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How do you factorise?

  1. Jan 28, 2010 #1
    I know that factorising means expanding an equation and adding brackets but what is the process that you follow to do that?

    For example, how would you factorise:

    4x squared - 4

  2. jcsd
  3. Jan 28, 2010 #2
    2 (2 x^2-1)
  4. Jan 28, 2010 #3


    Staff: Mentor

    Factoring (as we on this side of the Atlantic call it) is not something that is done to an equation. Factoring an expression means writing the expression as a product of factors. For your example, 4x2 - 4 = 4(x2 - 1) = 4(x - 1)(x + 1). The original expression is now factored. Alternatively, you could go one small step further to 22(x - 1)(x + 1), so that every factor is now broken down into prime factors.

    Note that 4x2 - 4 [itex]\neq[/itex] 2(2x2 - 1), as Tevion asserted in another post in this thread. If you expand (i.e., multiply) the two factors shown on the right side, you get 4x2 - 2, which is different from 4x2 - 4.

    If you factor an expression, you can always check that your factors are correct by multiplying the factors. If the factors are correct, multiplying them together should get you the original unfactored expression.
  5. Jan 28, 2010 #4


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    The easy part of factoring is identifying common factors and pulling them out. In your example, for instance, the number 4 is common to both factors, so you get


    The hard part is factoring polynomials. Recognizing a few common patterns helps in many situations. One pattern you need to recognize is the difference of squares:

    [tex]x^2-a^2 = (x+a)(x-a)[/tex]

    Identifying this pattern in your sample problem allows you to complete its factorization into 4(x+1)(x-1). You should also be able to recognize a perfect square:

    [tex]x^2+2bx+b^2 = (x+b)^2[/tex]

    For example, x2-6x+9 = [x+(-3)]^2.

    If you're asked to factor something like x2+8x+15, however, you essentially end up making educated guesses. This polynomial factors into (x+3)(x+5). Multiplying it out, you see the 3 and 5 multiply to give the constant term 15, and the 8x results from the sum of 3x and 5x. What you're trying to do is reverse this process. So look at the constant term, including its sign, and figure out pairs of factors it has; then sum those pairs to see if it gives you the right middle term. If it doesn't, try a different pair.

    For example, suppose you're asked to factor x2-5x-6. The possible pairs of factors of -6 are: -1 and 6, 1 and -6, 2 and -3, and -2 and 3. Only the second pair, 1 and -6, sums to -5, the coefficient of the middle term, so you have

    [tex]x^2-5x-6 = (x+1)(x-6)[/tex]

    It may turn out that none of the pairs works, which just means you can't factor it by hand. There are also cases, like x2+1, which you can't factor without using complex numbers. In both cases, the quadratic equation can help in factoring or identifying that you can't factor it without using complex numbers.

    There are, obviously, more difficult cases, like higher-order polynomials and a coefficient on the x2 term that isn't equal to 1, but this should get you started.
  6. Jan 28, 2010 #5


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    You learn it by practice. Your textbook has surely shown you some specific methods to use, and some powerful relevant theorems. But factoring by hand is mainly exercise in manipulating expressions and to program your algebraic intuition to instantly recognize many useful facts.

    Actual honest-to-goodness algorithms for factoring are generally things you would probably never want to do by hand.

    In other words, your goal is to learn how to instantly recognize and apply the "easy" stuff. And pretty much every problem you are given has an "easy" solution.
  7. Jan 29, 2010 #6


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    In general, you learn to factor by doing a lot of multiplication!

    For example, after you have notice that (x- 2)(x+ 2)= [itex]x^2- 2x+ 2x+ 4= x^2- 4[/itex], [itex](a+ b)(a- b)= a^2+ ab- ab- b^2= a^2- b^2[/itex], and [itex](x- y)(x+ y)= x^2- xy+ xy- y^2= x^2- y^2[/itex], you start to realize that multiplying a "sum and difference" always gives a "difference of squares" and so you can factor a "difference of squares" as a "product of sum and difference": [itex]x^2- 1[/itex] is the difference of [itex]x^2[/itex] and [itex]1^2[/itex] so [itex]x^2- 1= (x- 1)(x+ 1)[/itex].

    More generally [itex](x- a)(x- b)= x^2- (a+b)x+ ab[/itex] so if I need to factor, say, [itex]x^2- 5x+ 6[/itex], I start looking for "a" and "b", factors of 6, which will add to 5. I would soon realize that 2(3)= 6 and 2+ 3= 5 so (-2)+ (-3)= -5 but still multiply to give 6. [itex]x^2- 5x+ 6= (x- 3)(x- 2)[/itex].

    Of course, if it were [itex]x^2+ 4x+ 6[/itex], I would realize that there are no integer factors of 6 which add to 2 so this cannot be factored with integer coefficients.

    Normally, when we talk about "factoring" a polynomial, we mean with integer coefficients. The polynomial above can be factored with irrational and/or complex coefficients but to find the factors I would first have to solve the equation [itex]x^2+ 4x+ 6= 0[/itex] by the quadratic formula or by completing the square.
  8. Jan 30, 2010 #7
    thanks a lot everybody. I'm getting the idea now.. slowly!
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