If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Simplifying rational expressions: common binomial factors

Given a rectangle with length a²+6a+27 and width a²-9, Sal writes the ratio of the width to the length as a simplified rational expression. Created by Sal Khan.

Want to join the conversation?

  • leaf yellow style avatar for user niyatiaditti
    Is (x+3)/(x+9) our final answer?
    (19 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Stanley Harms
    for the final answer at the end of the video a+3/a+9 couldn't you divide 3 and 9 so the final answer is a+1/a+3?
    (15 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Justin Cope
      No, it's a bit more complicated. Think about it this way:

      You want to simplify the fraction. One way to do this would be dividing by 3. So you have:

      ( ( a + 3 ) / ( a + 9 ) ) / 3

      You could then distribute the 3 over the numerator ( a + 3 ) and denominator ( a + 9 ):

      ( ( a + 3 ) / 3 ) / ( ( a + 9 ) / 3 )

      Now you have to distribute the 3 across each of the terms ( a + 3 ) and (a + 9 ):

      ( ( a / 3 ) + ( 3 / 3 ) ) / ( ( a / 3 ) + ( 9 / 3 ) )

      Simplifying, you get:

      ( ( a / 3 ) + 1 ) / ( ( a / 3 ) + 3 )

      Of course, this isn't a very simple way to express ( a + 3 ) / ( a + 9 ).
      (19 votes)
  • old spice man green style avatar for user Derek
    I'm still kind of confused on the values that the variable cannot be equal to. For example a can't be 3 or -9. Could anybody answer that area for me? Thanks a lot!
    (4 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Chuck Towle
      Derek,
      If x=-9 then the denominator would be 0
      and if x=3 then the denominator would also be 0
      And if the denominator is 0 the answer is undefined.
      So the domain cannot include -9 or 3.

      After factoring and reducing the expression by making the (a-3)/(a-3) = 1
      this left only (a+9) in the denominator.
      But because the original equation did not allow a to be -9 or 3, the factored and reduced equation also cannot be equal to -9 or 3.

      I hope that helps.
      (7 votes)
  • piceratops tree style avatar for user university.rug
    Near the end It first was a not equal to -9 or 3. Then when only (a+3) / (a+9) was left, he wrote: a is not equal to -9 or 3. But if you fill in a=3. That would make a (3+3= )6 and not 0. Or am I missing something? I thought the answer at the final should be a -3 or -9.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Ghassan O. Najjar
    In the final expression a+3/a+9, i see that if a= 3 then that would result in a defined answer which is 6/12. That can be simplified to 3/4, too. So why "a" cannot be equal to 3 at the end of the video?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • starky ultimate style avatar for user Patrick
      It's because the original expression cannot have a = 3. In order to keep the original expression and the final expression the same, the restrictions in the original expression must be carried on to the final expression, or you don't really have the same expression.
      (2 votes)
  • leafers ultimate style avatar for user checkoutmybadges
    what if after i factor everything out, but nothing is able to be cancled out, would that mean that it is already in simplest form?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Kira Chand
    is there a difference between rationalising something and simplifying a rational expression?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • male robot johnny style avatar for user Noah
    If the width is over the length, does that always have to be the case? Can't you flip them, or is that illegal math?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Chuck Towle
      Noah,
      A ratio of two things can be writtern either way. It is perfectly acceptable (and often necessary to flip them. The only reason it would be restricted is because the problem required an answer in a specific manner.

      For instance, if you have a ratio of 12 miles to 1 hour, It could be written as
      12miles/1 hour or
      1 hour/12 miles.

      And depending on the problem you would need to figure out which way to use the ratio.

      For instance if you had a problem where you were give that a person traveled by bike at a ratio of 12 miles to 1 hour and he had traveled for 2 hours. You are asked how far he had traveled. You would use the equation.
      2 hour * 12 miles/1hour = 24 miles
      Notice the words (hours* miles/hours) results in hours in both the numerator and denominator whcih canel out giving you an answer in miles.

      But is you were given a problem that the person needed to travel 36 miles and you were asked how long it would take, you would use the equation
      36 miles * 1 hour/12 miles = 3 hours
      And notice this time the words (miles*hour/miles) results in miles in both the numerator and denominator which cancel out giving you an answer in hours.

      In one equation we used the ratio one way and in the other we flipped it over.

      The trick to using ratios is to realize they work either way and you need to construct an equation so that the words cancel out the correct way to give you the correct words in your answer.
      (4 votes)
  • piceratops ultimate style avatar for user Brendan D.
    How come we can't continue simplifying a+3/a+9? Couldn't we have crossed out the a's so we're left with 3/9 which is 1/3?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • hopper cool style avatar for user fun dip studios
    The video points out both values that a cannot be equal to, but in the exercises you only include the value the variable can't be in the factor you factored out when simplifying and not the value the variable cannot be in the remaining denominator. Why is that?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Jesse
      When you cancel a factor, you lose information about what the variable cannot be. So in your simplified form, you should retain that information explicitly. The information about the other values that the variable cannot be are included in the denominator of the simplified form.
      (3 votes)

Video transcript

Given a rectangle with length a-squared plus 6a minus 27 and a width a-squared minus 9, Write the ratio of the width of the rectangle to its length as a simplified rational expression. So we want the ratio of the width to the length to the length of the rectangle, and they give us the expressions for each of these. The width—the expression for the width of the rectangle—is a-squared minus 9. So, the width —let me do it in this pink—is a-squared minus 9, and we want the ratio of that to the length; the ratio of the width of the rectangle to its length. The length is given right over there; it is a-squared plus 6a minus 27, they want us to simplify this. And so the best way to simplify this, whether we're dealing with expressions in the numerator or denominator, or just numbers, is we want to factor them and see if they have common factors and if they do we might be able to cancel them out. So if we factor this top expression over here—which was the expression for the width—this is of the form a-squared minus b-squared, where b-squared is 9. So this is going to be the same thing as a plus the square root of 9 times a minus the square root of 9. So this is a plus 3 times a minus 3 and I just recognized that from just the pattern; if you ever see something a-squared minus b-squared, it's a plus b times a minus b and you can verify that for yourself; multiply this out, you'll get a-squared minus b-squared. So, the width can be factored into a plus 3 times a minus 3. Let's see if we can do something for the denominator. So here, if we want to factor this out, we have to think of two numbers that when we add them, I get positive 6, and when I take their product, I get negative 27. Let's see, if I have positive 9 and negative 3, that would work. So, this could be factored as a plus 9 and a minus 3. 9 times a is 9a, a times negative 3 is negative 3a, when you add those two middle terms together, you'll get 6a, just like that and then, 9 times negative 3 is negative 27—of course the a times a is a squared. So I've factored the two expressions and let's see if we can simplify it. And before we simplify it, because when we simplify it we lose information, let's just remember what are allowable a's here, so we don't lose that information. Are there any a values here that will make this expression undefined? Well, any a value that makes the denominator zero will make this undefined. So a cannot be equal to negative 9 or 3, 'cause if a was negative 9 or 3, then the denominator would be zero; this expression would be undefined. So we have to remember this, this is part of the expression; we don't want to change this domain; we don't want to allow things that weren't allowable to begin with, so let's just remember this right over here. Now with that said, now that we've made this constraint, we can simplify it more; we can say, look we have an a minus 3 in the numerator and we have an a minus 3 in the denominator and we're assuming that a is not going to be equal to 3, so it's not like we're dividing zero over zero. So a will not be equal to 3, any other number; this will be an actual number, you divide the numerator and denominator by that same value, and we are left with a plus 3 over a plus 9 and the constraint here—we don't want to forget the constraints on our domain— a cannot equal negative 9 or 3. And it's important that we write this here, because over here we lost the information that a could not be equal to 3, but in order for this to really be the same thing as this thing over here, when a was equal to 3 it wasn't defined so in order for this to be the same thing, we have to constrain the domain right over there; a cannot be equal to 3. Hopefully you found that useful.