Video: Using Polynomial Division to Solve Problems

Use polynomial division to simplify (2π‘₯Β³ + 5π‘₯Β² + 7π‘₯ + 4)/(π‘₯ + 1).

05:32

Video Transcript

Use polynomial division to simplify two π‘₯ cubed plus five π‘₯ squared plus seven π‘₯ plus four all divided by π‘₯ plus one.

We’re going to simplify this fraction using two methods. In the first method, we’re simply going to modify this fraction in place. The second method will involve laying this out as an algebraic long division problem. Hopefully, the first method will give some insight into the second.

So we first write out this big fraction. The first thing we’re going to do is to add and subtract two π‘₯ squared from the numerator of the fraction. Hopefully, you will agree that this doesn’t change the value of the fraction, but it’s not particularly clear why we’ve done it yet.

We can split this fraction into two, so it’s 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. The reason that we’ve 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.

This factor of the numerator cancels with the denominator of the fraction to leave just two π‘₯ squared. You can also see that we’ve 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.

So maybe you’re not convinced that this is simpler, but we’ll keep going for the moment. We’d like to perform the same trick again, adding something and subtracting something from the numerator, and then split the fraction into two.

Well, the numerator of the first traction is a multiple of π‘₯ plus one, allowing us to cancel. If you’d like to, you can pause the video and think about what the orange question mark should be to allow us to do this.

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.

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‘₯. The numerator of the second fraction can be simplified by combining like terms giving us just four π‘₯ plus four as a numerator.

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 we’ve made the fraction simpler.

We started with a cubic. After one step, the numerator of our fractional part is now a quadratic. And of the two steps where we are now, we have a linear expression as the numerator.

So far we’ve simplified to two π‘₯ squared plus three π‘₯ plus four π‘₯ plus four over π‘₯ plus one. Notice that the numerator of the fraction is a multiple of the denominator already without doing any work.

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; that’s as simple as you can get.

Now let’s see how we can set this problem out using polynomial long division. 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.

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.

The answer is that it goes two π‘₯ squared times. Now we need to subtract two π‘₯ squared times π‘₯ plus one, which is two π‘₯ cubed plus two π‘₯ squared.

Performing this subtraction, we get three π‘₯ squared plus seven π‘₯ plus four. We can now ask how many times does π‘₯ go into three π‘₯ squared. The answer is three π‘₯ times.

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.

We can either finish off by noticing that π‘₯ plus one goes into four π‘₯ plus four exactly four times. Or we can do the same thing as before saying that π‘₯ goes into four π‘₯ four times.

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.

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.

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