Polynomial Long DivisionDate: 03/17/2004 at 18:01:48 From: Angela Subject: Long divison!! Use long divion to divide (2x - 3) into 4x^4 - x^2 - 2x + 1. I really need help in doing this. Date: 03/18/2004 at 11:03:29 From: Doctor Roberts Subject: Re: Long divison!! Dear Angela, There are three things that tend to give students problems when it comes to polynomial long division. I'll address all three, but if your confusion is about something else, let me know. First, when you write the expressions down to divide them, make sure to include _every_ power of x. If there is a power missing in the given problem, fill it in with a 0 to hold that place. For example, if you were dividing x - 1 into 2x^3 + x - 5, you'd write it like this: ______________________ x - 1 ) 2x^3 + 0x^2 + 1x - 5 Adding the zero term keeps the columns in order, as we will see when we do the problem shortly. Also, since the answer may include the second power of x, this provides a column to write that part of the answer. And note that nothing really changed about the polynomial, since zero times x^2 is zero. Adding zero terms is sometimes needed in the divisor as well. For example, if you had been dividing by x^2 - 1 instead of x - 1, you would write the divisor as x^2 + 0x - 1. OK, let's start doing this division, and I'll point out the other two common problems as we come to them. To start the division, look only at the first term of the divisor and the first term of the dividend, so that gives us 'x' and '2x^3'. What times 'x' will make it 2x^3? Or, maybe you prefer to think about what you would get if you divide 2x^3 by x. Either way, the answer is 2x^2. So that's the first term of the quotient, or answer. Let's put it on top in the x^2 column (good thing we added that 0x^2!): 2x^2 ______________________ x - 1 ) 2x^3 + 0x^2 + 1x - 5 Now multiply the 2x^2 times the (x - 1) and write the answers in the correct columns: 2x^2 ______________________ x - 1 ) 2x^3 + 0x^2 + 1x - 5 2x^3 - 2x^2 As with regular long division, once you have determined how many times the divisor goes in, written that number up top, and multiplied, the next step is to subtract. And this is the second common mistake that people make with polynomial long division. Remember that you are subtracting the whole expression of 2x^3 - 2x^2. The subtraction will change the signs on _both_ of those terms since -(2x^3 - 2x^2) is equal to -2x^3 + 2x^2. So what we get when we subtract is: 2x^2 ______________________ x - 1 ) 2x^3 + 0x^2 + 1x - 5 -2x^3 + 2x^2 ----------- 0 + 2x^2 Can you see why we got +2x^2? What you really had for the subtraction was 0x^2 - (-2x^2) which became 0x^2 + 2x^2 which is just 2x^2. Some people find it helpful to immediately change the signs after they multiply and then just add, while others prefer to do the subtraction and just remember that they are subtracting and think about what happens if you subtract a negative. Whatever makes the most sense to you is fine, but in my experience this is the place where most students make their mistakes in the long division process. Continuing, we can "bring down" the next term, which would be the 1x, and then proceed the same way as the first cycle. What's 2x^2 divided by x? Sounds like 2x to me! Put that up top in the x column and multiply it by x - 1, then subtract: 2x^2 + 2x ______________________ x - 1 ) 2x^3 + 0x^2 + 1x - 5 -2x^3 + 2x^2 ----------- 2x^2 + 1x 2x^2 - 2x --------- 3x Note that this time I did not change the signs and add, I just remembered that when I subract I have 1x - (-2x) which is 1x + 2x or 3x. Now bring down the -5 and continue by dividing 3x by x to get 3: 2x^2 + 2x + 3 ______________________ x - 1 ) 2x^3 + 0x^2 + 1x - 5 -2x^3 + 2x^2 ----------- 2x^2 + 1x 2x^2 - 2x --------- 3x - 5 3x - 3 ------ - 2 Again, this time I had -5 - (-3) which is -5 + 3 or -2. Since there is nothing left to "bring down", and because the degree of the remainder (there is no x in it, so it's degree zero) is less than the degree of the divisor (with the x^1 it's degree one) we're finished and the -2 is the remainder. This brings us to the third idea that sometimes causes problems--what to do with the remainder. To figure what to do, let's look at how remainders work with regular division. __ Say we want to divide 3 )5. The answer is 1 rem. 2, since 3 goes into 5 one time with 2 left over. Another way to write this is __ 3 )5 = 1 + 2/3. The remainder is divided by the divisor and added to the answer. Checking the answer, 3(1 + 2/3) = 3 + 6/3 = 3 + 2 = 5. So, in synthetic division, if there's a remainder, just write it as a fraction, with the remainder as the numerator and the divisor as the denominator, and add it to the rest of the answer. If it's negative, as in our example, you can write it one of two ways: -2 2 2x^2 + 2x + 3 + ----- OR 2x^2 + 2x + 3 - ----- x - 1 x - 1 Note that the two choices are either to write it as adding a negative (on the left) or subtracting a positive (on the right). Either is fine, though the right hand one is slightly more elegant. So, the three things to keep in mind for long division are: 1. Write the divisor and dividend in descending order of powers of the variable and fill in zero terms where needed. 2. Be careful in the subtraction step, especially when subtracting a negative. 3. If you have a reminder, write it as a fraction by putting it over the divisor. Can you take a shot at your problem now? If you get stuck, write back and show me what you've done and I'll give you some help. - Doctor Roberts, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/