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Factoring two-variable quadratics: rearranging

Sal rearranges 30x^2+11xy+y^2 as y^2+11xy+30x^2 and then factors it as (y+5x)(y+6x). Created by Sal Khan.

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  • leaf red style avatar for user Jack McClelland
    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)
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  • aqualine ultimate style avatar for user Toni
    I dont understand the way Sal's doing these. I paused at and factored it by grouping, because it seemed like the obvious way to go. Like this:
    30x^2 + 11xy + y^2
    30x^2 + 5xy + 6xy + y^2
    5x(6x+y) + 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?
    (17 votes)
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    • duskpin ultimate style avatar for user learn
      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)
  • leaf red style avatar for user Ty Anandaraj
    how would you factor an expression like this: x^3-x^2*y-x*y^2+y^3? please reply!!
    (4 votes)
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    • stelly blue style avatar for user Kim Seidel
      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.
      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)
      (9 votes)
  • leaf green style avatar for user Hudjefa
    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)
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  • blobby green style avatar for user Map1211
    What do you do with that left over Y?
    (5 votes)
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    • purple pi purple style avatar for user doctorfoxphd
      If you mean the y in the 11xy, it is not left over.
      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
      (3 votes)
  • winston baby style avatar for user Meerah Tahir
    how do we know in 11xy in the example whether 11x or 11y is the coefficient?
    (1 vote)
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  • leaf green style avatar for user krishna chivukula
    is there a way to factor this by grouping?
    (2 votes)
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  • aqualine seedling style avatar for user Alex Snyder
    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)
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  • spunky sam blue style avatar for user Macoy
    What if you have two variables like y^2+5x+4 / y^2-3x-4 ?
    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...
    (1 vote)
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  • primosaur tree style avatar for user Melissa
    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)
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    • leaf green style avatar for user Hudjefa
      This is how I would factorize 2x^2 - 8y^2 + 16y - 8
      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.
      (2 votes)

Video transcript

Let's see if we can use our existing factoring skills to factor 30x squared plus 11xy plus y squared. And I encourage you to pause the video and see if you can handle it yourself. Now, the first hint I will give you-- and this might open up what's going on here-- is to maybe rearrange this a little bit. We could rewrite this as y squared plus 11xy plus 30x squared. And my whole motivation for doing that-- there are ways to factor a quadratic where your first coefficient, your coefficient on this first term, is something other than 1. But we haven't seen that yet. And so rearranging it this way, this got us a little bit more into our comfort zone. Now our coefficient is a 1 on the y squared term. So now we can start to think of this in the same form that we've looked at some of the other factoring problems. Can we think of two numbers whose product is 30x squared and whose sum is 11x? Notice, 11x is the coefficient on y. We have y squared, some coefficient on y. And then in terms of y, this isn't in any way dependent on y. So one way to think about this, if you knew what x was, then this would be a quadratic in terms of y. And that's how we're really thinking about it here. So can we find two numbers whose product is 30x squared and two numbers whose sum is the coefficient on this y term right here, whose sum is 11x? So let's just think about all of the different possibilities. If we were just thinking about two numbers whose product was 30 and whose sum was 11, we would be thinking of 5 and 6. 5 times 6 is 30. 5 plus 6 is 11. It's some trial and error. You could have tried 3 and 10. Well, that would have been-- 13 would be their sum. You could have tried 2 and 15. That wouldn't have worked. But 5 and 6 does work here, so we've already seen that multiple times. So 5 and 6 would work for 30, but we have 30x squared. So what if we have 5x and 6x? Well, 5x times 6x is 30x squared, and 5x plus 6x is 11x. So this actually works. So then our factoring or our factorization of this expression is just going to be y plus 5x times y plus 6x. And I'll leave it up to you to verify that this does indeed, when you multiply it out, equal this up here.