2.6: Polynomial Long Division
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Polynomial long division has many similarities to numerical long division, so it is important that we understand how and why numerical long division works the way it does before discussing polynomial long division. First the HOW?
Given the numerical problem 87,462÷38, the first step is to determine the highest place value in the answer.
Often the first step in numerical long division is to say "Does 38 divide into 8?" "No." "Does 38 divide into 87?" "Yes." This tells us that the first digit in the answer will be over the 7 in 87,462, and consequently will be in the thousands place. Once we know that the first digit in the answer will be in the thousands place, the next question is "How many thousands?" We can determine that 2∗ 38=76 but 3∗38=114 (too big), so we know that the first digit in the answer will be 2. Then we subtract, include the 4 and examine what is left over to continue.
Here, we see that 114÷38=3, so we know that the next digit in the answer will be 3
After including the 6, we can see that 38 does not divide evenly into 6, so we put a zero as the next digit in our answer and proceed:
Now that we have included all the digits from our original number 87,462, the last step is to divide 38 into 62. This goes one time with 24 left over.
So, now we have the solution to the original problem 87,462÷38=2,301R24 or
2,3012438
The WHY? of the long division algorithm is somewhat hidden by the HOW? In the first step, we are determining which place value will hold the first digit of our answer. When we determine that 38 does divide into 87, this indicates that the first digit in our answer will be the thousands place. Dividing 38 into 87 tells us how many thousands there will be. Then we subtract:
87,462−76,00011,462
Now we need to determine how many times 38 will divide into 11,462. We decided on 300∗38=11,400, then we subtract to see how much is left over:
11,462−11,40000,062
We can see that we won't need any tens in our answer, and that 38 divides into 62 one time with 24 left over, thus the answer is 2 thousands, 3 hundreds, no tens, 1 and a remainder of 24. To check the answer, we multiply 38∗2301 and add 24:
2,301×3818408690387438+2487462
Polynomial long division works in much the same way that numerical long division does. Given a problem A÷B, the goal is to find a quotient Q and remainder R so that A=B∗Q+R
Let's look at this with the example 2x4+7x3+4x2−2x−1÷x2+3x+1 or:
2x4+7x3+4x2−2x−1x2+3x+1
So, we are looking to answer the question:
A=B∗Q+R2x4+7x3+4x2−2x−1=(x2+3x+1)∗(?+?+?)+?
If we want to mulitply x2+3x+1 times something and end up with 2x4+7x3+ 4x2−2x−1, then what we multiply by is going to have to start with 2x2, because
x2∗2x2=2x4
Now we're working with this:
A=B∗Q+R2x4+7x3+4x2−2x−1=(x2+3x+1)∗(2x2+?+?)+?
But the 2x2 doesn't just get multiplied by the x2, it will also get multiplied by the 3x and the 1. So now we have:
A=B∗Q+R2x4+7x3+4x2−2x−1=(x2+3x+1)∗(2x2+?+?)+?=2x4+6x3+2x2+?????
The issue this raises is that the next multiplication (?∗x2+?∗3x+?∗1) needs to add only 1x3 to our answer, because we need 7x3 and we already have 6x3 from the previous multiplication. That means we're going to want to multiply next by
1x:
A=B∗Q+R2x4+7x3+4x2−2x−1=(x2+3x+1)∗(2x2+1x+?)+?=2x4+6x3+2x2=1x3+3x2+x=2x4+7x3+5x2+1x+???
In the next round of mutiplication, we're going to want to bring the 5x2 down to 4x2, so we'll need to multiply by -1
A=B∗Q+R2x4+7x3+4x2−2x−1=(x2+3x+1)∗(2x2+1x−1)+0=2x4+6x3+2x2=1x3+3x2+x=−1x2−3x−1=2x4+7x3+5x2−2x−1
Now we also know that the remainder is zero, because x2+3x+1 divides evenly into 2x4+7x3+4x2−2x−1 and so:
2x4+7x3+4x2−2x−1=(x2+3x+1)∗(2x2+1x−1)
This method makes the reasoning behind dividing polynomials somewhat more apparent than the long division process, but it is more cumbersome. The way that polynomial long division is usually approached is as follows:
Then, just as we did in the other method, we question "What should we multiply x2 by to get 2x4?′′ Answer: ′′2x2′′ This is the first term in our answer:
Then we multiply 2x2(x2+3x+1) and change all the signs to see what we'll be left with:
This indicates that we'll have the 2x4 we'll need in our answer, as well six of the seven x3 's and two of the four x2 's. We'll now need 1x3 next:
This means we'll need to multiply by 1x :
Here, we still need to pick up a −1x2, which means that our next multiplication will be with -1:
Because x2+3x+1 divides evenly into 2x4+7x3+4x2−2x−1 we have a zero remainder. In the next example there will be a non-zero remainder:
Example
Divide:
3x4−8x3+19x2−15x+10x2−x+4
First, we set up the problem:
Then, we question: "What do we need to multiply x2 by to get 3x4?" Answer:
Then, we multiply, change signs (subtract) and combine like terms:
Now we'll need to multiply by −5x, change signs and combine like terms:
We need 2x2 so we'll need to multiply by 2 , change signs and combine like terms:
Because there is no positive power of x that we can multiply x2 by to get 7x, then this is our remainder: 7x+2
So:
A=B∗Q+R3x4−8x3+19x2−15x+10=(x2−x+4)∗(3x2−5x+2)+(7x+2)
Exercises 2.6
Find the quotient in each problem.
1) y3−4y2+6y−4y−2
2) x3−5x2+x+15x−3
3) x3−4x2−3x−10x2+x+2
4) 2x3−3x2+7x−3x2−x+3
5) x4+2x3−x2+x+6x+2
6) x4+x3+5x2+3x+6x2+x−1
7) 2z3+5z+8z+1
8) x5+3x+2x3+2x+1
9) x4+2x3+4x2+3x+2x2+x+2
10) 2x4+3x3+3x2−5x−32x2−x−1
11) 2y5−3y4−y2+y+4y2+1
12) 3x5−4x3+3x2+12x−10x2+2x−1
13) 5x4−3x2+2x2−3x+5
14) 3y3−4y2−3y2+5y+2