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Dividing Polynomials using Synthetic Division - Concept
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When **dividing polynomials**, we can use either long division or synthetic division to arrive at an answer. We can use synthetic division to divide polynomials if the degree in the divisor is equal to 1 and if the coefficient of the variable in the divisor is equal to 1. After we do this, we can write the coefficients of the divisor and dividend and use synthetic division to determine the answer.

Using synthetic division to divide polynomials. Synthetic division is a special way that we can divide polynomials and we can only use it if a couple factors are occuring. So first and foremost we have to have our we look at our divisor, okay? The degree of our divisor has to be 1 and what that means is that we can only have a single x down there. There can't be any x squareds, x cubed anything like that.

The other rule that we need is there can't be a coefficient in front of the x. So the coefficient has to be 1, okay? So really what you need is your divisor, the denominator to be able to form x plus or minus a number, okay?

Once you have those restrictions we could then use synthetic division. For this particular example we could actually simplify this even easier by using factoring. So let's just take a look at that, see what our answer should be and then try synthetic division. So looking at this we know that our numerator can factor. We look and we say x+4, x-2 over x-2. Our x-2's cancel and we're left with x+4. Okay. That's easy enough and that actually will be easier than synthetic division but I just want to show you that synthetic division actually works and we can compare our answers when we're done.

How synthetic division operates is what you need to do is look at your divisor, look at your denominator and figure out what number will make that zero, okay. So here we have x-2. We want this to be equal to zero, we'll have to plug in 2. Take that number and write it down and then make a sort of upside down bracket. Okay? From there in the top of this bracket, we want to take the coefficients from our dividend, the new number in the numerator. So our first coefficient is 1, our second coefficient is 2 and our last coefficient is -8. We need every single power of x to be accounted for, so if we're missing something, if this 2x wasn't here, we would still need a place over there so we need to put in a zero just to make sure it was accounted for. Okay? Then what we do is from this bracket take your first number and just drop it down, okay? So we just drop it down. From here take the number on the outside, the number that made the zero, multiply by the number we just wrote down. So 2 times 1 is 2 and that goes here. So that will just go up once you multiply it. Then we add, okay? 2+2 is 4 and then repeat. 2 times 4, this becomes 8, -8+8 is zero.

That last spot represents our remainder. And then going up from our remainder, the ext term over is our constant term and then we increase on our x's. So this is our constant term so that's just going to represent the number 4, then going up this will be 1x or just x. We're adding and so we divide this out using synthetic division. Okay?

For this particular example factoring would probably be easier but hopefully in doing this we saw how synthetic division works and how we can apply it to more complicated problems.

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