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Homework Help: Factoring xsquared - 1

  1. Dec 5, 2011 #1
    I know this is real simple stuff but can someone show me how you get xsquared -1 to be (x+1) (x-1)?
  2. jcsd
  3. Dec 5, 2011 #2


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    Think of the graph y=x2-1 and consider where it cuts the x-axis - which means to let y=0. We get [itex]x=\pm 1[/itex] right?

    So these would be the roots of the quadratic and we can always factorize a polynomial by finding its linear roots and each factor being x-a where a is the root, thus (x-(1))(x-(-1))=(x-1)(x+1)
  4. Dec 5, 2011 #3
    There has to be a simpler answer than that.
  5. Dec 5, 2011 #4


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    Sure, expand (x+1)(x-1) and observe that it works :wink:

    I reckon it was pretty simple, and it works for all quadratics that cut the x-axis. The quadratic formula is pretty useful, so you should practice using it to the point that it does become simple and easy.
  6. Dec 5, 2011 #5
    That is right.

    I just have forgotten so much Math I need to catch up.
  7. Dec 6, 2011 #6


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    x^2 - 1

    = x^2 + 1 - 2

    = x^2 + 2x + 1 - 2 - 2x

    = (x+1)(x+1) - 2(x+1)

    = (x+1)(x+1-2)

    = (x+1)(x-1)
  8. Dec 6, 2011 #7


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    With all due respect, I don't think this is helpful to the OP, who said that he/she has "forgotten so much Math I need to catch up." If he/she wishes to verify a factoring problem, he/she can just FOIL the two binomials, as Mentallic said.
  9. Dec 6, 2011 #8
    Right, I have no idea what is going on in Curious's example. FOIL method seems to be ok for now.
  10. Dec 6, 2011 #9


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    The OP should memorize the standard identity:

    a2-b2 = (a-b)(a+b)

    This comes in handy. Try to do the following examples:

    x2 - 16 = ?
    4x2 - 1 = ?
    x4 - 100 = ?

    If you can do these three, you're in pretty good shape.
  11. Dec 6, 2011 #10
    Curious was making x^2 into (x+1)^2 and seeing what was left over.

    More interestingly, try mulitplying out and simplifying (x-1)(x2+x+1)

    and then (x-1)(x3+x2+x+1)
  12. Dec 6, 2011 #11
    I can do these but I am doing them more through intuition of the formula than a real understanding at this point.

    It's easier for me to factor (a-b)(a+b) into a2-b2 than to do the reverse.
  13. Dec 6, 2011 #12


    Staff: Mentor

    It would be helpful to you to get the terminology straight. You aren't "factoring" (a - b)(a + b). You already have the factors and are multiplying them to get a2 - b2. Factoring and multiplying (expanding) are opposite operations.

    When you factor an expression, you rewrite it as the product of two or more expressions that are being multiplied. The expressions x2 - 9 and y2 + 4y + 4 can be factored into, respectively, (x - 3)(x + 3) and (y + 2)(y + 2). Part of being able to factor is to have enough practice at multiplying two binomials to recognize certain patterns.
  14. Dec 6, 2011 #13


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    If you want some of an understanding of why we do it, then just look back at my first post in this thread. It makes it easy to find where the x-axis is cut by a parabola, thus to get a rough idea of sketching a pretty accurate parabola.
  15. Dec 6, 2011 #14
    That sure is taking the long way around. :tongue2:
  16. Dec 7, 2011 #15


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    Staff: Mentor

    Yes. But it's a handy identity to memorize, and may be useful if you are returning to any maths study.
  17. Dec 7, 2011 #16


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    Of course. Expanding is always taught before factorizing for exactly that reason.

    Squaring is taught before square roots because the former is easier to do.
    Adding before subtracting.
    Multiplying before dividing.

    But as you can see, all of these techniques are handy to learn for their own reasons, whether or not they're harder than some crude substitute.
  18. Dec 8, 2011 #17
    Here is a lot easier method :

    = x2 + x -x - 1
    Now factorize the above as given below ->
    = x(....)-1(....)

    You can proceed similarly.

    Here is the same method used again :

    =a2 - ab + ab -b2
    Now factorize this as done above.

    This method is handy according to me.:wink:
  19. Dec 8, 2011 #18


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    Or even...
    Think about a square of side length x that's missing 1 "cross". So we have something like this:

    X X X X X
    X X X X X
    X X X X X
    X X X X X
    X X X X

    In this case x=5, but I can't be bothered making it look like an unknown variable integer :tongue:

    So this represents x2-1, and now as you can see, the bottom row can be taken and moved to become another column.

    X X X X X X
    X X X X X X
    X X X X X X
    X X X X X X

    So what is this? It's a rectangle! And it's dimensions are x+1 columns and x-1 rows. Thus, there are (x+1)(x-1) crosses, which is equivalent to the x2-1 crosses we began with.
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