Lesson Video: Dividing Polynomials by Monomials | Nagwa Lesson Video: Dividing Polynomials by Monomials | Nagwa

# Lesson Video: Dividing Polynomials by Monomials Mathematics • First Year of Preparatory School

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In this video, we will learn how to divide polynomials by monomials.

15:36

### Video Transcript

In this video, weβre going to learn how to divide polynomials by monomials. Letβs begin by recalling what we mean by a polynomial and a monomial.

A polynomial is an expression that consists of a number of terms either being added or subtracted. Each of those individual terms could be a constant. They could have a variable like π₯ or π¦. And they can have exponents. Those exponents, however, do need to be positive whole numbers. And they canβt be an infinite number of terms.

Then, a monomial is an algebraic expression that consists of just one term. Once again, any powers do need to be positive whole numbers. So an example of a monomial could be two π₯ or three π₯π¦. And to divide a polynomial by a single term, we simply divide term by term. Letβs see what that looks like at first with a really simple example.

Simplify negative 43π₯ to the seventh power plus 12π₯ cubed minus six π₯ squared over π₯.

In order to simplify our algebraic fraction, letβs recall what the fraction line actually means. It tells us that these two expressions are dividing one another. That is, weβre dividing negative 43π₯ to the seventh power plus 12π₯ cubed minus six π₯ squared by π₯. In this instance, we call the numerator of our fraction the dividend. The denominator, that expression that weβre going to divide by, is the divisor. And to divide our dividend by π₯, we simply do it term by term.

One way we could represent this is to use a bus stop method. Using a bus stop method to divide algebraic expressions is just like using a bus stop method to divide numbers. Instead, however, of dividing digit by digit, we divide algebraic term by algebraic term. And so weβre going to begin by dividing negative 43 times π₯ to the seventh power by π₯. This is the same as dividing negative 43π₯ to the seventh power by one π₯ to the power of one.

Negative 43 divided by one is negative 43. And then we recall that to divide terms with exponents, as long as the base is the same, we subtract the exponents. So π₯ to the power of seven divided by π₯ to the power of one is π₯ to the sixth power. Meaning that negative 43π₯ to the seventh power divided by π₯ is negative 43π₯ to the sixth power.

Next, we repeat this process for 12π₯ cubed. Weβre going to divide it by π₯. This is once again the same as dividing by one π₯ to the power of one. 12 divided by one is 12. And using the same rules for dividing with power terms, we see that π₯ cubed divided by π₯ to the power of one is π₯ squared. And so 12π₯ cubed divided by π₯ is 12π₯ squared.

Weβll do this one more time. Now, weβre going to divide negative six π₯ squared by π₯. Negative six divided by one is negative six. Then, π₯ squared divided by π₯ to the power of one is π₯ to the power of one. But we know π₯ to the power of one is just π₯. And so the final term in our quotient, thatβs the result we get when we divide, is negative six π₯. And so when we simplify this algebraic fraction, we get negative 43π₯ to the sixth power plus 12π₯ squared minus six π₯.

But of course this wasnβt the only way to answer the question. Letβs go back to some of our working out. Each time, we perform the division by considering it as a fraction. And so what we could alternatively have done is essentially reverse the process of adding fractions. And we could have written our fraction as negative 43π₯ to the seventh power over π₯ plus 12π₯ cubed over π₯ minus six π₯ squared over π₯ and gone from there. Either method is perfectly valid. Itβs very much personal preference.

Now, this is all fine and well for simple divisors like π₯. But what do we do if we have a multivariate term? Thatβs a term with more than one variable. Letβs see.

Simplify 23π₯ to the fifth power π¦ cubed plus 49π₯ to the fourth power π¦ cubed plus 41π₯ cubed π¦ cubed over negative π₯π¦.

We recall that the fraction line means divide. And so one way we have to simplify this fraction is to use, say, the bus stop method and divide each term by negative π₯π¦. Alternatively, we can essentially reverse the process of adding fractions. And we can split our fraction up. We write it as 23π₯ to the fifth power π¦ cubed over negative π₯π¦ plus 49π₯ to the fourth power π¦ cubed over negative π₯π¦ plus 41π₯ cubed π¦ cubed over negative π₯π¦.

And since weβve already established that simplifying is just like dividing, letβs think about how we simplify fractions where the numerator and denominator is simply a number. Well, we divide through by greatest common factors. And so to simplify each of these individual fractions, letβs begin by looking for greatest common factors in the numerator and denominator.

To make this easier, weβre going to begin by adding a coefficient of one. And so negative π₯π¦ is the same as negative one π₯π¦. And then weβre going to begin by looking for greatest common factors in the numerical parts and then the π₯-parts and then separately the π¦-parts. And so letβs begin by looking for the greatest common factor of 23 and one. Well, these two numbers are coprime. And this means that their only common factor is one. And so we can only divide 23 and one by one. And so the numerical part of our numerator and denominator is just 23 over one, which is 23.

And what about the greatest common factor of π₯ to the fifth power and π₯? We ask ourselves, whatβs the highest power of π₯ that divides evenly into both of these terms without leaving a remainder? Well, we can only divide both of them by π₯. And so letβs divide π₯ to the fifth power by π₯ and π₯ by π₯. If we start with our denominator, we know π₯ divided by π₯ is simply one. And then we can use one of our laws for dividing indices to divide π₯ to the fifth power by π₯ or π₯ to the power of one.

To divide these kinds of terms, as long as their bases are equal, we simply subtract their powers. So π₯ to the fifth power divided by π₯ to the power of one is π₯ to the fourth power. Weβll repeat this process one more time, this time finding the greatest common factor of π¦ cubed and π¦. Their greatest common factor is π¦. And so we divide π¦ by π¦ to get one. And when we divide π¦ cubed by π¦, we get two.

Weβre not quite finished though. We have to spot that we were dividing a positive by a negative. And so the result is going to be a negative term. Itβs negative 23π₯ to the fourth power π¦ squared. Remember, we no longer need one on the denominator. So letβs move on to our second fraction.

Once again, 49 and one are coprime. Then we find the greatest common factor of π₯ to the fourth power and π₯. Well, itβs π₯. So we divide our denominator by π₯ to get one and our numerator by π₯ to get π₯ cubed. Then, we find the greatest common factor of π¦ cubed and π¦, which we already established to be π¦. So we divide our denominator and our numerator by π¦. And once again, spotting that weβre dividing a positive by a negative, we get negative 49π₯ cubed π¦ squared.

We can do this one more time. 41 and one are coprime. And so we find the greatest common factor of π₯ cubed and π₯. Itβs π₯ again. So we divide both the numerator and denominator by π₯. We calculated the greatest common factor of π¦ cubed and π¦ to be π¦. So we divide both the numerator and denominator by π¦. And in our third term, we are also dividing a positive by a negative. So our third term is negative. Itβs negative 41π₯ squared π¦ squared.

And so in simplifying our fraction, weβve divided a polynomial by a monomial. The result is negative 23π₯ to the fourth power π¦ squared minus 49π₯ cubed π¦ squared minus 41π₯ squared π¦ squared. Now, we also saw that we can perform this process using the bus stop method. Once again, itβs important to realize that either method is perfectly valid. Itβs very much personal preference.

Weβll now consider an example where we can simplify before dividing term by term.

Simplify 12π to the fifth power times 11π to the 13th power π to the 13th power minus 12π to the fifth power π to the 13th power over two π to the seventh power π squared.

Remember, this fraction line means to divide. So essentially, here weβre dividing a polynomial by a monomial. Our first instinct might be to distribute the parentheses on our numerator. However, we can save ourselves a little bit of time. Weβre going to begin by looking for the greatest common factor of 12π to the fifth power and two π to the seventh power π squared.

Now, weβre allowed to do this and divide by this greatest common factor because we can see that any factor of 12π to the fifth power must be a factor of the entire numerator. So letβs find the greatest common factor of these two terms. Now, we can do this by first considering the numerical parts, then considering the π-part and then the π-part. The greatest common factor of 12 and two is two. Two is the largest number that divides evenly into 12 and two without leaving a remainder.

Then the greatest common factor of π to the fifth power and π to the seventh power is π to the fifth power. There is no π in 12π to the fifth power. So weβve found the greatest common factor. And before we do anything, weβre going to divide the numerator and the denominator of our fraction then by two π to the fifth power.

To divide 12π to the fifth power by two π to the fifth power, we first divide 12 by two. Thatβs six. Then π to the fifth power divided by π to the fifth power is one. So 12π to the fifth power divided by two π to the fifth power is six. Similarly, we divide two π to the seventh power π squared by two π to the fifth power. Two divided by two is one. Then, π to the seventh power divided by π to the fifth power is π squared. π squared divided by one is π squared. So two π to the seventh power π squared divided by two π to the fifth power is π squared π squared.

This means we can rewrite our fraction as six times 11π to the 13th power π to the 13th power minus 12π to the fifth power π to the 13th power over π squared π squared. Now, weβre ready to distribute our parentheses. We do so by multiplying six by the first term and then six by the second. Six times 11π to the 13th power π to the 13th power is 66π to the 13th power π to the 13th power. And then six times negative 12π to the fifth power π to the 13th power is negative 72π to the fifth power π to the 13th power. And this is all over π squared π squared.

Now, there are a number of ways we can evaluate the division at this point. Since that fraction line actually means divide, we could use the bus stop or we can reverse the process for adding fractions. And we can separate this into two individual fractions. Then, much as we did earlier, weβll divide by the greatest common factor of our numerator and denominator. And if we wish, we can do this by the numerical part, then by the π-part, then by the π-part separately.

Letβs make the denominator one π squared π squared. And then the greatest common factor of 66 and one is one. So we leave these for now. Then, what about the greatest common factor of π to the 13th power and π squared? Well, itβs π squared. So we divide both the numerator and denominator by π squared. This leaves π to the 11th power on our numerator and then simply one on our denominator.

Similarly, the greatest common factor of π squared and π to the 13th power is π squared. So when we divide through by π squared, weβre simply left with π to the 11th power on our numerator. And so our first term is 66π to the 11th power π to the 11th power over one, or simply 66π to the 11th power π to the 11th power.

Letβs repeat this process for our second fraction. 72 and one have a greatest common factor of one, so we leave them. Then, the greatest common factor of π to the fifth power and π squared is π squared. Dividing through, and weβre left with π cubed on the numerator. Then, the greatest common factor of π squared and π to the 13th power is π squared. And as before, that leaves us with π to the 11th power. And so we see that weβre left with 66π to the 11th power π to the 11th power minus 72π cubed π to the 11th power.

In our final example, weβll see how we can link dividing polynomials by monomials to geometrical questions.

The area of a triangle is 12π₯ squared plus four π₯ square centimeters, and its base is four π₯ centimeters. Write an expression for its height.

Letβs begin by recalling how we find the area of a triangle. For a triangle whose base is π units and whose perpendicular height is β units, its area is a half base times height. And the area will be given in square units. Now, it doesnβt matter that weβre working with algebraic expressions. We can still substitute these into this formula.

Letβs let the height be equal to β or β centimeters. Weβre told the area is 12π₯ squared plus four π₯ and its base is four π₯. So we can write 12π₯ squared plus four π₯ equals a half times four π₯ times β. Now, because we can find a half of four π₯ quite easily, we should. This will make the next step a little bit easier. But if we couldnβt, what we could do is multiply both sides by two.

Weβre not going to do that though. Weβre going to write the right-hand side as two π₯ times β. And then since weβre looking to find the value of β or certainly an expression for β, we solve for β by dividing through by two π₯. So β is 12π₯ squared plus four π₯ all over two π₯.

There are a number of ways we can simplify this fraction or divide the numerator by the denominator. One way is to use the bus stop method. So letβs see what that looks like. Since weβre dividing by a monomial, we donβt need to use long division. Weβre simply going to divide each term in our dividend, thatβs the quadratic here, by the divisor, thatβs two π₯. So 12 divided by two is six, and π₯ squared divided by π₯ is π₯. So we see that 12π₯ squared divided by two π₯ is six π₯. Then we divide four π₯ by two π₯. Well, four divided by two is two, and π₯ divided by π₯ is one. So when we divide 12π₯ squared plus four π₯ by two π₯, we get six π₯ plus two. And so thatβs our expression for β.

It is, of course, worth noting that weβre working in centimeters and square centimeters. So the units for the height β are in centimeters also. The height is six π₯ plus two centimeters.

In this video, we saw that to divide a polynomial by a monomial, we divide term by term. One way we have to do this is to use a bus stop style method. Another method we have to do this is to split the problem up into fractions whose numerators are also monomials. And then we simplify by dividing both the numerator and denominator by their greatest common factor.

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