Log in Jack McClelland 11 years agoPosted 11 years ago. Direct link to Jack McClelland's post “What if you have a non 1 ...” What if you have a non 1 coefficient and you need to solve a problem like 18x^2 + 3xy - 10y^2? I'm not asking for how to solve that specific problem, I'm asking how to solve problems of its nature. • (20 votes) Wrath Of Academy 10 years agoPosted 10 years ago. Direct link to Wrath Of Academy's post “You can do it with factor...” You can do it with factoring by grouping. Starting with for example See the https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-by-grouping series for the basics of that technique. Also see http://tinyurl.com/nvzstcv where I did another example from another user's question. (32 votes) Toni 10 years agoPosted 10 years ago. Direct link to Toni's post “I dont understand the way...” I dont understand the way Sal's doing these. I paused at • (17 votes) learn 8 years agoPosted 8 years ago. Direct link to learn's post “I think you are correct. ...” I think you are correct. Sal's reference to not knowing another way to do this is probably because this video was originally in a different order. If they move the videos around, the learning sequence doesn't follow exactly. I think the important thing is that you learned that skill, and you are able to apply it. I think it is good to see the different ways to do these problems, because it helps you to remember to keep analyzing these problems. Sal teaches us to really look at the problem so we don't get in the habit of always just plugging in formulas... rather we learn to analyze problems - Really look at them and understand them so we can choose the best strategy for the given situation. (10 votes) Ty Anandaraj 10 years agoPosted 10 years ago. Direct link to Ty Anandaraj's post “how would you factor an e...” how would you factor an expression like this: x^3-x^2*y-x*y^2+y^3? please reply!! • (4 votes) Kim Seidel 10 years agoPosted 10 years ago. Direct link to Kim Seidel's post “1) Always look to see if...” 1) Always look to see if there is a common factor contained in all terms. In this case there isn't. If there had been, factor out the common factor as your 1st step. (9 votes) Hudjefa 10 years agoPosted 10 years ago. Direct link to Hudjefa's post “Is there a way to check i...” Is there a way to check if an expression is not factorizable? Like in arithmetic we have a check for divisibility by numbers e.g. if the sums of the digits of a number don't add up to a multiple of 3, that number is not divisible by 3. • (4 votes) Megan Morgan 10 years agoPosted 10 years ago. Direct link to Megan Morgan's post “Everything is factorable,...” Everything is factorable, the question is: Is it pretty, and is it a real number? (6 votes) Map1211 10 years agoPosted 10 years ago. Direct link to Map1211's post “What do you do with that ...” What do you do with that left over Y? • (5 votes) doctorfoxphd 10 years agoPosted 10 years ago. Direct link to doctorfoxphd's post “If you mean the y in the ...” If you mean the y in the 11xy, it is not left over. (3 votes) Meerah Tahir 6 years agoPosted 6 years ago. Direct link to Meerah Tahir's post “how do we know in 11xy in...” how do we know in 11xy in the example whether 11x or 11y is the coefficient? • (1 vote) MinervaGao 6 years agoPosted 6 years ago. Direct link to MinervaGao's post “A coefficient is always ...” A coefficient is always the number in front of the variable. (4 votes) krishna chivukula 6 years agoPosted 6 years ago. Direct link to krishna chivukula's post “is there a way to factor ...” is there a way to factor this by grouping? • (2 votes) Kim Seidel 6 years agoPosted 6 years ago. Direct link to Kim Seidel's post “Yes, you can. You find f...” Yes, you can. You find factors of 30 that add to 11. There are 6 and 5 Hope this helps. (2 votes) Alex Snyder 7 years agoPosted 7 years ago. Direct link to Alex Snyder's post “How would you factor two-...” How would you factor two-variable quadratics when both the x^2 and y^2 terms don't have a coefficient of one? • (1 vote) Kim Seidel 7 years agoPosted 7 years ago. Direct link to Kim Seidel's post “You can use factoring by ...” You can use factoring by grouping. When you expand the middle term into 2 terms, make sure both terms carry "xy" variables. (3 votes) Macoy 5 years agoPosted 5 years ago. Direct link to Macoy's post “What if you have two vari...” What if you have two variables like y^2+5x+4 / y^2-3x-4 ? • (1 vote) Kim Seidel 5 years agoPosted 5 years ago. Direct link to Kim Seidel's post “Your polynomials are not ...” Your polynomials are not factorable. So, your fraction is fully reduced in its current form. (2 votes) Melissa 10 years agoPosted 10 years ago. Direct link to Melissa's post “How do you factor an expr...” How do you factor an expression like 2x^2-8y^2+16y-8? I can figure out the answer through trial & error, but if there's a logical way to get to it, I'm not seeing it. • (1 vote) Hudjefa 10 years agoPosted 10 years ago. Direct link to Hudjefa's post “This is how I would facto...” This is how I would factorize 2x^2 - 8y^2 + 16y - 8 (2 votes)Want to join the conversation?
18x^2 + 3yx - 10y^2, you pretend the y terms are the numerical portions of the grouping. (I rewrote 3xy as 3yx to make this more obvious.) So you need 2 terms that multiply together to make -18*10y^2, and add up to 3y. Well, looking at the factors of 180, -12 and 15 work, so -12y and 15y are the factors. That yields 18x^2 - 12yx + 15yx - 10y^2, which undistributes to 6x(3x-2y) + 5y(3x-2y), and then factoring out the 3x-2y gives (3x-2y)(6x+5y). And we are done!™
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30x^2 + 11xy + y^2
30x^2 + 5xy + 6xy + y^2
5x(6x+y) + y(6x+y)
=(5x+y)(6x+y)
Multiplying it out, i get back to the beginning.
I did the problem in the previous video the same way. Is this correct? Or am i doing it wrong and getting the right answers just by accident?
2) Next - as there are 4 terms, factor by using grouping. Look for common factors in the 1st 2 terms, then look for common factors in last 2 terms. Here are the steps for grouping:
x^3-x^2*y-x*y^2+y^3 =
x^2 (x - y) -y^2 (x - y) =
(x - y) (x^2 - y^2)
3) The new expression is not completely factored as the 2nd binomial is a difference of 2 squares, so it can be factored further. This gives you:
(x - y) (x - y) (x + y)
The fast way is look at the rules for synthetic division and Descartes's Rule of Signs.
Sal is ignoring it a while because it is easiest and best to ignore it when you are factoring. The y part of the quadratic is the variable that is easier to factor.
(The only way you can get y² in one of these factoring and multiplication problems is to multiply y times y.)
After you have figured out the more difficult variable (30x²) and written down the
(y + 5x)(y + 6x),
then you can and should check by multiplying the two binomials. You can either FOIL or distribute. First the y
( y + 5x)(y + 6x) → y² + 6xy
Next the 5x
( y + 5x)(y + 6x) → y² + 6xy + 5xy + 30x²
Combine the like terms to complete your check
y² + 11xy + 30x²
Ta da! That means your (y + 5x)(y + 6x) binomial factors were the correct factors for this polynomial in two variables. And notice that there were no extra y's
In this case, 11 is with the variable xy.
So 11 is coefficient
Expand the quadratic: 30x^2 + 6xy + 5xy + y^2
Then use grouping: 6x (5x + y) + y (5x + y) = (5x + y)(6x + y)
I'm not sure if I can answer this. It's part of my assignment in my math book. I'm asking how to solve this problem...
2(x^2 - 4y^2 + 8y - 4)
2[x^2 - 4(y^2 - 2y + 1)]
2[x^2 - 4(y - 1)^2]
2[x^2 - {2(y - 1)}^2]
2[x^2 - (2y - 2)^2]
2[x - (2y - 2)][x + (2y - 2)]
2(x - 2y + 2)(x + 2y - 2)
TRIAL AND ERROR is a valid method in mathematics. This method:
1. gives us insight into the problem
2. it is used when there are only a few possible answers
3. it is used when you can systematically check all possibilities
4. it is used when there is no other way to solve the problem
Proof of the four color map theorem is a type of trial and error proof. It is also called brute force search.
I think this is a bit off-topic but I thought you'd like to know. If there is a logical method to approach your question I don't know.
