Video: Using Polynomial Division to Solve Problems

Use polynomial division to simplify (6π‘₯Β³ + 5π‘₯Β² βˆ’ 20π‘₯ βˆ’ 21)/(2π‘₯ + 3).

03:53

Video Transcript

Use polynomial division to simplify six π‘₯ cubed plus five π‘₯ squared minus 20π‘₯ minus 21 over two π‘₯ plus three.

One method we do have for simplifying algebraic fractions is to factor where necessary. It’s not particularly straightforward to factor this cubic on our numerator. So instead, we’re going to recall that this line in a fraction actually just means divide. And we’re going to use polynomial long division. Our dividend, that’s the numerator of our fraction, goes inside the bus stop. The divisor, that’s the denominator, goes on the outside.

And then, we remember the first thing that we do is we take the first term in our dividend, that’s six π‘₯ cubed, and we divide it by the first term in our divisor, that’s two π‘₯. Six divided by two is three. Then, if we consider π‘₯ as being π‘₯ to the power of one, we know that we can subtract these exponents. And π‘₯ cubed divided by π‘₯ to the power of one is π‘₯ squared. This means that six π‘₯ cubed divided by two π‘₯ must be three π‘₯ squared.

Our next step is to multiply three π‘₯ squared by each term in our divisor. Three π‘₯ squared times two π‘₯ is six π‘₯ cubed. Notice that this is the same as the first term in our dividend, so we know we’ve probably started this correctly. We then calculate three times three π‘₯ squared. Well, that’s nine π‘₯ squared. Our next step is to subtract six π‘₯ cubed plus nine π‘₯ squared from six π‘₯ cubed plus five π‘₯ squared. Six π‘₯ cubed minus six π‘₯ cubed is zero. We don’t really need to write this zero. And then, five π‘₯ squared minus nine π‘₯ squared is negative four π‘₯ squared. We bring down the next term. Some people bring down all of the terms, but I prefer to keep things a little bit simpler.

And we’re now going to divide negative four π‘₯ squared by two π‘₯. Negative four divided by two is negative two. π‘₯ squared divided by π‘₯ to the power of one is π‘₯. So, negative four π‘₯ squared divided by two π‘₯ is negative two π‘₯. And we add negative two π‘₯ above five π‘₯ squared in our problem. Now, we multiply negative two π‘₯ by each term in our divisor. Two π‘₯ multiplied by negative two π‘₯ is negative four π‘₯ squared, and negative two π‘₯ times three is negative six π‘₯.

We then subtract each of these terms from negative four π‘₯ squared minus 20π‘₯. Negative four π‘₯ squared minus negative four π‘₯ squared is negative four π‘₯ plus four π‘₯ squared. So that’s zero, and we don’t really need to write that. We then do negative 20π‘₯ minus negative six π‘₯. That’s negative 20π‘₯ plus six π‘₯, which is negative 14π‘₯. We bring down negative 21. And we’re now going to divide negative 14π‘₯ by two π‘₯. π‘₯ divided by π‘₯ is just one. So, we get negative 14 divided by two, which is negative seven. So, we add negative seven here. And once again, we divide this number by each term in our divisor. Negative seven times two π‘₯ is negative 14π‘₯, and negative seven times three is negative 21.

We do one final subtraction, and this is a really important step to do because it tells us whether there’s a remainder or not. In fact, negative 14π‘₯ minus 21 minus itself is just zero. And so, we’ve completed the division. When we simplify our algebraic fraction, we’re left with three π‘₯ squared minus two π‘₯ minus seven.

Now, at this stage, it’s really useful just to discuss briefly how we might check our solution. We perform an inverse operation. We take our quotient, here that’s the solution to the division, and multiply that by the divisor, remembering, of course, that the divisor is the algebraic expression here that we’re dividing by. We multiply three π‘₯ squared minus two π‘₯ minus seven by two π‘₯ plus three. And when we do, we should get the numerator, or the dividend.

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