POLYNOMIAL PATTERNS Learning Task:
(nomial means name or term.)
1. In the activation activity, we looked at four different polynomial functions.
a. Let’s break down the word: poly- and –nomial. What does “poly” mean?
b. A monomial is a numeral, variable, or the product of a numeral and one or more
variables. For example, -1, ½, 3x, and 2xy are monomials. Give a few examples
of other monomials: 7, -3, 5x, 8xy
c. What is a constant? Give a few examples: 3, -6, .4, ½, 26
d. A coefficient is the numerical factor of a
Monomial
Coefficient
monomial or the _constant_ in front of
4x
4
the variable in a monomial. Give some
3ab
3
examples of monomials and their
12xy
12
coefficients.
e. The degree of a monomial is the sum of the exponents of its variables.
The monomial 5x3y has degree 4. Why? 3+1 = 4
What is the degree of the monomial 3? Why? x0 = 1
f. The examples shown below are all polynomials. Based on these examples and the
definition of a monomial, define polynomial .
2x2 + 3x
4xy3
7x5
18x3 + 2x2 – 3x + 5
A polynomial function in one variable is defined as a function of the form
f(x)= anxn + . . . a2x2 + a1x + a0 , where the coefficients are real numbers.
1. a. y = x2
y = x3
y = x4
y = x5
3.
Example
Degree
2
2x2 + 3
-x3
x4 + 3x2
3x5 – 4x + 2
Name (based on
degree)
No. of
terms
Constant
Quadratic
Cubic
Quartic
Quintic
Name (based on
no. of terms)
Monomial
Binomial
Trinomial
Example
Degree
Name
No. of
terms
2
0
Constant
1
Monomial
2x2 + 3
2
Quadratic
2
Binomial
-x3
3
Cubic
1
Monomial
x4 + 3x2
4
Quartic
2
Binomial
3x5 – 4x + 2
5
Quintic
3
Trinomial
Name
Handout of Graphs of Polynomial Functions :
f(x) = x2 + 2x
f(x)= x ( x + 2)
k(x) = x4 – 5x2 +4
k(x) = (x – 1)(x + 1)(x – 2)(x + 2)
l(x) = –(x4 – 5x2 +4)
l(x) = -(x-1)(x+1)(x-2)(x+2)
g(x) = – 2x2 + x
g(x) = x(-2x+1)
m(x) = ½(x5 + 4x4 – 7x3 – 22x2 + 24x)
m(x) = ½x(x - 1)(x - 2)(x + 3)(x + 4)
h(x) = x3 – x
h(x) = x(x – 1)(x + 1)
j(x) = – x3 + 2x2 + 3x
j(x) = -x(x - 3)(x + 1)
n(x) = – ½(x5 + 4x4 – 7x3 – 22x2 + 24x)
n(x) = -½x(x – 1)(x – 2)(x + 3)(x + 4)
5. a. For each graph, use the graph to find the x-intercepts of the
functions. On the table below, write the x-intercepts of the
function in the first column, then write the linear factors of the
function in the second column.
Function
f(x) = x2 + 2x
g(x) = –2x2 + x
h(x) = x3 – x
j(x) = –x3 + 2x2 + 3x
k(x) = x4 – 5x2 + 4
l(x) = –(x4 – 5x2 + 4);
x-intercepts Linear factors
–2, 0
0, ½
–1, 0, 1
–1, 0, 3
f(x)= x(x + 2)
g(x) = x(–2x + 1)
h(x) = x(x – 1)(x + 1)
j(x) = –x(x – 3)(x + 1)
–2, –1, 1, 2 k(x) = (x – 1)(x + 1)(x – 2)(x +2)
1,-1,2,-2
l(x) = –(x - 1)(x + 1)(x - 2)(x + 2)
m(x) = ½(x5 + 4x4 – 7x3 – 22x2 + 24x) 1,2,-3,-4
m(x) = ½x(x – 1)(x – 2)(x + 3)(x + 4)
n(x) = –½(x5 + 4x4 – 7x3 – 22x2 + 24x) 1,2,-3,-4
n(x) = –½x(x – 1)(x – 2)(x + 3)(x + 4)
The x-intercepts are
(0,0), (3,0) and (-1,0). We can see this graphically, or we know that at x-intercepts,
the value of the function is zero. We can find the x-values when j(x) equals zero
using the equations: –x = 0, (x-3) = 0, and (x+1) = 0.
Why might it be useful to know the linear factors of a function? You can
find the zeros of the function, which are the x-intercepts. Knowing these will help
you graph the function, and zeros are real solutions to the polynomial equations.
How are the intercepts related to the linear factors?
5.
a. Although we will not factor higher order polynomial functions in this unit, you
have factored quadratic functions in Math I and Math II. For review, factor the
following second degree polynomials, or quadratics.
i. y = x2 – x – 12
y = (x + 3)(x – 4)
ii. y = x2 + 5x – 6
y = (x - 1)(x + 6)
iii. y = 2x2 – 6x – 10
y=
b. Using these factors, find the roots of these three equations.
6
i. -3,4
ii. 1,-6
iii.
,
ii.
4
2
c. Use the relationship between the
linear factors and x-intercepts to
sketch graphs of the three
quadratic equations above.
-10
-5
5
10
5
10
-2
-4
6
-6
i.
6
4
4
2
2
-10
-5
5
10
iii.
-10
-5
-2
-2
-4
-4
-6
-6
d. Although you will not need to be able to find all of the roots of higher order
polynomials until a later unit, using what you already know, you can factor some
polynomial equations and find their roots in a similar way.
Factor y = x5 + x4 – 2x3.
f(x) = x3(x – 1)(x + 2)
What are the roots of this fifth order polynomial function?
How many roots are there? 0, 1, -2
Why are there not five roots since this is a fifth degree polynomial?
0 is a multiple root.
e. For other polynomial functions, we will not be able to draw upon our knowledge
of factoring quadratic functions to find zeroes. For example, you may not be able
to factor y = x3 + 8x2 + 5x - 14, but can you still find its zeros by graphing it in your
calculator? How? What are the zeros of this polynomial function?
* Yes, you can find the zeros on a graphing calculator.
* You graph the function on the calculator and use the “zero” function
on the calculator to find where the graph crosses the x-axis.
* -7, -2, 1