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Factoring a polynomial

  1. May 29, 2014 #1
    hi all

    I am stuyding how to factor equations such as :-

    7x^2-9X-6

    The problem i have is that it takes me too long to find 2 numbers whose sum is D and the same numbers whoes product is E.

    Is there any way/tips/guide on how i can achieve this quickly?
     
  2. jcsd
  3. May 29, 2014 #2

    Ray Vickson

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    Have you studied how to solve quadratic equations? If so, you will know that the
    quadratic equation ##7 x^2 - 9 x - 6 = 0## has roots ##r_1## and ##r_2##, and that means that the expression ##7 x^2 - 9 x - 6 ## factors as ## 7(x - r_1)(x-r_2)##. You can use the standard quadratic root formulas to find ##r_1## and ##r_2##.

    BTW: the thing you are factoring is not an equation; it is an expression. An equation must have an '=' sign in it, and yours does not.
     
  4. May 29, 2014 #3
    what about:-

    X^2+xy-2y^2
    6x^2+5xy-6y^2

    What is the process of factoring such expressions?
     
  5. May 29, 2014 #4

    Ray Vickson

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    Exactly the same as what I already explained.
     
  6. May 29, 2014 #5

    HallsofIvy

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    [itex]x^2+ xy- 2y^2= 0[/itex]
    [itex]x^2+ xy+ (1/4)y^2- (1/4)y^2- 2y^2= 0[/itex]
    [itex](x+ (1/2)y)^2- (9/4)y^2= 0[/itex]
    That's of the form "[itex]a^2- b^2= (a- b)(a+ b)[/itex]with a= x+ (1/2)y and b= (3/2)y
    so [itex]x^2+ xy- 2y^2= (x+ (1/2)y+ (3/2)y)(x+ (1/2)y- (3/2)y)= (x+ 2y)(x- y)[/itex]

    That's one way to do it- very "algorithmic". Another way, more "heuristic", is to note that the only way to factor [itex]x^2[/itex] (with integer coefficients) is (x)(x) while there are two ways to factor [itex]-2y^2[/itex] (again with integer coefficients) is (-2y)(y) or (2y)(-y). We try (-2y)(y): [itex](x- 2y)(x+ y)= x^2- 2xy+ xy- 2y^2= x^2- xy- 2y^2[/itex]. No, that's not what we want. So try (-y)(2y): [itex](x+ 2y)(x- y)= x^2+ 2xy- xy- 2y^2= x^2+ xy- 2y^2[/itex]. Yes, that's what we want.

    [itex]6x^2+ 5xy- 6y^2= 6(x^2+ (5/6)xy- y^2)= 6(x^2+ (5/6)xy+ (25/144)y^2- (25/144)y^2- y^2)= 6((x+ 5/12)^2- (169/144)y^2)= 6((x+ 5/12)^2- ((13/12)y)^2)= 6(x+ 5/12+ 13/12)(x+ 5/12- 13/12))= 6(x+ 18/12)(x- 8/12)= 6(x+ 3/2)(x- 2/3)= (2x+ 3)(3x- 2)[/itex].

    Or: 6=(3)(2) so we must have one of (3x- 3y)(2x+ 2y), (2x- 2y)(3x+ 3y), (3x- 2y)(2x+ 3y), or (2x- 3y)(3x+ 2y). Multiplying each of those out, we see that
    [itex](3x- 3y)(2x+ 2y)= 6x^2+ 6xy- 6xy- 6y^2= 6x^2- 6y^2[/itex]
    [itex](2x- 2y)(3x+ 3y)= 6x^2+ 6xy- 6xy- 6y^2= 6x^2- 6y^2[/itex]
    [itex](3x- 2y)(2x+ 3y)= 6x^2+ 9xy- 4xy- 6y^2= 6x^2+ 5xy- 6y^2[/itex]
    [itex](2x- 3y)(3x+ 2y)= 6x^2+ 4xy- 9xy- 6y^2= 6x^2- 5xy- 6y^3[/itex]
    so that (3x- 2y)(2x+ 3y) is the correct factoring.

    And we see that
     
    Last edited: May 29, 2014
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