WEBVTT
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Use polynomial division to simplify two π₯ cubed plus five π₯ squared plus seven π₯ plus four all divided by π₯ plus one.
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Weβre going to simplify this fraction using two methods.
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In the first method, weβre simply going to modify this fraction in place.
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The second method will involve laying this out as an algebraic long division problem.
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Hopefully, the first method will give some insight into the second.
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So we first write out this big fraction.
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The first thing weβre going to do is to add and subtract two π₯ squared from the numerator of the fraction.
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Hopefully, you will agree that this doesnβt change the value of the fraction, but itβs not particularly clear why weβve done it yet.
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We can split this fraction into two, so itβs two π₯ cubed plus two π₯ squared over π₯ plus one plus negative two π₯ squared plus five π₯ squared plus seven π₯ plus four over π₯ plus one.
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The reason that weβve chosen to do this is because the numerator of our first traction is equal to two π₯ squared times the denominator of the fraction, π₯ plus one.
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This factor of the numerator cancels with the denominator of the fraction to leave just two π₯ squared.
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You can also see that weβve combined the like terms in the numerator of the second fraction, so we get two π₯ squared plus three π₯ squared plus seven π₯ plus four over π₯ plus one.
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So maybe youβre not convinced that this is simpler, but weβll keep going for the moment.
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Weβd like to perform the same trick again, adding something and subtracting something from the numerator, and then split the fraction into two.
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Well, the numerator of the first traction is a multiple of π₯ plus one, allowing us to cancel.
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If youβd like to, you can pause the video and think about what the orange question mark should be to allow us to do this.
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We choose three π₯, so now we have two π₯ squared plus three π₯ squared plus three π₯ over π₯ plus one plus negative three π₯ plus seven π₯ plus four over π₯ plus one.
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The numerator of the first fraction can be factored to three π₯ times π₯ plus one, and this factor of the numerator cancels with the denominator to leave just three π₯.
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The numerator of the second fraction can be simplified by combining like terms giving us just four π₯ plus four as a numerator.
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So we get two π₯ squared plus three π₯ plus four π₯ plus four over π₯ plus one Notice that in each of these steps weβve made the fraction simpler.
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We started with a cubic.
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After one step, the numerator of our fractional part is now a quadratic.
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And of the two steps where we are now, we have a linear expression as the numerator.
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So far weβve simplified to two π₯ squared plus three π₯ plus four π₯ plus four over π₯ plus one.
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Notice that the numerator of the fraction is a multiple of the denominator already without doing any work.
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And so the factor of π₯ plus one in the numerator cancels with the π₯ plus one in the denominator, and we get two π₯ squared plus three π₯ plus four, which is our final answer; thatβs as simple as you can get.
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Now letβs see how we can set this problem out using polynomial long division.
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We set out a polynomial long division in the same way as whole number long division: the divisor goes on the left, and the dividend goes on the right underneath the long division symbol.
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We take the highest degree term of the divisor, in our case π₯, and ask how many times it can go into the highest degree term of the dividend, in our case two π₯ cubed.
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The answer is that it goes two π₯ squared times.
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Now we need to subtract two π₯ squared times π₯ plus one, which is two π₯ cubed plus two π₯ squared.
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Performing this subtraction, we get three π₯ squared plus seven π₯ plus four.
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We can now ask how many times does π₯ go into three π₯ squared.
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The answer is three π₯ times.
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Now we have to subtract three π₯ times π₯ plus one or three π₯ squared plus three π₯, which gives us four π₯ plus four.
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We can either finish off by noticing that π₯ plus one goes into four π₯ plus four exactly four times.
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Or we can do the same thing as before saying that π₯ goes into four π₯ four times.
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So we add a four to our quotient and then subtract four times π₯ plus one, which is four π₯ plus four, to get a remainder of zero.
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And we are happy to see that this method of polynomial long division gives the same answer of two π₯ squared plus three π₯ plus four.