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Thread: I forgot how to reverse foil (such as X^2 + 2X + 2)

  1. #11
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    Quote Originally Posted by Quick View Post
    But I was thinking along these lines:

    X^2+40X+6

    x^2+2(20x+3)

    But I don't think this actually changes the equation to the point that you can Factor from that point...
    Right; that's useless for your goal.

    As I said previously, "This polynomial, like the first, can't be factored over the integers (that is, factored into factors containing only integers)." In other words, you can't do what you want to do. More than that, I can't do what you want to do -- no one can do it, unless you expand the goal to allow irrational numbers.

    The fact is, factoring is not possible in general. If you make up a polynomial randomly, it will most likely be unfactorable (in the sense you are asking for).

  2. #12
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by mmm4444bot View Post
    There's a shortcut for determining whether a quadratic polynomial (Ax^2+Bx+C) factors nicely. Determine its Discriminant (B^2-4AC). If the Discriminant is not a perfect square, then the polynomial does not factor nicely.
    X^2 + 40X + 6

    A = 1
    B = 40
    C = 6

    Discriminant = 40^2 - 4(1)(6) = 1576

    sqrt(1576) ≈ 39.6989

    The Discriminant is not a perfect square. Hence, the given polynomial does not factor nicely.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  3. #13
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    Quote Originally Posted by Dr.Peterson View Post
    Right; that's useless for your goal.

    As I said previously, "This polynomial, like the first, can't be factored over the integers (that is, factored into factors containing only integers)." In other words, you can't do what you want to do. More than that, I can't do what you want to do -- no one can do it, unless you expand the goal to allow irrational numbers.

    The fact is, factoring is not possible in general. If you make up a polynomial randomly, it will most likely be unfactorable (in the sense you are asking for).
    Thanks for answering.

    I would just like to add that in this case, though it doesn't forward my goal to try and simplify part of the equation to solve the equation, it is my goal to learn, and that is what I have done in this thread.

    I know I am maybe a bit toooo experimental in my approach in learning what works and what doesn't, but this is the way I learn best...

    I had a goal to learn math from Khan Academy (and it's still a valuable recourse if I have something specific I want to look up), but if I can keep asking questions here, I think this is a way to learn that suits me a lot better. I would like to be at the point where I know basic algebra fairly well by the time I enroll in college, and I think I am on that pace so far. So that said, I really do appreciate the mods here - especially considering they are all volunteers.

  4. #14
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by Quick View Post
    I really do appreciate the mods here
    Quick note: you've been confusing "mods" with "volunteer tutors".

    We have well over a dozen tutors who volunteer regularly. The forum currently has four active moderators. The moderators handle housekeeping tasks and policy issues, in addition to tutoring.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  5. #15
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    Quote Originally Posted by mmm4444bot View Post
    Quick note: you've been confusing "mods" with "volunteer tutors".

    We have well over a dozen tutors who volunteer regularly. The forum currently has four active moderators. The moderators handle housekeeping tasks and policy issues, in addition to tutoring.
    Oh. Thanks for explaining that. It makes more sense.

  6. #16
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Quick View Post
    Thanks, I think I have the basic concept down now.

    You have to take the equation as a whole and you can't split it up into different sections. That is what I was trying to figure out.
    I'm not sure what you mean by "splitting into different sections"...? Also, a quadratic, such as what you've posted, is not an "equation". An equation is a statement of equality; thus, it has an "equals" sign (and, possibly, can be solved). What you have posted are "expressions", which can be "simplified", "factored", etc.

    To learn about finding the greatest common factor in polynomial expressions, please try here. To learn various methods for factoring quadratics, please try here. Thank you!

  7. #17
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    Quote Originally Posted by mmm4444bot View Post
    X^2 + 40X + 6

    A = 1
    B = 40
    C = 6

    Discriminant = 40^2 - 4(1)(6) = 1576

    sqrt(1576) ≈ 39.6989

    The Discriminant is not a perfect square. Hence, the given polynomial does not factor nicely.
    I see.

    how do you get the 4 in there?

    As I understand it if you have anything that could equal a perfect square from the formula above then it it factorable correct?

    So if you got something that ended up equaling 40^2 (or any number squared) then it is factorable. I assume it's required that B squared has to be low enough for the remaining -4(a)(c) to get a sum to another higher square, right? So then this would mean that B has to be lower than C unless you are dealing with negatives, right?

  8. #18
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by Quick View Post
    how do you get the 4 in there?
    The factor 4 is a part of the formula for the Discriminant.

    Quote Originally Posted by mmm4444bot View Post
    There's a shortcut for determining whether a quadratic polynomial (Ax^2+Bx+C) factors nicely. Determine its Discriminant (B^2-4AC). If the Discriminant is not a perfect square, then the polynomial does not factor nicely.



    Quote Originally Posted by Quick View Post
    As I understand it if you have anything that could equal a perfect square from the formula above then it it factorable correct?
    If the value of the Discriminant is a perfect square, then, yes, the quadratic polynomial will factor nicely.


    Quote Originally Posted by Quick View Post
    So if you got something that ended up equaling any number squared then it is factorable.
    If you got a Discriminant that ends up equaling any Rational number squared, then the polynomial factors nicely.

    For me, a nice factorization doesn't necessarily mean that only Integers appear with the factors. There can be nice factorizations involving Rational numbers. (Plus, the definition of a perfect square IS the square of a Rational number.) Here's a quick example:

    x^2 + x/6 - 1/3 = (x - 1/2)(x + 2/3)

    The Discriminant for this quadratic polynomial is 49/36. That's a perfect square.


    Quote Originally Posted by Quick View Post
    I assume it's required that B squared has to be low enough for the remaining -4(a)(c) to get a sum to another higher square, right? So then this would mean that B has to be lower than C unless you are dealing with negatives, right?
    I've got to leave for dinner. I will think about this, when I return. Cheers
    Last edited by mmm4444bot; 01-28-2018 at 08:24 AM. Reason: added definition of 'perfect square'
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  9. #19
    Quote Originally Posted by Quick View Post
    Thanks, I think I have the basic concept down now.

    You have to take the equation as a whole and you can't split it up into different sections. That is what I was trying to figure out.
    You can, however, deduce a few important things about your factorization by looking at the equation and thinking about it logically, to reduce the work to find the answer. Let's reverse engineer 2x^2 + 17x + 21, since it was provided above.

    First obvious thing is x^2. It means we're going to have an X on the same side of both equations. (x...something)(x...something)
    Then we look at the signs in the equation above. + and +. The ONLY way we're going to see this is if both equations have + operators in them. (x + something)(x + something)
    Now we look at the 2 in the 2x^2. The ONLY factors for this are 1 and 2, so we just plug them into the equations. (2x + something)(1x + something)

    All that's left at this point is to sort out how we can get +21. The possible factors here are (1 and 21) and (3 and 7). Since we know we're only adding, we can't add anything more to 21 to get a sum of 17, so we can discard it as a possible solution, which means 3 and 7 have to be the factors. We just need to try and place them with the 2 and 1 so that they add up to 17.

    (2x + 7)(x + 3).. if we just mentally plug them in like this, we can quickly see that 6x + 7x isn't going to give us 17x... All we have to do is swap them and then try again.
    (2x + 3)(x + 7).. 14x + 3x = 17x...

    PING! We've deduced the solution with just a little logical thinking and very little hard math-stuff needed at all.

    Just break it down step by step and you can often find the solution to these type problems in your head, without even needing a piece of paper to scribble notes down on. (Not that your teacher will approve of you just writing the answer, if you calculate it mentally, as that makes it hard to complete the "show your work" requirement.)

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