1. 6.5 The Remainder and Factor Theorems
OBJ: use long division to divide polynomials
Do Now:
Use long division to divide. Round to the nearest hundredth.
1) divided by 36 2) 12 divided by 5

2. Polynomial Long Division
If a term is missing in standard form, fill it in with a 0 coefficient.
Ex 1:
2x4 + 3x3 + 5x – 1
x2 – 2x + 2
2x2
2x4 = 2x2 x2

3. -( ) 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

4. The answer is written: Quotient + Remainder over Divisor
2x2 + 7x x – x2 – 2x + 2
Ex 2:
y4 + 2y2 – y y2 – y + 1
y2 + y y2 – y + 1

5. Remainder Theorem:
If a polynomial is divisible by (x – k), then the remainder is f(k)
ONLY when you divide by x-k,
check if the remainder is f(k) by directly substituting k back into the eqn!!!
On your homework, this is possible on problems #6 and # 7
HW:

6. Divide f(x)= 3x3 – 2x2 + 2x – 5 by x - 2
continued…
Do Now:
Use long division to divide f(x)= 3x3 – 2x2 + 2x – 5 by x - 2
3x2 + 4x x – 2
Synthetic Division
(like synthetic substitution)
Ex 3:
Divide f(x)= 3x3 – 2x2 + 2x – 5 by x - 2

7. f(x)= 3x3 – 2x2 + 2x – 5 Divide by x - 2 6 8 20 3 4 10
R: 15
+ 15
x-2
3x2
+ 4x
+ 10

8. Divide x3 + 2x2 – 6x -9 by x-2 1 2 -6 -9 2 Ex 4: 2 8 4 1 4 2 R:-5 2 8 4 1 4 2
R:-5
x2 + 4x x-2

9. Divide x3 + 2x2 – 6x -9 by x+3 1 2 -6 -9 -3 Ex 5: -3 3 9 1 -1 -3 R: 0 -3 -3 3 9 1 -1 -3
R: 0
x2 – x - 3

10. Factor Theorem: A polynomial has the factor x-k if f(k)=0
Factor Theorem: A polynomial has the factor x-k if f(k)=0 ** if the remainder is zero, x-k is a factor since it divides evenly ** it also means that k is a “zero” (a solution) of the function!
HW:

11. Factoring a polynomial
continued…
Factoring a polynomial
Do Now:
Use synthetic division to divide f(x)= x4 -6x3 -40x +33 by x - 7
x3+ x2 + 7x
x – 7
Factor f(x) = 2x3 + 11x2 + 18x + 9
Given f(-3)=0
*Since the remainder is zero, x – k is a factor!
x-(-3) or x+3
*So use synthetic division to find the others!!
Ex 6:

12. -3
-15 -9 -6 2 5 3
R: 0
(x + 3)(2x2 + 5x + 3)
*keep factoring (bustin ‘da ‘b’) gives you:
(x+3)(2x+3)(x+1)

13. Factor f(x)= 3x3 + 13x2 + 2x -8 Given f(-4)=0 (x + 1)(3x – 2)(x + 4)
Ex 7:
Factor f(x)= 3x3 + 13x2 + 2x -8
Given f(-4)=0
(x + 1)(3x – 2)(x + 4)

14. Finding the zeros of a polynomial function
Ex 8:
One zero of f(x) = x3 – 2x2 – 9x +18 is x=2. Find the others.
*if x=2 is a zero, then x-2 is a factor
*Use synthetic division to factor completely.
*then solve
(x-2)(x2-9)
(x-2)(x+3)(x-3)
x=2 x=-3 x=3

15. One zero of f(x) = x3 + 6x2 + 3x -10 is x=-5. Find the others.
Ex 9:
One zero of f(x) = x3 + 6x2 + 3x -10 is x=-5. Find the others.
(x-2)(x+1)(x+5)
x=2 x=-1 x=-5
HW: