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This module covers the division of polynomials as rational expressions.

Simplifying, multiplying, dividing, adding, and subtracting rational expressions are all based on the basic skills of working with fractions. Dividing polynomials is based on an even earlier skill, one that pretty much everyone remembers with horror: long division.

To refresh your memory, try dividing 745 3 size 12{ { {"745"} over {3} } } {} by hand. You should end up with something that looks something like this:

So we conclude that 745 3 size 12{ { {"745"} over {3} } } {} is 248 with a remainder of 1; or, to put it another way, 745 3 = 248 1 3 size 12{ { {"745"} over {3} } ="248" { {1} over {3} } } {} .

You may have decided years ago that you could forget this skill, since calculators will do it for you. But now it comes roaring back, because here is a problem that your calculator will not solve for you: 6x 3 8x 2 + 4x 2 2x 4 size 12{ { {6x rSup { size 8{3} } - 8x rSup { size 8{2} } +4x - 2} over {2x - 4} } } {} . You can solve this problem in much the same way as the previous problem.

Polynomial division
6x 3 8x 2 + 4x 2 2x 4 size 12{ { {6x rSup { size 8{3} } - 8x rSup { size 8{2} } +4x - 2} over {2x - 4} } } {} The problem
The problem, written in standard long division form.
Why 3 x 2 ? This comes from the question: “How many times does 2 x go into 6 x 3 ?” Or, to put the same question another way: “What would I multiply 2 x by, in order to get 6 x 3 ?” This is comparable to the first step in our long division problem: “What do I multiply 3 by, to get 7?”
Now, multiply the 3 x 2 times the ( 2 x 4 ) and you get 6 x 3 12 x 2 . Then subtract this from the line above it. The 6 x 3 terms cancel—that shows we picked the right term above! Note that you have to be careful with signs here. –8 x 2 ( –12 x 2 ) gives us positive 4 x 2 .
Bring down the 4 x . We have now gone through all four steps of long division—divide, multiply, subtract, and bring down. At this point, the process begins again, with the question “How many times does 2 x go into 4 x 2 ?”
This is not the next step...this is what the process looks like after you’ve finished all the steps. You should try going through it yourself to make sure it ends up like this.
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So we conclude that 6x 3 8x 2 + 4x 2 2x 4 size 12{ { {6x rSup { size 8{3} } - 8x rSup { size 8{2} } +4x - 2} over {2x - 4} } } {} is 3x 2 + 2x + 6 size 12{3x rSup { size 8{2} } +2x+6} {} with a remainder of 22, or, to put it another way, 3x 2 + 2x + 6 + 22 2x 4 size 12{3x rSup { size 8{2} } +2x+6+ { {"22"} over {2x - 4} } } {} .

Checking your answers

As always, checking your answers is not just a matter of catching careless errors: it is a way of making sure that you know what you have come up with . There are two different ways to check the answer to a division problem, and both provide valuable insight

The first is by plugging in numbers. We have created an algebraic generalization:

6x 3 8x 2 + 4x 2 2x 4 = 3x 2 + 2x + 6 + 22 2x 4 size 12{ { {6x rSup { size 8{3} } - 8x rSup { size 8{2} } +4x - 2} over {2x - 4} } =3x rSup { size 8{2} } +2x+6+ { {"22"} over {2x - 4} } } {}

In order to be valid, this generalization must hold for x = 3 size 12{x=3} {} , x = 4 size 12{x= - 4} {} , x = 0 size 12{x=0} {} , x = ϖ size 12{x={}} {} ,or any other value except x = 2 size 12{x=2} {} (which is outside the domain). Let’s try x = 3 size 12{x=3} {} .

Checking the answer by plugging in x = 3 size 12{x=3} {}

6 ( 3 ) 3 8 ( 3 ) 2 + 4 ( 3 ) 2 2 ( 3 ) 4 = ? 3 3 2 + 2 3 + 6 + 22 2 3 4 size 12{ { {6 \( 3 \) rSup { size 8{3} } - 8 \( 3 \) rSup { size 8{2} } +4 \( 3 \) - 2} over {2 \( 3 \) - 4} } { {}={}} cSup { size 8{?} } 3 left (3 right ) rSup { size 8{2} } +2 left (3 right )+6+ { {"22"} over {2 left (3 right ) - 4} } } {}
162 72 + 12 2 6 4 = ? 27 + 6 + 6 + 22 6 4 size 12{ { {"162" - "72"+"12" - 2} over {6 - 4} } { {}={}} cSup { size 8{?} } "27"+6+6+ { {"22"} over {6 - 4} } } {}
100 2 = ? 39 + 22 2 size 12{ { {"100"} over {2} } { {}={}} cSup { size 8{?} } "39"+ { {"22"} over {2} } } {}
50 = ? 39 + 11 size 12{"50" { {}={}} cSup { size 8{?} } "39"+"11"} {}

The second method is by multiplying back. Remember what division is: it is the opposite of multiplication! If 745 3 size 12{ { {"745"} over {3} } } {} is 248 with a remainder of 1, that means that 248 3 size 12{"248" cdot 3} {} will be 745, with 1 left over. Similarly, if our long division was correct, then 3x 2 + 2x + 6 2x 4 + 22 size 12{ left (3x rSup { size 8{2} } +2x+6 right ) left (2x - 4 right )+"22"} {} should be 6x 3 8x 2 + 4x 2 size 12{6x rSup { size 8{3} } - 8x rSup { size 8{2} } +4x - 2} {} .

Checking the answer by multiplying back

3x 2 + 2x + 6 2x 4 + 22 size 12{ left (3x rSup { size 8{2} } +2x+6 right ) left (2x - 4 right )+"22"} {}
= 6x 3 12 x 2 + 4x 2 8x 24 + 22 size 12{ {}= left (6x rSup { size 8{3} } - "12"x rSup { size 8{2} } +4x rSup { size 8{2} } - 8x - "24" right )+"22"} {}
= 6x 3 8x 2 + 4x 2 size 12{ {}=6x rSup { size 8{3} } - 8x rSup { size 8{2} } +4x - 2} {}

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Source:  OpenStax, Advanced algebra ii: conceptual explanations. OpenStax CNX. May 04, 2010 Download for free at http://cnx.org/content/col10624/1.15
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