Multiplying and Dividing Polynomials Chapter 5 Sections 5.4-5.7.
Dividing polynomials
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Transcript of Dividing polynomials
Dividing Polynomials
Warm Up• Without a
calculator, divide the following
113277323
Solution: 49251
This long division technique can also be used to divide polynomials
POLYNOMIALS – DIVIDINGEX – Long division
• (5x³ -13x² +10x -8) / (x-2)
5x³ - 13x² + 10x - 8x - 25x²
5x³ - 10x²-( )-3x²+ 10x
- 3x
-3x² + 6x- ( )4x - 84x - 8- ( )
+ 4
0
R 0
So in other words…
5x³ -13x² +10x - 8x-2
= 5x² -3x + 4
(5x² -3x + 4)(x-2) = 5x³ -13x² +10x - 8
OR
POLYNOMIALS – DIVIDINGEX#2 – Long division
• (x² +3x -12) / (x+6)
x² + 3x - 12x - 3
x
x² - 3x- ( )
6x -12
+ 6
6x - 18- ( )
6
R 6
• (2x² -19x + 8) / (x-8)
2x² - 19x + 8x - 8
Let’s Try One
• (2x² -19x + 8) / (x-8)
2x² - 19x + 8x - 8
Let’s Try One
EX – Synthetic division(5x³ -13x² +10x -8) / (x-2)
2 5 -13 10 -8
5
10
-3
-6
4
8
0
5x² -3x + 4
Oppositeof number indivisor
EX – Synthetic division(3x³ -4x² +2x -1) / (x+1)
-1 3 -4 2 -1
3
-3
-7
+7
9
-9
-10
3x2 -7x + 9 R-10
Oppositeof number indivisor
Let’s Try One(x³ -13x +12) / (x+4)
EX – Synthetic division(x³ -13x +12) / (x+4)Opposite
of number indivisor
A Couple of Notes• Use synthetic division when the coefficient in
front of x is 1(x- 2) (2x-3)122 xx 122 xx
YES NO
• 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.
• (2x² -19x + 8) / (x-8)
2x² - 19x + 8x - 8
From this example, x-8 IS a factor because the remainder is zero
In this case, x-3 is not a factor because there was a remainder of 6
x² + 3x - 12x - 3
x
x² - 3x- ( )
6x -12
+ 6
6x - 18- ( )
6
R 6
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