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Homework Help: Factorising - 2 brackets

  1. Feb 17, 2010 #1
    Can anyone help me with the following 2 factorising problems?

    x squared - 7 squared + 12

    6x squared + 5x - 4

    I know that the answers are:
    (x-3) (x-4)
    (3x + 4) (2x - 1)

    I guess that in the first one, the solution has something to do with the fact that 3 & 4 add together to make 7 amd multiply to make 21?
  2. jcsd
  3. Feb 17, 2010 #2

    Char. Limit

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    If by "multiply to make 21" you mean "multiply to make 12", yes. Simple typo, no problem.

    Of course, you are dealing with -3 and -4 here, the middle term (-7x, not -7 squared) is the sum of these two and the last term (12 here) is the product of these two.

    Of course, once you start using x coefficients it gets harder (3x and 2x) to see the solutions intuitively.
  4. Feb 17, 2010 #3


    Staff: Mentor

    From your answer, the expression you want to factor (as we say it here in the US) is x2 - 7x + 12.

    The idea is to factor this trinomial into a product of two binomial factors.
    x2 - 7x + 12 = (x + ?)(x + ?)
    What you need to do is find two factors of 12 that add up to -7

    You really should learn how to write the symbols. You can write exponents by using the X2 button in the menu bar above the input window. If that menu bar doesn't appear, click the Go Advanced button just below the input window.

    Also (and again) these -- () -- are parentheses, not brackets. These -- [] -- are brackets.
    You want two binomial factors that multiply to the above. Different possibilities are:
    (6x + ?)(x + ?)
    (3x + ?)(x + ?)

    Where I have the ? characters, you need to have factors of -4: -4 * 1, 4 * -1, -2 * 2, 2 * -2. One of these combinations combined with one of the two binomial pairs of factors above might be the one that works. Be advised, though, that it's not possible to factor all trinomials into binomial factors with integer coefficients.
  5. Feb 18, 2010 #4
    Can anyone help me with the following 2 factorising problems?


    x(x+7) - 21

    x(6x+5) - 4
  6. Feb 18, 2010 #5


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    You should make your own thread.

    Gringo123, maybe this will shed some light on why you need to do what you're doing.
    (by the way, this only works for the first problem where the coefficient of x2 is 1)

    We have (x+a)(x+b) where a and b are some numbers. If we expand this we get: x2+ax+bx+ab=x2+(a+b)x+ab

    Now, we have the problem x2-7x+12 to factorize.

    So, [itex]x^2+(a+b)x+ab\equiv x^2-7x+12[/itex] and equating coefficients, this means that


    If you can solve these simultaneously, you'll have your answers of a and b.
  7. Feb 18, 2010 #6
    Sorry I was just giving factorisations of those two expressions forgot that i left the question in the post
  8. Feb 18, 2010 #7


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    Aha. Sorry I didn't notice it was the same question as the original question the OP gave.
    But then, your answers aren't completely factorized.
  9. Feb 18, 2010 #8
    I believe my answers are more easiliy worked with rather then the double brackets, just showing there are two different ways of doing it.
  10. Feb 18, 2010 #9


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    I disagree.
    In what circumstances would your form of factorizing be easier to work with? You can't find the roots of the function y=x(x+7)-21 easily. Also it doesn't make things much easier to graph.

    However, completing the square is another way of factorizing that is useful when drawing the graph.

    e.g. [itex]x^2-7x+12=(x-7/2)^2-1/4[/itex] tells us the quadratic's minimum point is at (7/2,-1/4).
  11. Feb 18, 2010 #10

    Char. Limit

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    I find that an easier way to calculate the min/max is to find the two roots. The min/max is almost always halfway between them.

    Example, [tex]x^2-1 = (x+1)(x-1)[/tex].

    The root is halfway between -1 and +1, or zero. Look on the graph and this is the case.

    Of course, this wouldn't work for a graph such as [tex]x^2+1 = (x+i)(x-i)[/tex]... wait, it does. This can't be the case for every quadratic with no real roots, however. Or maybe it can. I just know that it does work for most every quadratic with one or two real roots.
  12. Feb 18, 2010 #11


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    I just realized it works as well! :smile:

    And yeah, for every real quadratic it can be done since the requirements for a real quadratic are that if one root is complex, then the other is the complex conjugate.

    e.g. [tex]x^2-2x+2=(x-1)^2+1=(x-(1+i))(x-(1-i))[/tex]

    The thing is though that you must first find the average of the roots to find the axis of symmetry. After this you need to plug it back into the formula to get the y-value. Completing the square does both of these steps in one... so the work load is less overall.

    Also, notice that to find the roots of the above example you need to use the quadratic formula, which is another hassle that's unnecessary.
  13. Feb 18, 2010 #12
    Yes, i've got to agree with you. completely forgot about roots of the functions. (:
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