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Dividing Polynomials using Long Division - Concept
When dividing polynomials, we can use either long division or synthetic division to arrive at an answer. Using long division, dividing polynomials is easy. We simply write the fraction in long division form by putting the divisor outside of the bracket and the divided inside the bracket. After the polynomial division is set up, we follow the same process as long division with numbers.
Dividing polynomials with long division. So we're going to talk about how we can divide these 2 polynomials using long division. Before we do that, I just want to profess it by you know drawing a parallel to something we already know how to do. Just say we're asked to divide 1,170 by 70, okay? What we end up doing is we draw our 7 in a little bracket with a number on the inside. We then start and say, okay. How many times does 7 go into 1, it doesn't so we go to the next [IB] How many times does 7 go into 11? Draw the 1, 1 times 7 is 7 and then we subtract dropping down the next term 47, continue from there. How many times does 7 go into 47, it's 6, 6 times 7 is 42,. Subtract again, leaving us 50 [IB] and repeat, we need 7, so this is in 49 and we're left with a remainder of 1. Okay.
So we know how to divide numbers. Okay? Dividing polynomials is exactly the same idea just with a little bit more x's and variables and terms involved. So what we're going to end up doing is exactly the same thing as this, okay? So x+1 goes on the outside and then our, remember this is called the dividend our dividend goes inside the bracket. Have a big bracket. 3x cubed and I think I forgot the square of my problem, let's throw that in there. 3x squared minus sorry 3x cubed minus 4x squared plus 2x minus 1, okay. So we basically are rewriting our problem as a fraction in long division. In logic it's exactly the same as we did right here, okay?
You say the first thing you want to get rid of is the 3x squared and we looked at our leading term on our divisor, the term in the bottom, okay? In order to get 3x squared from x we need to multiply it, sorry 3x cubed from x we need to multiply it by 3x squared. So I always try to line up my powers. So I'm going to put 3x squared right there. So I have all my squares in a column right here. Okay? Then like what we did over here we multiply the number on top by the number in front. The only difference now is we have some variables the same idea holds. Okay? So 3x squared times x, is 3x cubed. 3x squared times 1 is 3x squared. Okay? Just like we did over here with our numbers, we need to subtract, okay? So subtraction. 3x cubed minus 3x cubed, those cancel which is what we wanted and the -4x squared minus 3x, make sure you distribute that negative sign becomes negative 7x squared.
The next term we want to get rid of is -7x squared. If you want to bring this down, you can, you don't have to. Just remember that we need to include it next step when we subtract. Okay.
So we need to get rid of the -7x squared with our leading term of an x. So we need to multiply our x by -7x in order to cancel it out. -7x and then we just want to multiply and subtract once again. So -7x times x, -7x squared. -7x times 1 is -7x. Once again we want to subtract making sure we distribute that negative sign through. -7x squared minus -7x squared, those cancel which is what we wanted and then we still have these two up here. So it's 2x minus -7x, 2x plus 7x which turns into 9x and again we can bring this one down if we want, we don't need to. Just remember it's there. So this 9x is the last thing we need to get rid of. In order to get 9x from an x we need a 9 plus 9, that 9 gets distributed through. 9x plus 9 and once again we subtract, okay? 9x-9x. Those cancel. -1-9 plus a negative 9 so this is going to give us -10, which has a remainder [IB] we multiply x by will give us -10. Okay?
So there's different ways you can write this up. I'm going to take a second and come back over here and show us what this actually means. So, this was our process for numbers same exact process over there. Well, what we just really found out was that this statement is actually equal to our quotient. What comes up is the number that is above the sign, so if we come back over here. It's just a number up here. So that is 3x squared minus 7x plus 9 and then plus our remainder over our divisor. So this is then going to be plus -10 over our divisor here, x+1, okay? So long division is a little bit more in dealing with numbers but the process is exactly the same.