WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to take care of dividing polynomials.
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I like to break this down into two different parts.
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First, we will look at the division process when you have a polynomial divided by a monomial.
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We will look at what happens when you divide any two polynomials together.
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In the very end, I will also show you a very special technique to make the division process nice and clean.
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To get into the basics of understanding the division of polynomials, I like to take it back to looking at the division process for real-world fractions.
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Suppose that when you were adding fractions together you know how that process will go.
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One very important thing that you do when adding fractions is you would find a common denominator.
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Once you have a common denominator, then you can go ahead and just combine the tops of those fractions together.
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Think of a quick example like 2/3 and are looking to add 5/3.
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Since they have exactly the same bottoms then you will only add the 2 and 5 together and get 7/3 as your result.
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Since this has a giant equal sign in between it, it means you can also follow this process in the other direction.
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This may look a little unfamiliar, but it does work out if you go the other way.
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Suppose I had A + B and all of that was dividing by C.
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The way I could look at this is that both the A and B are being divided by C separately.
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This is the same equation I had earlier, I just turned it around.
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The reason why I show this is this will help us understand what happens when we have a binomial divided by a monomial.
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In fact, that is the example that I have written out.
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The top is a binomial and the bottom is an example of a monomial.
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You can see that the way we handle it is we split up that monomial under each of the different parts of the polynomial in the top.
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Let us see this process with numbers and see what polynomials and you will get an idea of how this works.
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Working the other direction, if I see (2 + 5) ÷ 3, I want to visualize that as the 2 ÷ 3 and also the 5 ÷ 3.
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For our polynomials if I have something like (x + 3z) ÷ 2y, then I will put that 2y under both of the parts.
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In addition, you will notice splitting up over both of the parts in the top, you always want to make sure that you simplify further, if possible.
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This means if you use your quotient rule for exponents, go ahead and do reduce those powers as much as possible.
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If you have just any two general polynomials, then you want to think of how the process works with numbers.
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In fact we are going to go over a long division process so that we can actually keep track of all the parts of what goes into what.
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To make this process easier, remember to write your polynomial in descending power.
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Start with the largest power and write it all the way down to the smallest power.
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An additional thing that will also help in the division process is to make sure you put placeholders for all the missing variables.
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If I’m looking at a polynomial like 5x² + 1, then I will end up writing it with a placeholder for the missing x.
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It is not missing but it will help me keep track of where just my x terms go.
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I'm going to show you this process a little bit later on
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so you can see how it works with numbers and make some good parallels too doing this with polynomials.
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Let us look at the division process for numbers.
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Suppose I gave you 8494 and I told you to divide it by 3 and furthermore I said okay, let us see if you can do this by hand.
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Some good news that you will probably tell me is that you do not have to take the 3 into that entire number all at once.
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No, you will just take the 3 into 8494 bit by bit.
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In fact, the first thing that you will look at is how many times this 3 go into the number 8.
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That will be the only thing you are worried about.
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How many times does the 3 go into 8?
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It goes in there twice and you will write that number on the top.
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Now after you have that number on the top, you do not leave it up there you go through a multiplication process,
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2 × 3 and you will write the result right underneath the 8.
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With the new number on the bottom, you will go ahead and subtract it away, so 8-6 would give you a 2.
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That would be like one step of the whole division process.
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You would continue on with the division process by bringing down more terms and doing the process again.
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At this next stage, we will say okay how many times this 3 go into 24?
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It goes in there 8 times.
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Then we could multiply the 8 and 3 together and get a new number for that 3 × 8 = 24.
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We can go ahead and subtract those away.
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We will not stop there, keep bringing down our other terms and see how many times 3 goes into 9.
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Write it onto the top and multiply it by your number out front.
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Subtract it away and continue the process until you have exhausted the number you are trying to divide.
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Let us see.
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3 goes into 4, it looks like it goes in there once.
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I will get 3, subtract them away and I have remainder of 1.
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That is a lengthy process but notice the Q components in there.
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You are only dividing the number bit by bit.
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You do not have to take care of it all at once.
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The way you take care of it is you are saying how many times 3 goes into that leading number.
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You write it on the top, you go through the multiplication process and then you subtract it away from the number.
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You will see all of those same components when we get into polynomials.
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We have our answer and I could say it in many different ways but I’m going to write it out.
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8494 if we divide this by 3 is equal to 2831 and it has a remainder down here of 1.
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We could say +1 and still being divided by 3.
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We have many different parts in here that are flying around, and you want to keep track of what these parts are.
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The part underneath your division bar is the dividend.
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What you are throwing in there this is your divisor.
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Your answer would be your quotient.
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This guy down here is our remainder.
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Notice how those same parts actually show up in our answer.
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Dividend, divisor, quotient, and remainder.
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We put the remainder over the divisor because it is still being divided.
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Now that we brushed up on the process with numbers, let us take a look at how we do this with polynomials.
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I want to divide 2x² + 10x + 12 and divide that by x +3.
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The good news is we do not have to take care of the entire polynomial all at once.
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We are going to take it in bits and pieces.
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We will first going to look at x and see how many times it will go into 2x².
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In fact one thing that I can do to help out the process is think to myself what would I have to multiply x by in order to get a 2x².
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1 × what would equal to 2? That would have to be 2.
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What would I have to multiply x by to get an x²?
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I have to multiply it by x.
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I will put that on top, just like I did with numbers.
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After I do that, we will run through a multiplication process.
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We will take the 2x to multiply it by x, I will multiply it by 3.
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We will record this new polynomial right underneath the other one.
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2x × x = 2x².
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2x × 3 = 6x.
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Now comes a very important step.
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Now that we have this new one, we want to subtract it away from the original.
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Notice how I put those parentheses on there, that will help me remember that I need to subtract away both parts and keep my signs straight.
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Starting over here, I have 10x – 6x = 4x.
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Then I have 2x² – 2x², that will cancel out and will give me 0x².
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If you do this process correctly, these should always cancel out.
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If they do not cancel out, it means we need to choose a new number up here.
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That is just one step of the division process, let us bring down our other terms and try this one more time.
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I want to figure out how many does x goes into 4x?
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What would I have to multiply x by in order to get 4x?
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I think I have to multiply it by 4 that is the only way it is going to work out.
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Now we have the 4, go ahead and multiply it by the numbers out front.
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4 × x = 4x and 4 × 3 = 12.
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Once you have them, put on a giant pair of parenthesis and we will go ahead and subtract it away.
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12 – 12 =0 and 4x – 4x = 0.
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What that shows is that there is no remainder and that it went evenly.
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Let us write this out.
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When I had 2x² + 10x + 12 and I divided it by x + 3, the result was 2x + 4.
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There was no remainder.
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We can label these parts as well.
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We have our dividend, divisor, quotient, and if I did have a remainder I will probably put it out here.
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Here is our dividend.
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Here is our divisor and our quotient.
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Now that we know a lot more about dividing polynomials, let us look at a bunch of examples.
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Some of them will take a polynomial divided by a monomial.
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Some of them will take two polynomials and divide them.
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We will approach both of those cases in two different ways.
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Example 1, divide a polynomial by a monomial.
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I will take (50m⁴ -30m³ + 20m) ÷ 10m³.
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Since we are dividing by a monomial, I will take each of my terms and put them over what I’m dividing them by.
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50m⁴ ÷ 10m³, 30m³ ÷10m³ and 20m ÷ 10m³.
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Now that I have done that, I will go through and simplify these one at a time.
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50 ÷ 10 = 5, m⁴ ÷ m³ = I can use my quotient rule and simply subtract the exponents and get m¹.
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Continuing on 30 ÷ 10 = 3, if I subtract my exponents for m³ and m³, I have m⁰.
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Onto the last one, 20m ÷ 10m³, there is 2.
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Let us see, if I subtract the exponents I will m⁻².
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Then I can go through and just clean this up a little bit.
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5m – anything to the 0 power is 1, 1 × 3 + and I will write this using positive exponents, 2/m².
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This will be the final result of dividing my polynomial by a monomial.
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Let us try this scheme with something a little bit more complicated.
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This one is (45x⁴ y³ + 30x³ y² - 60x² y) ÷ 50x² y.
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We have to put our thinking caps for this one.
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We will start off by taking all of our terms in the top polynomial and putting them over our monomial, only one term.
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Once we have this all written out we simply have to simplify them one at a time.
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Let us start at the very beginning.
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15 goes into 45 3 times, now I will simplify each of my variables using the quotient rule.
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4 – 2 =2 and 3 – 1= y².
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There is my first term, moving on.
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15 goes into 30 twice.
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Using my quotient rule on the x, 3 – 2 = 1 and y² – y¹ = y¹.
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Both of these have an exponent of 1 and I do not need to write it.
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One more, 15 goes into 60 four times.
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I have x²/x² which will be x⁰ and y/y will be y⁰.
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Anything to the 0 power is 1.
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I can end up rewriting that term 3x2 y² + 2xy – 4.
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There is my resulting polynomial.
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In this next example, we are going to take one polynomial divided by another polynomial.
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I would not be able to split it up quite like I did before, now we will have to go through that long division process.
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Be careful you want to make sure that you line up your terms in descending order.
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If you look at the powers of the top polynomial you would not mix up.
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I want to start with that 3rd power then go to the 2nd power, then go to the 1st power, just to make sure I have it all lined up.
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Let us write it out.
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2x³ + x² + 5x + 13, now it is in a much better order to take care of.
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We will take all of that and we will divide it by 2x + 3.
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That looks good.
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It is time to get into the division process.
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Our first terms there, and what would I need to multiply 2x by in order to get 2x³?
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The only thing that will work would be an x².
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Once we find our numbers up top, we will multiply them by the polynomial out front.
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2x × x² = 2x³, that is a good sign, it is the same as the number above it.
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X² × 3 =3x², now comes the part that is tricky to remember.
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Always subtract this away and do not be afraid to use this parenthesis to help you remember that.
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x² – 3x², 1 – 3 = -2x².
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Then I have 2x³ – 2x³ and those will be gone, cancel out like they should.
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The first iteration of this thing looks pretty good.
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Let us try another one.
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I want to figure out how may times does 2x go into -2x²?
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I have to multiply it by –x.
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Let us bring down some more terms.
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I will get onto our multiplication process.
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-x × 2x =-2x².
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-x × 3 =-3x.
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We have our terms in there, it is time to subtract it away and be careful with all of our signs.
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5x - -3x when we subtract a negative that is the same as addition.
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I’m looking at 5x + 3x = 8x.
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-2x² - -2x² = that is a lot of minus signs.
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That would be the same as -2x + 2x² and they do cancel out like they should.
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That was pretty tricky keeping track of all those signs but we did just fine.
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We have 8x + 13 and I’m trying to figure out what would I have to multiply 2x by in order to get that 8x.
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There is only one thing I can do I need to multiply by 4.
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We will multiply everything through and see what we get.
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4 × 2x = 8x, 4 × 3 =12.
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I can subtract this away and let us see what we get.
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13 – 12=1, 8x – 8x = 0.
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Here I have a remainder of 1.
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I could end up writing my quotient and I can put my remainder over the divisor.
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There is our answer.
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In some polynomials you want to make sure you put in those placeholders.
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That way everything lines up and works out good.
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It is especially what we will have to do for some of those missing powers in this one.
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I’m missing an x² and a single x.
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Let us write this side and put those in.
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x³, I have no x², no x – 8.
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All of that is being divided by x – 2.
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I’m going to go through and let us see what I need to multiply x by in order to get x³.
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I think I’m going to need x².
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Now that I found it, I will go ahead and multiply it through.
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x² × x =x³ and x² × -2 = -2x².
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We have our terms, let us go ahead and subtract it away.
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0x² – 2x², this is one of those situations where if we subtract a negative is the same as adding.
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0x² + 2x² =2x².
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x³ – x³, that is completely gone.
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Bring down our other term here and we will keep going.
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What would I have to multiply x by in order to get 2x²?
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I’m going to need 2x.
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Let us multiply that through, 2x × x =2x².
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2x × -2 =-4x.
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Now we found that, let us subtract that away.
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Starting on the end, 0 – -4, that is the same as 0 + 4x = 4x.
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Then 2x² – 2x², they are completely gone, you do not have to worry about it.
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We will bring down our -8 and continue.
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What would I have to multiply x by in order to get 4x?
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That will have to be 4.
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4 × x = 4x and 4 × -2 = -8.
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It looks like it is exactly the same as the polynomial above it.
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I know when I subtract, I would get 0.
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There is no remainder for this one.
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I will take (x³ – 8) ÷ x -2 and the result is x² + 2x + 4.
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Division process can take a bit but as long as you do the steps very carefully, you should turn out okay.
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Let us try this giant one.
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This is (2m⁵ + m⁴ + 6m³ – 3m² – 18) ÷ m² + 3.
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There are a lot of things to consider in here.
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One thing that I will be careful of is putting those placeholders for this guy down here.
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Notice that it is missing an m, let us give it a try.
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I will have m² + 0m + 3 and all of that is going into our other polynomial 2m⁵ + m⁴ + 6m³ – 3m² – 18.
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Lots of things to keep track of but I think we will be okay.
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What would I have to multiply the m² by in order to get a 2m⁵?
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That would be 2m³.
00:25:39.400 --> 00:25:42.900
I can run through the multiplication and write it here.
00:25:42.900 --> 00:25:48.800
2m³ × m² = 2m⁵.
00:25:48.800 --> 00:25:57.700
Now we can multiply it by our 0 placeholder but anything times 0 will give us 0 so 0m⁴.
00:25:57.700 --> 00:26:08.100
By my last one, let us see 2m³× 3 + 6m³.
00:26:08.100 --> 00:26:11.700
Let us subtract that away.
00:26:11.700 --> 00:26:15.300
6m³ – 6m³ those are gone.
00:26:15.300 --> 00:26:20.500
I have m⁴ – 0m⁴ I still have m⁴.
00:26:20.500 --> 00:26:24.400
2m⁵ – 2m⁵ those are gone.
00:26:24.400 --> 00:26:27.000
I dropped away quite a bit of terms.
00:26:27.000 --> 00:26:38.500
Let me go ahead and write in my 0m³ as one of those placeholders so I can keep track of it.
00:26:38.500 --> 00:26:40.200
Let us try this again.
00:26:40.200 --> 00:26:49.600
m² goes into m⁴, if I multiply it by another m².
00:26:49.600 --> 00:26:54.500
Multiplying through I have m² × m² = m⁴.
00:26:54.500 --> 00:27:06.600
m² × 0m = 0m³ and m² × 3 = 3m².
00:27:06.600 --> 00:27:16.000
We will take that and subtract it away.
00:27:16.000 --> 00:27:18.100
I need to go ahead and subtract these.
00:27:18.100 --> 00:27:20.800
Be very careful on the signs of this one.
00:27:20.800 --> 00:27:32.400
I have –m² and I’m subtracting -3m², the result here will be -6m².
00:27:32.400 --> 00:27:38.700
The reason why it is happening is because of that negative sign out there.
00:27:38.700 --> 00:27:49.100
Now I have 0m³ – 0m³, 0 – 0 =0, m⁴ – m⁴ = 0.
00:27:49.100 --> 00:27:55.900
It looks like I forgot an extra placeholder.
00:27:55.900 --> 00:28:04.600
I need 0m and then I need my 18.
00:28:04.600 --> 00:28:11.600
Let us bring down both of these.
00:28:11.600 --> 00:28:19.100
I need to figure out what I have to what would I have to multiply m² by in order to get -6m².
00:28:19.100 --> 00:28:38.100
-6 will do it, -6m² 6 × 0 = 0m and -6 × 3 =-18.
00:28:38.100 --> 00:28:42.900
It is exactly the same as the polynomial above it.
00:28:42.900 --> 00:28:48.700
Since they are exactly the same and I’m subtracting one from the other one, 0 is the answer.
00:28:48.700 --> 00:28:51.900
There is no remainder, it went evenly.
00:28:51.900 --> 00:29:04.000
The quotient for this one would be 2m³ + m² – 6.
00:29:04.000 --> 00:29:13.000
In this polynomial, I have (3x³ + 7x² + 7x + 11) ÷ 3x + 6.
00:29:13.000 --> 00:29:22.400
The reason why I put this one is because it can get a little bit difficult figuring out what you need to multiply to get into that second polynomial.
00:29:22.400 --> 00:29:27.500
I’m going to warn you, this involves a few fractions.
00:29:27.500 --> 00:29:29.700
Let us give it a shot.
00:29:29.700 --> 00:29:48.300
(3x³ + 7x² + 7x + 11) ÷ 3x + 6.
00:29:48.300 --> 00:29:50.200
Let us start off at the very beginning.
00:29:50.200 --> 00:29:57.400
What do I need to multiply my 3x by in order to get 3x³?
00:29:57.400 --> 00:30:03.700
The only thing that will work will be an x².
00:30:03.700 --> 00:30:09.000
I will go through and I will multiply and get the result.
00:30:09.000 --> 00:30:21.000
3x³ + 6x² and let us subtract that away.
00:30:21.000 --> 00:30:37.500
7x² – 6x² = 1x².
00:30:37.500 --> 00:30:45.100
Now comes the tricky part, I need to figure out what I need to multiply 3x by in order to get x².
00:30:45.100 --> 00:30:51.400
If I’m looking at just the variable part of this, I have to multiply and x by another x in order to get an x².
00:30:51.400 --> 00:30:54.500
We will go ahead and put that as part of our quotient.
00:30:54.500 --> 00:30:58.900
What do I have to multiply 3 by in order to get 1 out front?
00:30:58.900 --> 00:31:00.700
That is a little bit trickier.
00:31:00.700 --> 00:31:05.600
3 × what = 1.
00:31:05.600 --> 00:31:09.000
That is almost like an equation onto itself.
00:31:09.000 --> 00:31:13.800
What we see is that x would have 1/3, a fraction.
00:31:13.800 --> 00:31:20.700
It is okay, we can use fractions and end up multiplying by those.
00:31:20.700 --> 00:31:26.900
3x × 1/3x = 1x².
00:31:26.900 --> 00:31:28.200
Let us multiply that through.
00:31:28.200 --> 00:31:40.400
1/3x × 3x = 1x² I will write it down and 1/3x × 6 = 2x.
00:31:40.400 --> 00:31:44.400
Now we can take that and subtract it away.
00:31:44.400 --> 00:31:52.800
7x – 2x = 5x.
00:31:52.800 --> 00:31:57.500
Bringing down our extra terms and I think this one is almost done.
00:31:57.500 --> 00:32:02.200
What would I have to multiply 3x to get 5x?
00:32:02.200 --> 00:32:03.700
Let us see.
00:32:03.700 --> 00:32:13.600
he only way I can get an x into another x is to multiply by 1, but I’m going to think of how do I get 3 and turn it into 5?
00:32:13.600 --> 00:32:18.700
Let us do a little bit of scratch work on this one.
00:32:18.700 --> 00:32:25.400
3 × what = 5?
00:32:25.400 --> 00:32:31.300
If we divide both sides by 3 I think we can figure out it is 5/3.
00:32:31.300 --> 00:32:35.700
That I can write on top 5/3.
00:32:35.700 --> 00:32:38.100
We can go through multiplying.
00:32:38.100 --> 00:32:46.200
5/3 × 3 = 5x.
00:32:46.200 --> 00:32:57.600
5/3 × 6 =10.
00:32:57.600 --> 00:33:00.600
We will go ahead and subtract this away.
00:33:00.600 --> 00:33:05.500
11 – 10 = 1 and 5x – 5x = 0.
00:33:05.500 --> 00:33:12.900
We have a remainder of 1.
00:33:12.900 --> 00:33:18.300
Now that we have all of the quotient and the remainder, let us go ahead and write it out.
00:33:18.300 --> 00:33:29.800
We have (x² + 1/3x + 5/3 + 1) ÷ 3x + 6.
00:33:29.800 --> 00:33:37.800
Definitely do not be afraid some of those fractions to make sure it goes into that second polynomial.
00:33:37.800 --> 00:33:42.200
This process can get a little messy as you can definitely see from those examples.
00:33:42.200 --> 00:33:47.600
I have a nice clean way that you can go through the division process known as synthetic division.
00:33:47.600 --> 00:33:52.900
This is a much cleaner way for the division process so that you can keep track of all the variables.
00:33:52.900 --> 00:33:57.000
It is much clean but be very careful in how you approach this.
00:33:57.000 --> 00:34:03.500
It works good when dividing by polynomials of the form x + or – number.
00:34:03.500 --> 00:34:11.300
It will work especially with my little example right here (5x³ - 6x² +8) ÷ x -4.
00:34:11.300 --> 00:34:13.900
First watch how I will set this up.
00:34:13.900 --> 00:34:22.100
I'm going to create like a little upside down division bar and that is where I will end up putting the polynomial that I'm dividing.
00:34:22.100 --> 00:34:29.100
But I will not put the entire thing I’m only going to put the coefficients of all of the terms.
00:34:29.100 --> 00:34:33.200
The coefficient of the x³ is a 5.
00:34:33.200 --> 00:34:37.600
The coefficient of my x² is -6.
00:34:37.600 --> 00:34:45.700
I will put a 0 placeholder in for my missing x and then my last coefficient will be 8.
00:34:45.700 --> 00:34:50.500
Once I have all of those I will put another little line.
00:34:50.500 --> 00:35:04.900
I want to put in the value of x that would make this entire polynomial 0, if x was 4 that would be 0.
00:35:04.900 --> 00:35:08.800
I’m going to write 4 out here.
00:35:08.800 --> 00:35:18.100
That turns to be a tricky issue and many students remember what to put out over here because it will be the opposite of this one.
00:35:18.100 --> 00:35:21.000
If you see x – 4 put in a 4.
00:35:21.000 --> 00:35:27.600
If you see something like x + 7 then put in -7.
00:35:27.600 --> 00:35:31.500
We got that all set let us go through this synthetic division process.
00:35:31.500 --> 00:35:37.400
It tends to be quick watch very carefully how this works.
00:35:37.400 --> 00:35:46.200
The very first thing that you do in the synthetic division process is you take the first number here and you simply copy it down below.
00:35:46.200 --> 00:35:49.200
This will be a 5.
00:35:49.200 --> 00:36:01.000
Once you get that new number on the bottom, go ahead and multiply it by your number out front, 4×5 = 20.
00:36:01.000 --> 00:36:04.700
That is one step of the synthetic division process.
00:36:04.700 --> 00:36:12.700
To continue from there simply add the column -6 + 20 and get the result.
00:36:12.700 --> 00:36:17.300
This would be a 14.
00:36:17.300 --> 00:36:21.400
Once you have that feel free to multiply it out front again.
00:36:21.400 --> 00:36:31.300
14 × 4 = 56.
00:36:31.300 --> 00:36:40.000
There are 2 steps now we will take the 56 and we will add 0, 56.
00:36:40.000 --> 00:36:45.200
When we get our new number on the bottom, go ahead and multiply it right out front.
00:36:45.200 --> 00:36:51.900
4 × 50 = 200.
00:36:51.900 --> 00:37:02.300
4 × 6 = 24, 224.
00:37:02.300 --> 00:37:05.500
One last part to this we got to do some addition.
00:37:05.500 --> 00:37:13.800
8 + 224 = 232.
00:37:13.800 --> 00:37:16.500
It does not look like we did much of any type of division.
00:37:16.500 --> 00:37:22.800
We did a lot of adding and we did a lot of multiplying but do you know what these new numbers stand for on the bottom.
00:37:22.800 --> 00:37:29.400
That is the neat part, these new numbers I have here in green stand for the coefficients of our result.
00:37:29.400 --> 00:37:31.700
You know what happens after the division.
00:37:31.700 --> 00:37:40.600
The way you interpret these is the last number in this list will always be your remainder.
00:37:40.600 --> 00:37:44.200
I know that my remainder is 232.
00:37:44.200 --> 00:37:50.600
As for the rest of the values, the 5, 14, and 56, those are the coefficients on our variables.
00:37:50.600 --> 00:37:53.700
What should they be? Let me show you how you can figure that out.
00:37:53.700 --> 00:38:02.300
Originally we had x³ as the polynomial that we are dividing and these new ones will be exactly one less in power.
00:38:02.300 --> 00:38:13.500
That 5 goes with 5x² and the 14 goes with the 14x and 56 has no x on it.
00:38:13.500 --> 00:38:32.600
The result for this one is 5x² + 14x + 56 with a remainder of 232 which you can write over x – 4.
00:38:32.600 --> 00:38:35.100
Since it is still being divided
00:38:35.100 --> 00:38:38.500
It is a much cleaner and faster method for division.
00:38:38.500 --> 00:38:45.100
Let us go ahead and practice it a few times just to make sure got it down.
00:38:45.100 --> 00:38:52.200
We will go ahead and do (10x⁴ - 50x³ – 800) ÷ x – 6.
00:38:52.200 --> 00:38:54.100
It is quite a large problem.
00:38:54.100 --> 00:38:59.100
We will definitely remember to put in some of those placeholders to keep track of everything.
00:38:59.100 --> 00:39:03.700
First I will write in all of the coefficients of my original polynomial.
00:39:03.700 --> 00:39:16.500
I have a 10x⁴ - 50x³ I need to put in a placeholder for my x² and another placeholder for my x.
00:39:16.500 --> 00:39:24.300
We will go ahead and put in that -800.
00:39:24.300 --> 00:39:27.300
What shall we put on the other side?
00:39:27.300 --> 00:39:35.900
Since I'm dividing by x – 6, the value of 6 will be divided that would make that 0.
00:39:35.900 --> 00:39:41.600
I will use 6 and now onto the synthetic division process.
00:39:41.600 --> 00:39:46.700
The first part we will drop down to 10, just as it is.
00:39:46.700 --> 00:39:56.600
Then we will multiply by the 6 out front 60.
00:39:56.600 --> 00:40:04.800
Now that we have that, let us add the -50 and the 60 together, 10 again.
00:40:04.800 --> 00:40:09.300
We will multiply this out front.
00:40:09.300 --> 00:40:17.300
That result will be 60.
00:40:17.300 --> 00:40:29.800
We will go ahead and add 0 + 60 = 60 and multiply 60 × 6= 360.
00:40:29.800 --> 00:40:35.400
Now we will add 0 + 360 = 360.
00:40:35.400 --> 00:40:42.000
I have to take 360 × 6.
00:40:42.000 --> 00:40:43.600
think I have to do a little bit of scratch work for that one.
00:40:43.600 --> 00:40:55.300
6 × 0 = 6 × 6 = 36 and then 3 × 6 = 18 + 3 = 21.
00:40:55.300 --> 00:41:05.200
I have 2160, let us put it in.
00:41:05.200 --> 00:41:18.200
Only one last thing to do is we need to add -800 to the 2160 and then let us do a little bit of scratch work to take care of that.
00:41:18.200 --> 00:41:29.100
0 – 0= 0, 6 – 0= 6, 20 and we will breakdown and will say 11 – 8 =3.
00:41:29.100 --> 00:41:33.300
I have 1, I have 1360.
00:41:33.300 --> 00:41:37.700
Now comes the fun part, we have to interpret exactly what this means.
00:41:37.700 --> 00:41:45.100
Keep in mind that this last one out here that is our remainder.
00:41:45.100 --> 00:41:53.600
Originally our polynomial was x⁴ so we will start with x³ in our result.
00:41:53.600 --> 00:42:11.900
10 x³ + 10x² + 60x + 360 and then we have our remainder 1360.
00:42:11.900 --> 00:42:15.900
It is all still being divided by an x – 6.
00:42:15.900 --> 00:42:20.500
That was quite a bit of work, but it was a lot clean than going through the long division process.
00:42:20.500 --> 00:42:25.200
Let us see this one more time.
00:42:25.200 --> 00:42:35.200
In this last one we will use (5 - 3x + 2x² – x³) ÷ x + 1.
00:42:35.200 --> 00:42:40.500
We will first go ahead and put the coefficients of our top polynomial in descending order.
00:42:40.500 --> 00:42:44.100
Be very careful as you set this one up.
00:42:44.100 --> 00:42:50.000
On that side you can write out the polynomial first in descending order and then go ahead and grab its coefficients.
00:42:50.000 --> 00:42:58.900
-x³ would be the largest, then I have 2x² and then I have 3x, and 5.
00:42:58.900 --> 00:43:09.100
I need -1, 2, -3 and 5.
00:43:09.100 --> 00:43:16.700
I’m dividing by x + 1 so the number I will use off to the left will be -1.
00:43:16.700 --> 00:43:20.500
I think we have it all set up now let us run through that process.
00:43:20.500 --> 00:43:27.900
The first I’m going to bring down is -1 then we will go ahead and multiply.
00:43:27.900 --> 00:43:35.500
Negative × negative is positive.
00:43:35.500 --> 00:43:50.500
2 + 1 =3, 3 × -1= -3.
00:43:50.500 --> 00:44:07.600
-3 + -3 = - 6, -6 × -1 = 6 and 5 + 6 =11.
00:44:07.600 --> 00:44:16.700
Now we have our remainder, we can finally write down the resulting polynomial.
00:44:16.700 --> 00:44:22.500
We started with x³ so I know this would be x².
00:44:22.500 --> 00:44:36.400
-x² + 3x – 6 and I still have my 11 being divided by x + 1.
00:44:36.400 --> 00:44:40.500
Now you know pretty much everything that there is to know about dividing polynomials.
00:44:40.500 --> 00:44:44.900
If you divide by a monomial, make sure you split it out among all the terms.
00:44:44.900 --> 00:44:49.800
If you divide a monomial by a polynomial, you can go through the long division process
00:44:49.800 --> 00:44:53.800
or use this synthetic division process to make it nice and clean.
00:44:53.800 --> 00:44:56.000
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