Simplifying rational expressions 2 | Polynomial and rational functions | Algebra II | Khan Academy

Simplifying rational expressions 2 | Polynomial and rational functions | Algebra II | Khan Academy

Simplify the rational
expression and state the domain. Once again, we have a trinomial
over a trinomial. To see if we can simplify
them, we need to factor both of them. That’s also going to help us
figure out the domain. The domain is essentially
figuring out all of the valid x’s that we can put into this
expression and not get something that’s undefined. Let’s factor the numerator
and the denominator. So let’s start with the
numerator there, and since we have a 2 out front, factoring
by grouping will probably be the best way to go, so let’s
just rewrite it here. I’m just working on the
numerator right now. 2x squared plus 13x plus 20. We need to find two numbers,
a and b, that if I multiply them, a times b, needs to be
equal to– let me write it over here on the right. a times b needs to be equal to
2 times 20, so it has to be equal to positive 40. And then a plus b has
to be equal to 13. The numbers that jump out
at me immediately are 5 and 8, right? 5 times 8 is 40. 5 plus 8 is 13. We can break this 13x into a
5x and an 8x, and so we can rewrite this as 2x squared. It’ll break up the 13x into–
I’m going to write the 8x first. I’m going to
write 8x plus 5x. The reason why I wrote the 8x
first is because the 8 shares common factors with the
2, so maybe we can factor out a 2x here. It’ll simplify it
a little bit. 5 shares factors with the 20, so
let’s see where this goes. We finally have a plus
20 here, and now we can group them. That’s the whole point of
factoring by grouping. You group these first two
characters right here. Let’s factor out a 2x, so this
would become 2x times– well, 2x squared divided by 2x
is just going to be x. 8x divided by 2x is going
to be plus 4. Let’s group these
two characters. And if we factor out a
5, what do we get? We get plus 5 times x plus 4. 5x divided by 5 is x, 20
divided by 5 is 4. We have an x plus 4 in
both cases, so we can factor that out. We have x plus 4 times two
terms. We can undistribute it. This thing over here will be x
plus four times– let me do it in that same color–
2x plus 5. And we’ve factored
this numerator expression right there. Now, let’s do the same thing
with the denominator expression. I’ll do that in a different–
I don’t want to run out of colors. So the denominator is right over
here, let’s do the same exercise with it. We have 2x squared
plus 17x plus 30. Let’s look for an a and a b. When I multiply them, I get
2 times 30, which is 60. And an a plus a b, when
I add them, I get 17. Once again, 5 and
12 seem to work. So let’s split this up. Let’s split this up
into 2x squared. We’re going to split up the 17x
into a 12x plus a 5x and that adds up to 17x. When you multiply 12 times 5,
you get 60, and then plus 30. Then on this first group right
here, we can factor out a 2x, so if you factor out a 2x, you
get 2x times x plus 6. In that second group, we can
factor out a 5, so you get plus 5 times x plus 6. Now, we can factor out an x plus
6, and we get we get x plus 6 times 2x plus 5. We’ve now factored the numerator
and the denominator. Let’s rewrite both of these
expressions or write this entire rational expression with
the numerator and the denominator factored. The numerator is going to
be equal to x plus 4 times 2x plus 5. We figured that out
right there. And then the denominator is
x plus 6 times 2x plus 5. It might already jump out at you
that you have 2x plus 5 in the numerator and the
denominator, and we can cancel them out. We will cancel them out. But before we do that, let’s
work on the second part of this question. State the domain. What are the valid x values
that we could put in here? A more interesting question is
what are the x values that will make this rational
expression undefined? It’s the x values that will make
the denominator equal to 0, and when will the denominator
equal to 0? Well, either when x plus 6 is
equal to 0, or when 2x plus 5 is equal to 0. We could just solve
for x here. Subtract 6 from both sides,
and you get x is equal to negative 6. If you subtract 5 from both
sides, you get 2x is equal to negative 5. Divide both sides by 2. You get x is equal
to negative 5/2. We could say the domain– let
me write this over here. The domain is all real numbers
other than or except x is equal to negative 6 and x is
equal to negative 5/2. The reason why we have to
exclude those is those would make this denominator– either
way you’re right. It’s going to go make the
denominator equal to 0, and it would make the entire rational
expression undefined. We’ve stated the domain. Now let’s just simplify the
rational expression. We’ve already said that x cannot
be equal to negative 5/2 or negative 6, so let’s just
divide the numerator and the denominator by 2x plus 5. Or just looking at the 2x plus
5, we know that 2x plus 5 won’t be 0, because x won’t be
equal to negative 5/2, and so we can cancel those out. The simplified rational
expression is just x plus 4 over x plus 6.

5 thoughts on “Simplifying rational expressions 2 | Polynomial and rational functions | Algebra II | Khan Academy

  • May 20, 2011 at 2:43 am


  • April 9, 2012 at 8:16 pm

    Thank you for the help.(P.S. do you read the youtube comments?)

  • November 5, 2014 at 5:40 pm

    Dear Sal!

    It doesnt matter which order you use when grouping. So you can write the expression as 2x^2 + 5x + 8x + 20 instead of 2x^2 + 8x + 5x + 20. After factoring the first and the second term, and factoring the third and the fourth term, you'll get the same result. And its true in every case. 
    High respect anyway! Great videos!

  • December 16, 2015 at 11:45 pm

    Can you not simplify the 4/6 in the end to 2/3? Or would that change the something?

  • March 8, 2016 at 1:30 pm



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