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Find the remainder π of π₯ and the quotient π of π₯ when three π₯ cubed plus two π₯ squared minus three π₯ minus five is divided by π₯ plus four.
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This is a division problem.
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Weβre dividing one polynomial, three π₯ cubed plus two π₯ squared minus three π₯ minus five, by another polynomial, π₯ plus four.
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And we do this by using polynomial long division.
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Here is our polynomial long division sign.
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π₯ plus four, which is the polynomial weβre dividing by, goes on the left of this symbol like so.
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And the dividend which is three π₯ cubed plus two π₯ squared minus three π₯ minus five, which is the polynomial that weβre dividing by π₯ plus four, goes underneath the long division sign like so.
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We can note here that we weβre lucky that both our polynomials were already simplified and already written in order from highest degree term first to lowest degree term last.
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So for example, if we did instead had to divide by four plus π₯, we first wouldβve had to swap those two terms around to get it in a canonical form with the highest degree term π₯ first.
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Okay, so weβve laid out our long division problem, but we still have to do it.
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So whatβs the first step?
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We identify the highest degree term of the dividend, which is in our case three π₯ cubed, and the highest degree term of the divisor, which is just π₯.
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And we divide them to get three π₯ squared.
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And we take that three π₯ squared and we put it on top of the long division symbol because itβs the first term of our quotient.
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What we do now?
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We take our term three π₯ squared and we multiply it by our divisor π₯ plus four.
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Okay, so we need to multiply out this bracket.
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Three π₯ squared times π₯ plus four is three π₯ squared times π₯ plus three π₯ squared times four.
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And now weβve expanded this bracket, we can simplify.
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Three π₯ squared times π₯ is just three π₯ cubed.
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Of course, this isnβt particularly surprising given that we found three π₯ squared by dividing three π₯ cubed by π₯.
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And three π₯ squared times four is just 12π₯ squared.
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And now weβve performed the multiplication, we want to make sure that itβs aligned nicely with the dividend above.
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So now the 12π₯ squared is nicely under the two π₯ squared of the dividend.
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So now we found this product and weβve aligned it nicely under the dividend, we can subtract it from the dividend.
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Now we can see why it was important to line up the terms.
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So first the π₯ cubed terms, three π₯ cubed minus three π₯ cubed is zero.
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And I suppose we could write that in if we wanted to; we donβt need to.
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Now onto the π₯ squared terms, we have two π₯ squared and 12π₯ squared.
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And remember we are subtracting, so we get negative 10π₯ squared like so.
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And there arenβt any π₯ terms or constant terms to subtract, so we just carry those down from the dividend.
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Here weβre really subtracting zero π₯ plus zero, so we just get negative three π₯ minus three.
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And if we get rid of the zero at the front, we see that weβre left with negative 10π₯ squared minus three π₯ minus five.
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The procedure for polynomial long division is just the same as the procedure for normal whole number long division, except itβs slightly easier because you donβt have to carry anything.
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And we repeat the process for the next step.
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Now our dividend is negative 10π₯ squared minus three π₯ minus five and its highest degree term is negative 10π₯ squared.
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And we want to divide that by the highest degree term of our divisor.
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Our divisor is still π₯ plus four and its highest degree term is still π₯.
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And the answer is of course negative 10π₯.
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So we write minus 10π₯ next to the three π₯ squared; thatβs the next term of our quotient.
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So now we have to work out what negative 10π₯ times the divisor π₯ plus four is.
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It is of course negative 10π₯ squared minus 40π₯.
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And we have to subtract that from the line above.
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This might be slightly tricky: negative 10π₯ squared minus negative 10π₯ squared is of course zero, so we donβt need to write anything.
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And negative three π₯ minus negative 40π₯ is the same as negative three π₯ plus 40π₯, which is 37π₯.
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And thereβs nothing to subtract from the minus five, so we just take that down.
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And we get 37π₯ minus five.
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Okay, so thereβs one last step.
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The highest degree term of the dividend, 37π₯, minus the highest degree term of the divisor, π₯, is 37.
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37 times the divisor π₯ plus four is 37π₯ plus 148.
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We subtract this to get negative 153.
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And we canβt really divide negative 153 by our divisor π₯ plus four, so weβre done with the procedure.
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Negative 153 is our remainder π of π₯.
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And three π₯ squared minus 10π₯ plus 37 is our quotient π of π₯.
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So there we have it.
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Thatβs using a polynomial long division to find the remainder π of π₯ and the quotient π of π₯ when three π₯ cubed plus two π₯ squared minus three π₯ minus five is divided by π₯ plus four.