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## SAT

### Course: SAT > Unit 10

Lesson 2: Passport to advanced mathematics- Solving quadratic equations — Basic example
- Solving quadratic equations — Harder example
- Interpreting nonlinear expressions — Basic example
- Interpreting nonlinear expressions — Harder example
- Quadratic and exponential word problems — Basic example
- Quadratic and exponential word problems — Harder example
- Manipulating quadratic and exponential expressions — Basic example
- Manipulating quadratic and exponential expressions — Harder example
- Radicals and rational exponents — Basic example
- Radicals and rational exponents — Harder example
- Radical and rational equations — Basic example
- Radical and rational equations — Harder example
- Operations with rational expressions — Basic example
- Operations with rational expressions — Harder example
- Operations with polynomials — Basic example
- Operations with polynomials — Harder example
- Polynomial factors and graphs — Basic example
- Polynomial factors and graphs — Harder example
- Nonlinear equation graphs — Basic example
- Nonlinear equation graphs — Harder example
- Linear and quadratic systems — Basic example
- Linear and quadratic systems — Harder example
- Structure in expressions — Basic example
- Structure in expressions — Harder example
- Isolating quantities — Basic example
- Isolating quantities — Harder example
- Function notation — Basic example
- Function notation — Harder example

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# Operations with rational expressions — Harder example

Watch Sal work through a harder Operations with rational expressions problem.

## Want to join the conversation?

- *Hello!* 😢 Now I know my brain takes a little while to process but, I only in the 9th grade and I starting to practice for the SAT. I just made a 900 on my eighth practice test on Khan Academy. At the paste I am going I hope to make 1000 by May. My question to anybody is there any way you can break this down more. Especially, the factoring part!

- please and thank you 💕(14 votes)- Hi! There are a lot of useful videos and exercises on Khan Academy that should help you understand the topics. Use the search bar to find the topic you're struggling with. :)

The SAT section assumes you have knowledge about these topics, and focuses more on*how*to solve a question.(6 votes)

- What happened to the 3 in 3- 7x-56(4 votes)
- Sal combined like terms.

3 - 7x - 56 became -7x + (-56 + 3) = -7x - 53(7 votes)

- why did he times the (x+8) on the numerator??(5 votes)
- (x+8)/(x+8) = 1 and anything multiplied with one is itself. [Ex. (1)(2) = 2](3 votes)

- still don't understand how to factor(4 votes)
- Sal made everything look easy(3 votes)
- this question is with calculator ?(2 votes)
- A: The question that Sal is solving is the "no calculator" type.

When you see questions with variables such as x in the answer, they are usually problems that do not allow the use of calculators.(2 votes)

- at1:56, where did he get the 56 from?(2 votes)
- He distributed 7(x+8) and got 7*x or 7x and 7*8 or 56.(2 votes)

- Why didnt he multiply the other numerator by x-3?(0 votes)
- He wanted both fractions to have the same denominator. He wanted to keep the second fraction the same.

5/4; 5/4 *4/4 = 20/16

If i multiply the top by 4 and the bottom by 4 it's the same fraction. It's the same thing here.(6 votes)

- yooo he talks so slow, like usually i put 1.5, but i had to put 2 and stilll slow

great stuff tho(2 votes) - What if it is not able to be factored?(1 vote)
- If it is unable to be factored, the two denominators are multiplied. [Ex. a/b + a/c = (ac+ab)/b(c)](3 votes)

## Video transcript

- [Instructor] We're asked, "Which of the following is equivalent to 6x squared plus 5x over 3x plus one?" Pause this video and see if
you can work through this before we do it together. All right, now let's work
through this together. There's two ways that
you could approach this. One is to do algebraic long division. So another way of
rewriting this is saying, we are going to divide 3x
plus one into 6x squared plus 5x, and if algebraic
long division is unfamiliar to you, I encourage you to look it up on the non-SAT part of Khan Academy. But the way we look at it, we look at the highest degree term, which here is the first degree term. And we see, all right, how many times does it go into 6x squared? Well, 3x goes into 6x squared, 2x times. And so we write that in the
first degree space here. So 2x and then 2x times 3x is 6x squared, and then 2x times one is equal to 2x. And then we subtract these from above. So it's very much like the long division that you learned in elementary school. It's just, we're doing with
algebraic expressions now. And so these cancel out. 5x minus 2x is equal to 3x. Now, how many times does
3x plus one go into 3x? Well, one way to think about
it, 3x goes into 3x one time. So let's just write that in
the zeroth degree column. You could view that as a constant column. one times 3x plus one is 3x and one. Now we wanna subtract this and
we are going to be left with, those go away, and you're
just left with a negative one. And you can't divide 3x
anymore into negative one. So you could view that as the remainder. And so this whole thing is going
to be equal to 2x plus one, and then minus one over 3x plus one. Minus one over 3x plus one. And you can see that this is
choice D, right over here. Now, the other way that you
could do it if you forgot how to do algebraic long division, or you think this is just taking too long. Given that you have multiple choices here, you could just try out a simple number that's easy to compute. I wouldn't try out zero or one, because whether you just
multiply by zero one or whether you square them, zero squared is zero, one squared is one. So it might not
differentiate itself so much. So what I would do is
maybe use a number like two and then see which of
these other expressions are the same when you evaluate two. So for example, if I say
six times two squared plus five times two, over
three times two plus one, this is equal to, see, six
times four, this is 24. Five times two is 10. This is six. So we get 34/7. 34/7. Now, if you put two in
2x, you just get four. That's not 34/7. Rule that one out. You put two over here,
three times two is six, plus four is 10. That's not 34/7. Rule that out. You put two here, you
get four minus one over, let's see, it looks like one over, three times two is six plus one is seven. So minus 1/7, and let's see, four is the same thing as 28/7 minus 1/7, which is equal to 27/7, still not 34/7. So if you're doing this
on a standardized test and you feel confident in your math so far you might be able to do sets
D, but we can verify that. Two times two, and once again,
I've picked the number two just for simple computation, plus one minus one over
three times two, plus one. So this is going to give
us five right over here. And this is 1/7. So five minus 1/7. So that's the same
thing as 35/7 minus 1/7, this is five, that is equal to 34/7. So when you use the number two you also see that this one
evaluates to the same value. That's possible you try out some number, a two or a three, and several of them come to the same value, in which
case you could rule out some, but that's where the algebraic
long division is more useful. You know for sure what
the choice is going to be.