1. 5.5 Apply the Remainder and Factor Theorems
Algebra II

2. When you divide a Polynomial f(x) by a divisor d(x), you get a quotient polynomial q(x) with a remainder r(x) written: f(x) = q(x) + r(x) d(x) d(x)

3. The degree of the remainder must be less than the degree of the divisor!

4. Polynomial Long Division:
You write the division problem in the same format you would use for numbers. If a term is missing in standard form …fill it in with a 0 coefficient.
Example:
2x4 + 3x3 + 5x – 1 =
x2 – 2x + 2

5. 2x2
2x4 = 2x2 x2

6. -( ) 2x2 +7x +10 2x4 -4x3 +4x2 7x3 - 4x2 +5x -( ) 7x3 - 14x2 +14x
-( )
2x4
-4x3
+4x2
7x3
- 4x2
+5x
-( )
7x x2 +14x
10x2 - 9x -1
7x3 = 7x x2
-( )
10x2 - 20x +20
11x - 21
remainder

7. The answer is written: 2x2 + 7x + 10 + 11x – 21 x2 – 2x + 2
Quotient + Remainder over divisor

8. Now you try one! y4 + 2y2 – y + 5 = y2 – y + 1
Answer: y2 + y y2 – y + 1

9. Remainder Theorem:
If a polynomial f(x) is divisible by (x – k), then the remainder is r = f(k).
Now you will use synthetic division (like synthetic substitution)
f(x)= 3x3 – 2x2 + 2x – 5
Divide by x - 2

10. f(x)= 3x3 – 2x2 + 2x – 5 Divide by x - 2
Long division results in ?......
3x2 + 4x x – 2
Synthetic Division:
f(2) = 6 8 20 3 4 10 15
Which gives you:
+ 15
x-2
3x2
+ 4x
+ 10

11. Synthetic Division Practice 1
Divide x3 + 2x2 – 6x -9 by (a) x-2 (b) x+3
(a) x-2 8 2 4 1 4 2 -5
Which is x2 + 4x x-2

12. Synthetic Division Practice cont.
(b) x+3 3 -3 9 1 -1 -3
x2 – x - 3

13. 5-5B Factor Theorem: A polynomial f(x) has factor x-k iff f(k)=0
note that k is a ZERO of the function because f(k)=0

14. Factoring a polynomial
Factor f(x) = 2x3 + 11x2 + 18x + 9
Given x + 3
Since f(-3)=0
x-(-3) or x+3 is a factor
So use synthetic division to find the others!!

15. Factoring a polynomial cont. -3
-15 -9 -6 2 5 3
So…. 2x3 + 11x2 + 18x + 9 factors to:
(x + 3)(2x2 + 5x + 3)
Now keep factoring gives you:
(x+3)(2x+3)(x+1)

16. Ex. 3 ) Your turn! Factor f(x)= 3x3 + 13x2 + 2x -8 given x+4

17. Finding the zeros of a polynomial function
f(x) = x3 – 2x2 – 9x +18.
One zero of f(x) is x=2
Find the others!
Use synthetic div. to reduce the degree of the polynomial function and factor completely.
(x-2)(x2-9) = (x-2)(x+3)(x-3)
Therefore, the zeros are x=2,3,-3!!!

18. Your turn! f(x) = x3 + 6x2 + 3x -10 X=-5 is one zero, find the others!
The zeros are x=-2,1,-5
Because the factors are (x+2)(x-1)(x+5)

19. Assignment