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