1. Long Division and Synthetic Division
Dividing Polynomials
Long Division and Synthetic Division

2. Example 1: x + 5 x − 3 x2 + 2x − 15 -x2 + 3x x2 – 3x 5x – 15 -5x + 15x
+ 5
x − 3
x2 + 2x − 15
-x2 + 3x
x2 – 3x
5x – 15
-5x + 15
5x – 15

3. Example 2: 2x + 5 x − 3 2x2 − x − 4 -2x2 + 6x 2x2 – 6x 5x – 4 -5x + 152x
+ 5
x − 3
2x2 − x − 4
-2x2 + 6x
2x2 – 6x
5x – 4
-5x + 15
5x – 15 11

4. Example 3: x2 – x – 3 x4 − x3 + 4x + 2 x2 + 3 –x4 – 3x2 x4 + 3x2x2
– x
– 3
x4 − x3 + 4x + 2
x2 + 3
–x4 – 3x2
x4 + 3x2
− x3 – 3x2 + 4x + 2
+ x x
– x – 3x
– 3x2 + 7x + 2
+3x
– 3x – 9
7x + 11

5. There is a shorthcut method…
… called Synthetic Division
a shorthand, or shortcut, method of polynomial division 
Uses just the Coefficients
Only when dividing by a linear factor. For example (x+2) or (x-5)
Synthetic division is generally used for finding zeros (or roots) of polynomial functions

6. Synthetic Division Change the sign If the remainder is 0, then
Change the sign
If the remainder is 0, then
(x-3) is a factor
x=3 is a zero.
remainder
The first term’s exponent is 1 less than the original.

7. Synthetic Division:
Change the sign

8. Synthetic Division:
Change the sign

9. f(x) = 3x3 + 2x2 − x + 3 is divided by g(x) = x − 4.
Example 4:
Use synthetic division to find the quotient Q(x) and the remainder R(x) when:
f(x) = 3x3 + 2x2 − x + 3 is divided by g(x) = x − 4.

10. Example 5:
Put in missing exponents and insert 0’s if needed.

11. Remainder Theorem:
If the polynomial P(x) is divided by D(x) = x − c, then
becomes
Plugging in x=c to the above equation one sees that

12. Example 6: Use the Remainder Theorem to find
f(−2) when f(x) = x3 + 2x2 − 7
Same!

13. Factor Theorem:
The number c is a zero of P(x) if and only if (x−c) is a factor of P(x); that is, P(x) = Q(x) · (x − c) for some polynomial Q(x).
In other words, when using Synthetic division if
the remainder = 0, you found a factor that “goes in evenly”

14. Example 7:
Use the Factor Theorem to determine whether x + 2 is a factor of f(x) = 3x6 + 2x3 − 176.
YES!!!!!!!!!!