Section 2-1: Polynomials
Polynomials
are expressions whose exponents are non-negative integers and the coefficients
are real numbers. The leading term
is the term with the highest power. The coefficient of the leading
term is the leading coefficient.
The power of x in the leading term is called the degree
of the polynomial! Here are some examples of polynomials:
| Degree |
Name |
Example |
| 0 |
constant |
11 |
| 1 |
linear |
4x + 7 |
| 2 |
quadratic |
3x2 + 4x + 5 |
| 3 |
cubic |
4x3 + 3x + 7 |
| 4 |
quartic |
3x4 + 2x2 + 4 |
| 5 |
quintic |
x5 + 3x + 1 |
Any value of x such that
P(x) = 0 is a root
of the equation and a zero
of the function.
Review of synthetic
substitution
To use synthetic substitution,
do the following:
1) Write the polynomial in descending order.
2) Make sure any missing terms are replaced with zero!
3) Bring down the leading coefficient.
4) Multiply by the divisor and add to the next term.
5) Repeat the process until the last coefficient is reached.
Example
Use synthetic division to find
P(2) for P(x) = 2x3 - 9x2 + 27
2|
2 -9
0 27
4 -10 -20
2 -5
-10 7
The value of P(2) = 7, the last
number!!
Sample Problems
1) State whether
each is a polynomial function. If yes, find the degree and the zeros.
Problem
Polynomial?
Degree
zeros
a)
f(x) = 5 - 4x
yes
1
5/4
b)
f(x) = 3x2 - 6x
yes
2
0, 2
c) f(x) = x - 3/x
no
d) f(x) = 18
yes
0
none
e) f(x) = (x-3)/(x+2)
no
2) Use synthetic
substitution to find P(3) for each function:
a) P(x) = 4x3 - 5x2 + 3
3|
4
-5
0 3
12
21 63
4
7
21 66
Therefore, P(3) = 66
b) P(x) = 3x4 + 2x3 - 5x - 2
3|
3 2
0 -5
-2
9 33
99 282
3 11
33 94
280
Therefore, P(3) = 280
3) If 4 is a zero of
f(x) = 3x3 + kx - 2, find the value of k.
Since 4 is a zero, the value of f(4) = 0. Use synthetic substitution.
4| 3
0 k
-2
12 48
192 + 4k
3 12 48 + k
190 + 4k
Now 190 + 4k = 0, solve for k
4k = -190
k = -190/4 = -47.5
That's it for this first
section. On to synthetic division!!
