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## Integrated math 3

### Course: Integrated math 3>Unit 3

Lesson 2: Dividing quadratics by linear factors

# Dividing quadratics by linear expressions with remainders

Learn how to simplify complex math problems by dividing quadratics by linear factors with remainders. Discover two methods for simplifying these kinds of math problems: factoring the numerator or using algebraic long division. This process helps break down tricky equations, making them easier to solve.

## Want to join the conversation?

• If someone came up to me on the street and told me to do a random math problem, I would probably tell them off.
• In this question, would the remainder just be 2, or would it be (2/x+2)? Because there is a difference between the two, and it seems to me that (2/x+2) would make more sense, but in the video, Sal says 2 is the remainder.
• You're right, I guess sal was assuming everyone knew that when you find the remainder with long division it isknown that you take the result and divide it by the divisor. The first time he got the answer by messing witht he numerator, or dividend if you wanna call it that, and canceled out the terms. He was just showing another way to do it.
• is there a video where he divides by (AX+-B) where a and b are constants and x is the variable. If so, may someone please tell me the title so I can see it( I can't manage to find it. And if not, it would be a good topic to put into video form
• The formula you just wrote is exactly the one of linear equations! you might want to search ''Linear equations'' seach bar. Sal has some REALLY old videos on that, you might start your search from there
• I like how the guy says "Quick" and it takes 4 minutes to do the problem. Anyone agree with me?
• I was wondering about when Sal gets the remainder. Why can't he divide the 2 from the numerator with the 2 in the denominator?
• so in 2/(x+2)? Wellyou would need some multiple of 2 multiplying each part of the denominator. I assume you are saying you would simplify 2/(x+2) to 1/(x+1) Let me know if that is incorrect in my assuming.

before aything else, f you check these two exprressions you will see they are not the same. Canceling out numbers int he numerator and denominator will always lead to the same result. for instance if you had 2/4 you could divide both numerator and denominator by 2 and get 1/2. Both fractions are the same. Or with functions, if you had x/(x^2) this could be rewritten as 1/x. You get the same number no matter what x you plug in.

This is actually something important to know. dividing something fromt he numerator or denominator means yu have to divide the whole numerator and denominator. with 2/(x+2) if you divided a 2 from both you would get 1/(x/2 + 1). Let me know if that didn't help or was not what you were asking.
• At , Sal cancels out the denominator of his fraction with an identical part of the numerator, ex. (x+2)(x+3)/(x+2). However, just a few videos ago, he broke up a fraction so that each part of the numerator was divided by the denominator seperately, ex. (18x^4-3x^2+6x-4)/6x, (18x^4/6x)+(-3x^2/6x)+(-4/6x). When the equation (x+2)(x+3)/(x+2) is seperated this way, a different answer is found from Sal's. I'm just not sure what about the situations differed. Was it maybe that in the first situation, (x+2)(x+3)/(x+2), the numerator is multiplied instead of added, (18x^4-3x^2+6x-4)/6x? If so, why did that alter the method?
• Can you show me how you separated (x+2)(x+3)/(x+2) because I feel as though there may have been a slip up.

Anyway, you are basically 100% correct about the multiplication and addition making a difference. If you expanded out the (x+2)(x+3) you could use the then divide each term.

(x+2)(x+3)
x^2 + 2x + 3x + 6
x^2 + 5x + 6

So then (x^2 + 5x + 6)/(x+2)COULD be rewritten as x^2/(x+2) + 5x/(x+2) + 6/(x+2).

The point here is if you have two terms being multiplied and then divide it by a third term it is actually like a third multiplication, just one over it. So for example
(x*y)/z = x*y*(1/z) or x*y*z^-1

Meanwhile if you have two terms added or subtracted and you divide them all by a third term it distributs just like multiplication.
(x-y)/z = x/z-y/z
Again, because you can rewrite a denominator as multiplication.
(x-y)(z^-1) = xz^-1 - yz^-1 = x/z - y/z
• for the remainder, could i write (x+3)(x+2)+2?
• I'm pretty sure you can. That's what I would do. 😉
• Does math have any other thing besides variables and numbers?
• Mathematics is the concept of using variables and numbers in a mathematical context. There are therefore functions and other concepts that help you use those variables and numbers.