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### Course: Algebra (all content) > Unit 10

Lesson 3: Adding & subtracting polynomials: two variables- Adding polynomials: two variables (intro)
- Subtracting polynomials: two variables (intro)
- Add & subtract polynomials: two variables (intro)
- Subtracting polynomials: two variables
- Add & subtract polynomials: two variables
- Finding an error in polynomial subtraction
- Add & subtract polynomials: find the error
- Polynomials review
- Adding and subtracting polynomials with two variables review

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# Adding polynomials: two variables (intro)

Sal simplifies (4x²y - 3x² - 2y) + (8xy - 3x² + 2x²y + 4). Created by Sal Khan.

## Want to join the conversation?

- Would it be considered more "simplified" by factoring out the multiple "2" making it:

2 (3x^2y - 3x^2 - y + 4xy + 2 ) ? Is either way considered more "simplified" or are both ways equally accepted? Thanks.(44 votes)- Well, technically the way you typed in your question is the most simplified version the answer can get. The version that Sal had written at2:20isn't wrong, but it can be simplified, and that is the way most teachers and test markers will accept, especially for CXC exams, SAT exams, ACT exams, and any other exams. Some teachers may accept Sal's answer, but would most likely write a note next to the answer saying that it can be simplified even more than it is, or they might write the most simplified version, which is your answer.(33 votes)

- aren't the exponents typically in decreasing order? if so, then why is 8xy after -2y in his answer?(8 votes)
- Well, those numbers have the same exponent, so it doesn't really matter. You don't
*need*to have numbers organized by their exponents, but doing that sometimes helps visual people.(8 votes)

- I have a question that says: (x + y)^2 - (9xy - 6x^2) = 7x^2 - 7xy + y^2

Can someone explain how they got this answer?(2 votes)- (x + y)² - (9xy - 6x²) = 7x² - 7xy + y²

Let's first expand the square of sum (x + y)².

(x + y)² = x² + 2xy + y²

If you don't know how I got the above equation, you may want to watch some videos on squaring binomials but this is basically what I did:

(x + y)² = (x + y)(x + y) = x² + 2xy + y² (you distribute each term when multiplying).

So now we can substitute our expansion of (x + y)² into our original equation:

(x + y)² - (9xy - 6x²) = x² + 2xy + y² - (9xy - 6x²)

Now lets distribute the coefficient of -1 into (9xy - 6x²):

x² + 2xy + y² - (9xy - 6x²) = x² + 2xy + y² - 9xy + 6x²

Now we combine terms:

x² + 2xy + y² - 9xy + 6x² = 7x² + 2xy + y² - 9xy

7x² + 2xy + y² - 9xy = 7x² - 7xy + y²

So:

(x + y)² - (9xy - 6x²) = 7x² - 7xy + y²(4 votes)

- Usually teachers want the order of the simplified polynomials in order from greatest to lowest so how would I go about ordering an equation with more than 2 variables in one term or with different exponents like in this example?(2 votes)
- Your teacher may have a preference, but in general, we order in terms of degrees of x, then followed by y and then z, though it does not matter the degree of y or z, as long as the highest degree of x is leftmost. Here are some examples: ²³⁴

1)`xy² + x²y³ + x⁴ = x⁴ + x²y³ + y²`

and sometimes further as`x(x³ + xy³ + y³)`

2)`xy² + x²y³ + y⁴ = x²y³ + xy² + y⁴`

and sometimes further as`y²(x²y + x + y²)`

3)`xy² + z²x²y³ + z³y⁴ = x²y³z² + xy² + y⁴z³`

and sometimes further as`y²(x²yz² + x + y²z³)`

I hope that helps(4 votes)

- Why do you add numbers with the same exponents/variables together? For example, why would you add 5x^2 and 10x^2? Would this work for 5x^2 plus 10y^2?(2 votes)
- you can add like terms because no matter what value of x is used, x^2 will always be the same, so 5x^2 + 10 x^2 = 15x^2 (say x=1, 5 + 10 = 15, x = 2 gives 20 + 40 = 60, etc.).

You cannot combine 5x + 10x^2 because different values of x would give different results if you tried to somehow combine them. Likewise, you cannot combine different variables such as 5x^2 + 10y^2 because you could change both x and y independently and get different results. If you tried to say it was 15x^2y^2, if x = 1 and y = 1 or x=0 and y = 0, you would be fine since 5 + 10 = 15 and 0 + 0 = 0, but any other values would be incorrect. Lets say x = 0 and y =1, you would end up with 0 + 10 = 0 which is a false statement.(3 votes)

- Do we need to take out the common factor 2 out from the expression? I see that Sal didn't take it out in the video so I am a bit confused. Thank you.(1 vote)
- Yes, it can be simplified a bit more. Click the "Sort by" drop-down menu near the comment section, and choose "Top Voted." Someone has already asked your same question, and you will find a good answer in that comment thread. =)(4 votes)

- I thought polynomials could only have one variable. Someone please explain this to me. (Or rather in the same variable.) The definition of polynomial that I've learned is, " an algebraic expression made up of one term or the sum or difference of two or more terms in the same variable." So what makes 4x^2y - 3x^2 - 2y a polynomial? Also, what makes 2y^3 + x^2 + x + 1 not?(2 votes)
- Sal put the 2y in front of the 8xy. Does it matter which order? If so, why? Any rules I should know?(1 vote)
- No, in addition the order doesn't matter, whether it's a polynomial or not. But order keeps things less complicated, and it'll be easier to deal with ordered polynomials than a messy one, not to mention that you'll less likely make mistakes. Order can be really useful, especially in multiplying/dividing polynomials and other more complicated problems.(3 votes)

- So at2:05, Sal put -2y after the -6x^2. But when I was doing the problem myself, I put 8xy first and then -2y. Is the way I ordered the expression wrong and if it is...why? This was my answer:

( 6x^2y - 6x^2 + 8xy - 2y +4 ) This was Sal's answer:

( 6x^2y - 6x^2 - 2y + 8xy +4 )(1 vote) - This problem : 7(a+b)+ 4(a+b)- 5(a+b) is in my math book. They asked me to simplify. I said that the answer was 6a+ 16b. I got it incorrect. The true answer is 6(a+b). Where did I go wrong? I can't figure it out. I used the distributive property and my problem looked like this 7a+7b+4a+4b-5a+5b. So then I added all the like terms with the variable a ( 7a, 4a, -5a), I then added the like terms with the variable b (7b, 4b, 5b). Ending up with 7a+4a-5a= 6a. And 7b+4b+5b= 16b, therefore resulting in 6a+16b. PLEASE HELP!(1 vote)
- When you distribute the -5, you only distributed the negative to the a, not to both terms, so -5(a+b) = -5a - 5b, so you could have gotten 6a + 6b which is the same as 6(a+b). Since you have a+b in common, you could take this as common factor to get (7 + 4 - 5)(a+b) to get their answer.(2 votes)

## Video transcript

So let's get some practice
simplifying polynomials, especially in the
case where we have more than one
variable over here. So I have 4x squared y
minus 3x squared minus 2y. So that entire expression
plus the entire expression 8xy minus 3x squared
plus 2x squared y plus 4. So the first thing
that jumps out at me is that I'm just adding this
expression to this expression. So to a large degree, these
parentheses don't matter. So I can just rewrite it as
4x squared y minus 3x squared minus 2y plus 8xy minus
3x squared plus 2x squared y plus 4. Now we can try to group
similar terms or like terms. So let's think about
what we have over here. So this first term right
here is a 4x squared y. So can I add this to any
of the other terms here? Do we have any other
x squared y terms? Well, sure, this
one right over here is another x squared y term. If I have 4 of
something-- in this case, I have 4x squared y's and
I add 2x squared y's to it, how many x squared
y's do I now have? Well, 4 plus 2-- I now
have 6x squared y's. Now let's move on to this term. So I have negative 3x squareds. Do I have any more x
squareds in this expression right over here? Well, sure, I have another
negative 3x squared. So if I have negative
3 of something and then I have another negative
3, I end up with negative 6 of that thing. So it's negative 6x squared. Now let's think about
this negative 2y term. Are there any other
y's over here? Well, it doesn't
look like there are. This is an 8xy. This is a 4. There's no just y's. So I can't really
add that to anything. So I'll just rewrite
it, negative 2y. And then 8xy--
well, once again, it doesn't seem like that can
be added to anything else. So let's just write
that over again. And then finally, we just
have the constant term plus 4. And it pretty much
looks like we're done. We have simplified
this as much as we can.