How To Do Polynominal Division
At the end of the equation what must you watch out for?
Hi, I'm Dr. Shah. I was the National Lecture Competition winner in 1999, and I'm the Math Master at Math School. Now, ready for a new way of doing math? I've set up a division that I want to find out already, which is this cubic. And I want to divide it by this linear divisor. So I start by putting my cubic inside the "divided by" sign and my divisor outside. Now what I'm going to do, first of all, is ignore all of these and I'm going to ignore that. And I'm just looking at that "X" and that "X-cubed", so I'm just looking at these two. And I'm asking myself, "What do I have to multiply 'X' by to get 'X-cubed'?" And the answer, of course, is "X-squared". So I put "X-squared" on the top, there, and the next thing that I do is multiply that. So I do "X-squared" times "X", which is "X-cubed". And I do "X-cubed" times two, which is "two X squared". And then, once I've done that, I subtract: "X-cubed" minus "X-cubed" is going to give me zero; "five X squared" minus "two X squared" is going to give me "three X squared". Bring down the next term, "plus four X" and then, again, do exactly the same thing - so the procedure repeats itself now. Look at the first one of these and the first one of those. "What do I have to multiply 'X' by to get 'three X squared'?" The answer, of course, would be "three X". So I write "plus three X" on that and then I multiply this out into that. So "three X" times "X" is going to give me "three X squared" and "three X" times two is going to give me "plus six X". And again, put a line underneath that and I'm going to subtract them. Again, same as before, the first terms are going to just cancel out. Now I have "four X" minus "six X". Now "four X" minus "six X" is "minus 2 X". Bring down the next term which is also the last term, and then repeat the procedure all over again. Looking at the front two, here, "What do I have to multiply 'X' by to get 'minus two X'?" The answer is, "minus two". So I put my "minus two" on the top there and, again, I'm going to multiply that "minus two" with both of those. So "minus two" times "X" is going to give me "minus two X", and "minus two" times two is going to give me "minus four". And subtract, yet again. Again, the first two will cancel out, so all I need to do is "one subtract minus four - watch out for the double negative here. "One subtract minus four", which would be five. There are no more numbers to bring down here and, so, I have to stop here. And that's going to be called my remainder and that, at the top here, is going to be called my quotient. Now, because I had a remainder (the remainder wasn't zero), that means that "X plus 2" is not a factor of "X cubed plus 5 X squared plus four X plus one". When you do try and divide it, you get a remainder left of five. So, that's how to do polynomial division and, once you get good at that, a speedier way of doing the same thing is to use synthetic division. .