

Quarticsy = 4x4 - 5

Quadraticsy = 4x2 - 5

We know that factoring and then solving those factors set equal to zero allows us to find possible x intercepts.

TOOLS WE’VE USED

Factoring

Quadratic Formula

Long Division (works on all factors of any

degree)

Synthetic Division

(works only with factors of degree 1)

GCF

(x + )(x + )

The “6” step

Grouping

p/q

Cubic**

We know that solutions of polynomial functions can be rational, irrational or imaginary.

X intercepts are real.Zeros are x-intercepts if they are real

Zeros are solutions that let the polynomial equal 0

We have seen that imaginaries and square roots come in pairs ( + or -).

So we could CREATE a polynomial if we were given the polynomial’s zeros.

i and 2 :of solutionsget wouldWe

0 )1)(8(4x

: withup end and polynomial afactor we22

x

If

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1 and the

given zeros.

-1, 2, 4Step 1: Turn the zeros into factors.

(x+1)(x- 2)(x- 4)Step 2: Multiply the factors together.

x3 - 5x2 +2x + 8Step 3: Name it!

f(x) =x3 - 5x2 +2x + 8

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1 and the given zeros.

Step 1: Turn the zeros into factors.

3 ,2i

Must remember that “i”s and roots come in pairs.

3,3 ,2,2 ii

)3)(3-x)(2)(2( xixix

Step 2: Multiply factors.)3-x)(4( 22 x 12)( 24 xxxf

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1 and the given zeros.

Step 1: Turn the zeros into factors.

3-2 ,2 i

Must remember that “i”s and roots come in pairs.

32,3-2 ,2- ,2 ii

)32)(32-x)(2)(2( xixix

Step 2: Multiply factors.

)32)(32-x)(2)(2( xixix

x 2 ix

2

-i

x2 2x ix2i42x

-ix -2i -i2 1xxx

x(x2+ 4x + 5)

x -2 3

x

-2

3

x2 -2x

4-2x

-3

3x

3x 32

32

xxx

x(x2- 4x + 1)

x2

-4x

1

x2+ 4x + 5x4 4x3

-3f(x) = x4-10x2 -16x + 5

-4x3

x2

-16x2

5x2

-20x

54x