1. Dividing Polynomials

2. Warm Up
Without a calculator, divide the following
Solution: 49251

3. This long division technique can also be used to divide polynomials

4. Example: Divide & Check
Example: Divide x2 + 3x – 2 by x – 1 and check the answer. x 1.
+ 2 2.
x x 3. 2x
– 2
2x + 2 4.
– 4 5.
remainder 6.
Answer: x + 2 +
– 4
Check:
(x + 2)
quotient
(x + 1)
divisor
+ (– 4)
remainder
= x2 + 3x – 2
dividend
correct
Example: Divide & Check

5. POLYNOMIALS – DIVIDING EX – Long division
(5x³ -13x² +10x -8) / (x-2)
5x²
- 3x
+ 4
R 0
x - 2
5x³ x² x - 8 - (
5x³ - 10x² )
-3x²
+ 10x - (
-3x² x ) 4x
- 8 - (
4x - 8 )

6. Let’s Try One
(2x² -19x + 8) / (x-8)
x - 8
2x² x + 8

7. Example: Divide & Check
Example: Divide 4x + 2x3 – 1 by 2x – 2 and check the answer. x2
+ x
+ 3
Write the terms of the dividend in descending order.
2x3 – 2x2
Since there is no x2 term in the dividend, add 0x2 as a placeholder.
2x2
+ 4x
2x2 – 2x 1. 2. 6x
– 1 4. 3.
6x – 6 5 5.
Answer: x2 + x + 3 5 6. 7. 8.
Check: (x2 + x + 3)(2x – 2) = 4x + 2x3 – 1 9.
Example: Divide & Check

8. Example: Division With Zero Remainder
Example: Divide x2 – 5x + 6 by x – 2. x
– 3
x2 – 2x
– 3x
+ 6
– 3x + 6
Answer: x – 3 with no remainder.
Check: (x – 2)(x – 3)
= x2 – 5x + 6
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Example: Division With Zero Remainder

9. A Couple of Notes
To test so see if a binomial is a factor, you want to see if you get a remainder of zero. If yes, it is a factor. If you get a remainder, the answer is no.

10. From this example, x-8 IS a factor because the remainder is zero
(2x² -19x + 8) / (x-8)
x - 8
2x² x + 8

11. In this case, x-3 is not a factor because there was a remainder of 6
+ 6
R 6
x - 3
x² + 3x - (
x² - 3x ) 6x
-12 - ( 6x ) 6