Creating Polynomials
Given the Zeros.
What do we already know about polynomial functions?
They are
either
ODD
functions
Linear
y = 4x - 5
They are
either
EVEN
functions
Cubic
y = 4x3 - 5
Fifth Power
y = 4x5 –x + 5
Quadratics
Quadratics
y = 4x2 -2 5
y = 4x - 5
Quartics
y = 4x4 - 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
GCF
(x + )(x + )
The “6” step
Grouping
Cubic**
p/q
Long Division
(works on all
Factoring
factors of any
degree)
Quadratic
Formula
Synthetic
Division
(works only
with factors
of degree 1)
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 -).
If we factor a polynomial and end up with :
(4x  8)( x  1)  0
2
2
We would get solutions of :  2 and  i
So we could CREATE a polynomial if we were
given the polynomial’s zeros.
Write a polynomial function f of least degree that has
rational coefficients, a leading coefficient of 1 and the
given zeros.
-1, 2, 4
Step 1: Turn the zeros into factors.
(x+1)(x- 2)(x- 4)
Step 2: Multiply the factors together.
3
3
2
2
f(x)
x =x
- 5x- 5x
+2x+2x
+ 8+
Step 3: Name it!
8
Write a polynomial function f of least degree that has rational
coefficients, a leading coefficient of 1 and the given zeros.
2i,
3
Must remember
thatthe
“i”szeros
and roots
in pairs.
Step 1: Turn
intocome
factors.
2i,2i,
3 , 3
( x  2i )( x  2i )( x - 3 )( x  3 )
Step 2: Multiply factors.
3)
f ( x)( x x 4)(xx - 12
2
4
22
Write a polynomial function f of least degree that has rational
coefficients, a leading coefficient of 1 and the given zeros.
 2  i, 2 - 3
Step 1: Turn
intocome
factors.
Must remember
thatthe
“i”s zeros
and roots
in pairs.
 2  i, - 2  i, 2 - 3 ,2  3
( x  2  i )( x  2  i )( x - 2  3 )( x  2  3 )
Step 2: Multiply factors.
( x  2  i )( x  2  i )( x - 2  3 )( x  2  3 )
x
x
2
2
x
i
x
x
2
2x ix
2x 4 2i
-i
x x
-ix -2i -i12
2
(x +
4x + 5)
x
-2
3
x
x
x
x2 -2x x
-2
-2x 4
x x
 3 x 32 3
2
(x -
3
2 3
-3
4x + 1)
x2
2
x+
4x + 5
4
x
4x3 5x2
3
2 -20x
4x
-16x
-4x
2
x 4x 5
1
f(x) =
4
2
x -10x
-16x + 5
-3