Integrated math 3
- Reducing rational expressions to lowest terms
- Intro to rational expressions
- Reducing rational expressions to lowest terms
- Simplifying rational expressions: common monomial factors
- Reduce rational expressions to lowest terms: Error analysis
- Simplifying rational expressions: common binomial factors
- Simplifying rational expressions: opposite common binomial factors
- Simplifying rational expressions (advanced)
- Reduce rational expressions to lowest terms
- Simplifying rational expressions: grouping
- Simplifying rational expressions: higher degree terms
- Simplifying rational expressions: two variables
- Simplify rational expressions (advanced)
Simplifying rational expressions: grouping
Sal simplifies and states the domain of (2x^2+13x+20)/(2x^2+17x+30). Created by Sal Khan.
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- I still don't get this video, I understood the first one but the second one is confusing me. What is Sal doing? If he is doing (2xsq + 13x + 20)/(2xsq + 17x + 30) Wouldn't that just be
You can't just divide each term in the numerator by a corresponding term in the denominator, because each term in the numerator is being divided by the WHOLE denominator.
For example, the expression you provided (2x^2 +13x +20)/(2x^2+17x +30) is the same thing as 2x^2 / (2x^2+17x+30) + 13x/(2x^2+17x+30) + 20/2x^2+17x+30)
Instead, to simplify the expression you could try factoring the top and the bottom of the expression. This would be sort of like reducing a fraction like 45/90 into 1/2, except done with a more complicated expression:
For example, in the equation you listed the numerator can be factored into:
And the denominator can be factored into:
So that means the whole expression can be rewritten as:
Now, since there is a 2x+5 in the numerator divided by a 2x+5 in the denominator, the whole thing can be reduced to:
The only caveat is that the 2x+5 cannot be equal to 0, because then you would have a 0 in the denominator and the expression would be undefined. That means that x cannot equal -2.5. So, the best way to write the final answer would be:
(x+4)/(x+6) with x not equal to -2.5
Does that help it make sense?(36 votes)
Could someone tell me which video explains ( not how to use it )
the a*b, a+b and why it works ?
I know I have seen it but I still don't quite understand
I can derive the a+b, a*b for a normalized 2nd degree equation.
But not for any other.
I still miss the explanation why it works out and which coefficients
enter the equation.(5 votes)
- Suppose we have the following polynomial: x^2 + px + q, whose roots are x = a and x = b. We therefore know that it is equal to (x - a)(x - b). Expanding that gives us x^2 - ax - bx + ab = x^2 -(a+b)x + ab. So we know that p = -(a+b) and q = ab.
It also works when the coefficient of x^2 isn't 1. Suppose we have a polynomial px^2 + qx + r, whose roots are x = a and x = b. We therefore know that it's equal to p(x - a)(x - b) = p(x^2 - ax - bx + ab) = p(x^2 -(a+b)x + ab).
Set the two to be equal:
px^2 + qx + r = p(x^2 -(a+b)x + ab)
Divide by p:
x^2 + (q/p)x + r/p = x^2 -(a+b)x + ab
So we know that q/p = -(a+b) and r/p = ab.
Obviously, if a+b and ab are some ugly rationals, using this method is very impractical, but it works.(7 votes)
- Ok wait I am still very confused. In the video Sals factoring by grouping what is that? I need some further explanation(3 votes)
- These videos by Sal explain how to do grouping with polynomials: https://www.khanacademy.org/math/algebra/polynomial-factorization/factoring-quadratics-2/v/factor-by-grouping-and-factoring-completely Hope this helps!(2 votes)
- At2:52when he had two of the (x + 4), why did he only write one for the final equation? Wouldn't it be (x + 4) ^2 (2x + 5)?(3 votes)
- 2x(x+4)+5(x+4), let's write it like this: ac+bc. And a is 2x, b is 5, and since the two (x+4) are the same, let's call them both c.
Imagine that you have 4 apples, and your friend gives you 3 apples, you now have a total of 7 apples. That's the same here.
You have "a" much of "c" and "b" much of "c", and totally (since you need to add them) you have (a+b) much c.
And go back to the formula, you have "2x" much "(x+4)" and "5" much "(x+4)", And totally you have "(2x+5)" much of "(x+4)".
And "much = multiply", you got (2x+5)(x+4).
Hope that helps!(1 vote)
- I don't understand why at1:13and3:26he splits up the 13x and 7x. Couldn't you just factor the problem without splitting the coefficients? Do we have to do that for every factoring problem?(3 votes)
- Did you actually factor this without grouping? If you can, great! If you're like me and grouping is faster, then use grouping.(3 votes)
- I don't understand how this first expression isn't a quadratic?(3 votes)
- quadratics don't have x in the denominator(3 votes)
- I've been doing some practice on this and I usually get them wrong. The last one I did before giving up and coming here said the answer was x^2-2/x^2+2. I actually got that, but I kept cancelling out to get -1 (cancel out x^2 and get -2/2, reduce to -1). How do you know when to stop reducing? is there always supposed to be an x on both the numerator and the denominator?(2 votes)
- The thing to remember with cancelling is that we can only cancel factors (things being multiplied). In your example above, you cancelled terms (things being added/subtracted). We stop cancelling when there are no more common factors between the numerator & denominator.
Since the binomials in: (x^2-2)/(x^2+2) are not factorable and they do not match (aren't common factors), we can't reduce the fraction any further.
Hope this helps.(5 votes)
- when u factor the denominator wouldn't 2 and 15 work as a and b(2 votes)
- No, because:
For Ax^2 +Bx +C
a*b = AC
a + b = B
a+b has to be equal to 17 and 2+15 is 17 so that would work
However a*b has to be equal to AC = 60 and 2*15 is not equal to 60.(4 votes)
- how can this same solving method be used when one of the signs is minus such as the first one. The reason i ask is because on one of the little practice quizzes there is a problem :4y^2 −18y+18/y^3 −6y^2 +9y. This problem is making my brain hurt... its not the material its just this problem is really throwing me off. Thank you . Also what would be the "numbers" that make this problem undefined?(2 votes)
- First of all, move the common factors outside of the parentheses.
2 ( 2y^2 - 9y + 9 ) / [ y ( y^2 - 6y + 9 ) ]
Then factor both trinomials.
2 ( 2y - 3 ) ( y - 3 ) / [ y ( y - 3 ) ( y - 3 ] )
You can at this point easily notice the values that make the denominator zero ( 0 and 3 ), so those are the unallowed values of y that make the problem undefined.
Next, divide out the common factor from both numerator and denominator to get
2 ( 2y - 3 ) / [ y ( y - 3) ] as the final answer.
Hope this will help your brain relax!(3 votes)
- never mind. it's explained at6:07in the video(3 votes)
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.