0:00:01.070,0:00:04.850
In simplifying algebraic[br]fractions, we occasionally need
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a process known as.
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Polynomial.[br]Division.
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Before we do that, I want to[br]take you back to something
0:00:22.486,0:00:26.518
that you actually know very[br]well indeed, and that's
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ordinary long division.
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You know how to do long[br]division, but I want to go over
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it again. 'cause I want to point[br]out certain things to you.
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'cause the things that are[br]important about long division
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are also important in polynomial
0:00:45.466,0:00:52.369
division so. Let's have a look[br]at a long division. Some
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supposing I want to divide 25.
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Into Let's[br]say
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2675.
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When I would have to do is look[br]at 25 in tool 2.
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No way 25 into 26. It goes once[br]and write the one there.
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Add multiply the one by the 25.
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And subtract and[br]have one left.
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Hope you remember doing that.[br]You were probably taught how to
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do that at primary school or the[br]beginnings of Secondary School.
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Next step is to bring down the[br]next number, so we bring down
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17. Well, we bring down Seven to[br]make it 17 and now we say how
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many times does 25 going to 17.
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It doesn't go at all. It's not[br]enough, so we have to record the
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fact that it doesn't go with a
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0. Next we bring down the[br]five. So now we've got 175 and
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we say how many times does 25[br]go into that? And it goes 7
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and we can check that Seven 535,[br]five down three to carry. 7 twos
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are 14 and three is 17.
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Tracked, we get nothing left.
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So this is our answer. We've[br]nothing left there, no
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remainder, nothing left over.[br]And there's our answer.
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2675 divides by 25 and the[br]answer is 107. They just look at
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what we did. We did 25 into[br]26 because that went.
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We then recorded that once that[br]it went there, multiplied, wrote
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the answer and subtracted.
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We brought down the next number.
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Asked how many times 25 went[br]into it, it didn't go. We
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recorded that and brought down[br]the next number. Then we said
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how many times does 25 going[br]to that Seven we did the
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multiplication, wrote it down,[br]subtracted, got nothing left
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so it finished.
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What we're going to do now is[br]take that self same process and
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do it with algebra.
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So let us
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take. This[br]27 X cubed.
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+9 X squared.
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Minus 3X. Minus[br]10.
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All over.
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3X minus 2.
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We want to divide that into[br]that. We want to know how many
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times that will fit into there,[br]so we set it up exactly like a
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long division. Problem by[br]dividing by this. This is what
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we're dividing into 27 X cubed[br]plus nine X squared minus three
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X minus 10.
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So we ask ourselves, how many[br]times does well? How many times
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does that go into that? But[br]difficult what we ask ourselves
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is how many times does the[br]excpet go into this bit?
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Just like we asked ourselves how[br]many times the 25 went into the
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26, how many times does 3X?
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Go into 27 X cubed. The answer[br]must be 9 X squared because
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Nynex squared times by three X[br]gives us 27 X cubed and we need
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to record that. But we need to[br]record it in the right place and
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because these are the X[br]squared's we record that above
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the X squares.
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So now we do the multiplication.
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Nine X squared times 3X is 27
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X cubed. Nine X squared times[br]minus two is minus 18 X squared.
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Just like we did for long[br]division, we now do the
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Subtraction. 27 X cubed[br]takeaway 27 X cubed none of
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them, because we arrange for[br]it to be so Nynex squared
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takeaway minus 18 X squared[br]gives us plus 27 X squared.
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Now we do what we did before we[br]bring down the next one, so we
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bring down the minus 3X.
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How many times does 3X go into[br]27 X squared?
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Answer. It goes 9X times and[br]we write that in the X Column.
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So now we have 9X times 3[br]X 27 X squared 9X times, Y
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minus 2 - 18 X.
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And we subtract again.
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27 X squared takeaway, 27 X[br]squared, no X squared, but we
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arrange for it to be like that,[br]minus three X minus minus 18X.
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Well, that's going to give us[br]plus 15X altogether, and we
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bring down the minus 10.
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3X into 15X. This time it goes[br]five times, so we can say plus
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five there. And again it's in[br]the numbers. The constants
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column at the end.
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Five times by 15 times by three[br]X gives us 15X. Write it down
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there five times by minus two[br]gives us minus 10 and we can see
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that when we take these two[br]away. Got exactly the same
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expression. 15X minus 10[br]takeaway. 50X minus 10 nothing
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left. So there's our answer,[br]just as in the long division.
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The answer was there.
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It's there now so we can say[br]that this expression is equal to
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9 X squared plus 9X.
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Plus 5. Let's[br]take another one.
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So we'll take X to the 4th.
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Plus X cubed.
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Plus Seven X squared[br]minus six X +8.
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Divided by all over[br]X squared, +2 X
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+8. So this is what we're[br]dividing by and this is what
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we're dividing into is not[br]immediately obvious what the
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answer is going to be. Let's[br]have a look X squared plus 2X
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plus 8IN tool.
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All of this.
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Our first question is how many[br]times does X squared going to X
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to the 4th? We don't need to[br]worry about the rest, we just do
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it on the first 2 bits in each[br]one, just as the same as we did
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with the previous example. How[br]many times X squared going to X
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to the four will it goes X[br]squared times? So we write it
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there over the X squared's. Now[br]we do the multiplication X
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squared times. My X squared is X[br]to the 4th.
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X squared by two X is plus[br]2X cubed X squared by 8 is
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plus 8X squared.
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And now we do the Subtraction X.[br]The four takeaway X to the 4th
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there Arnold, but we arranged it[br]that way. X cubed takeaway 2X
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cubed minus X cubed. Seven X[br]squared takeaway, 8X squared
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minus X squared and bring down[br]the next term.
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Now we say how many times does X[br]squared going to minus X cubed,
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and it must be minus X, and so[br]we write it in the X Column.
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And above the line there, next[br]the multiplication minus X times
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by X squared is minus X cubed[br]minus X times 2X is minus two X
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squared and minus X times by 8[br]is minus 8X.
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Do the subtraction minus X cubed[br]takeaway minus X cubed. No ex
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cubes minus X squared minus[br]minus two X squared or the minus
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minus A plus, so that[br]effectively that's minus X
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squared +2 X squared just gives[br]us X squared.
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Minus six X minus minus 8X.[br]Well, that's minus 6X Plus 8X
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gives us plus 2X and bring down[br]the next one.
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X squared plus 2X plus a 12 X[br]squared goes into X squared
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once. And so X squared plus[br]2X plus eight. And again we
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can see these two are the[br]same when I take them away,
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I will have nothing left[br]and so this is my answer.
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The result of doing that[br]division is that.
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Well, the one that started[br]us off on doing this was if
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you remember.
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X cubed minus one over[br]X minus one.
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This looks a little bit[br]different, doesn't it? Because
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whereas the space between the X[br]Cube term and the constant term
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was filled with all the terms?
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This one isn't.
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How do we cope with the?
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Let's have a look. Remember, we[br]know what the answer to this one
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is already. So what we must do[br]is right in X cubed and then
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leave space for the X squared[br]term, the X term and then the
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constant term. So what I asked[br]myself is how many times does X
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go into XQ, and the answer goes[br]in X squared. So I write the
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answer there where the X squared[br]term would be.
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X squared times by X is X cubed.
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X squared times by minus one is[br]minus X squared.
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And subtract X cubed takeaway X[br]cubed no ex cubes.
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0 minus minus X squared is[br]plus X squared.
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Bring down the next term. There[br]is no next term to bring down.
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There's no X to bring down.
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So it's as though I got zero X.[br]There was no point in writing
0:12:33.200,0:12:39.575
it. If it's not there, so let's[br]carry on X in two X squared that
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goes X times. So record the X[br]there above where the X is would
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be. Let's do the multiplication[br]X times by X. Is X squared.
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X times Y minus one is minus X.[br]Do the subtraction X squared
0:12:57.307,0:12:59.502
takeaway X squared is nothing.
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Nothing takeaway minus[br]X. It's minus minus X.
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That gives us Plus X.
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Bring down the next term. We[br]have got a term here to bring
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down it's minus one.
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How many times does X going to[br]X? It goes once.
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Long times by XX. One times by[br]minus one is minus one. Take
0:13:24.868,0:13:29.626
them away and we've got nothing[br]left there and so this is my
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answer X squared plus X plus[br]one, and that's exactly the
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answer that we had before. So[br]where you've got terms missing?
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You can still do the same[br]division. You can still do the
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same process, but you just leave[br]the gaps where the terms would
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be and you'll need the gaps[br]because you're going to have to
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write something. Up here in[br]what's going to be the answer.