Notes 2.3: Polynomial and Synthetic Division, The Factor and Remainder Theorems
This section will cover the competency of arithmetic with polynomials.
What will you learn in this section?
1. How to divide using long and synthetic division.
2. How you can use these to help factor a polynomial and find the zeros.
3. How synthetic division can help us to evaluate a polynomial or determine if
something is a factor by using the remainder.
Long Division
What is the process to do the following:
Think about the steps required.
How can you check your answer?
4 286
We will use the same process for long division with polynomials.
Ex 1: Divide 6x3 – 19x2 + 16x – 4 by x – 2.
We can use the result to factor this completely.
Ex 2: Divide 6x4 – x3 – x2 + 9x – 3 by x2 + 1
Ex 3: Divide x4 – 1 by x + 1
Synthetic Division
-Only works for divisors of the form x – k
-Use the coefficients of the polynomial inside
-Remember place holders if there is a missing power!
-Last number is the _________________
Ex 4: Use synthetic division to divide x4 – 10x2 – 2x + 4 by x + 3
The Remainder Theorem:
If a polynomial f(x) is divided by x – k, the remainder is f(k) = r.
(Basically, if you divide a polynomial by a number, the remainder is the same as if you
had plugged that number into the function.)
Ex 5: Evaluate f(x) = 3x3 + 8x2 + 5x – 7 at x = - 2
The Factor Theorem:
A polynomial f(x) has a factor (x – k) if and only if f(k) = 0.
(Basically, you know if something is a factor of the function if it goes in evenly, meaning
no remainder!)
Ex 6: Show that (x – 2) and (x + 3) are factors of f(x) = 2x4 + 7x3 – 4x2 – 27x – 18
HW: pgs 233 – 234 #7 – 15 odds, 21 – 29 odds, 47, 51, 52