### Video Transcript

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.

This is a division problem. Weβre dividing one polynomial, three π₯ cubed plus two π₯ squared minus three π₯ minus five, by another polynomial, π₯ plus four. And we do this by using polynomial long division.

Here is our polynomial long division sign. π₯ plus four, which is the polynomial weβre dividing by, goes on the left of this symbol like so. 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.

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. 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. Okay, so weβve laid out our long division problem, but we still have to do it. So whatβs the first step?

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₯. And we divide them to get three π₯ squared. 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. What we do now? We take our term three π₯ squared and we multiply it by our divisor π₯ plus four.

Okay, so we need to multiply out this bracket. Three π₯ squared times π₯ plus four is three π₯ squared times π₯ plus three π₯ squared times four. And now weβve expanded this bracket, we can simplify. Three π₯ squared times π₯ is just three π₯ cubed. Of course, this isnβt particularly surprising given that we found three π₯ squared by dividing three π₯ cubed by π₯. And three π₯ squared times four is just 12π₯ squared.

And now weβve performed the multiplication, we want to make sure that itβs aligned nicely with the dividend above. So now the 12π₯ squared is nicely under the two π₯ squared of the dividend. So now we found this product and weβve aligned it nicely under the dividend, we can subtract it from the dividend.

Now we can see why it was important to line up the terms. So first the π₯ cubed terms, three π₯ cubed minus three π₯ cubed is zero. And I suppose we could write that in if we wanted to; we donβt need to. Now onto the π₯ squared terms, we have two π₯ squared and 12π₯ squared. And remember we are subtracting, so we get negative 10π₯ squared like so. And there arenβt any π₯ terms or constant terms to subtract, so we just carry those down from the dividend. Here weβre really subtracting zero π₯ plus zero, so we just get negative three π₯ minus three. 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.

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. And we repeat the process for the next step. Now our dividend is negative 10π₯ squared minus three π₯ minus five and its highest degree term is negative 10π₯ squared. And we want to divide that by the highest degree term of our divisor. Our divisor is still π₯ plus four and its highest degree term is still π₯. And the answer is of course negative 10π₯. So we write minus 10π₯ next to the three π₯ squared; thatβs the next term of our quotient.

So now we have to work out what negative 10π₯ times the divisor π₯ plus four is. It is of course negative 10π₯ squared minus 40π₯. And we have to subtract that from the line above. This might be slightly tricky: negative 10π₯ squared minus negative 10π₯ squared is of course zero, so we donβt need to write anything. And negative three π₯ minus negative 40π₯ is the same as negative three π₯ plus 40π₯, which is 37π₯. And thereβs nothing to subtract from the minus five, so we just take that down. And we get 37π₯ minus five.

Okay, so thereβs one last step. The highest degree term of the dividend, 37π₯, minus the highest degree term of the divisor, π₯, is 37. 37 times the divisor π₯ plus four is 37π₯ plus 148. We subtract this to get negative 153.

And we canβt really divide negative 153 by our divisor π₯ plus four, so weβre done with the procedure. Negative 153 is our remainder π of π₯. And three π₯ squared minus 10π₯ plus 37 is our quotient π of π₯. So there we have it. 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.