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Factoring Quadratics: The Hard Case:
     Examples of How to Use "Box"
(page 3 of 4)

Sections: The simple case, The hard case, The weird case


Note that "box" works only if you have first removed all common factors. For instance:

  • Factor 2x2 – 4x – 16
  • First, you would first need to remove the common factor of 2 from each term, to get 2x2 – 4x – 16 = 2(x2 – 2x – 8). Then factor the remaining quadratic: x2 – 2x – 8 = (x – 4)(x + 2). Just be careful not to forget the factored-out "2" when you write down your final answer:

      2x2 – 4x – 16 = 2(x – 4)(x + 2).

The one special case that often caused a bit of trouble is when you have a leading coeffiecient of negative one. In this case, factor out the –1. For instance:

  • Factor –6x2x + 2
  • First, you would first take out the minus one to get 6x2 x + 2 = 1(6x2 + x 2). (Remember that every sign changes when you multiply or divide by a negative. Don't fall into the trap of taking the 1 out of only the first term; take it out of all three!) Factoring what is left you get:

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

    Then:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

      6x2 x + 2 = 1(6x2 + x 2) = –1(2x – 1)(3x + 2)

Putting these two points together (factor out anything common, and take out a leading negative sign), you can handle such problems as this:

  • Factor –6x2 + 15x + 36
  • First, you would first remove the common factor of 3, taking the leading negative with it: –6x2 + 15x + 36 = –3(2x2 – 5x – 12). Then factor the remaining quadratic: 2x2 – 5x – 12 = (x – 4)(2x + 3):

      2x^2 - 5x - 12 = (2x + 3)(x - 4)

    When you write down your answer, remember to include the –3 factor:

      –6x2 + 15x + 36 = –3(x – 4)(2x + 3).

A disguised version of this factoring-out-the-negative case is when they give you a backwards quadratic where the squared term is subtracted. For example, if they give you something like 6 + 5x + x2, you would just reverse the quadratic, to put it back in the "normal" order, and then factor: 6 + 5x + x2 = x2 + 5x + 6 = (x + 2)(x + 3). You can do this because order doesn't matter in addition. However, in subtraction, order does matter, and you need to be careful with signs. For instance:

  • Factor 6 + xx2
  • First, you will want to reverse the quadratic, but you'll need to take care with the signs: 6 + xx2 = –x2 + x + 6. Then factor out the –1, and factor the remaining quadratic:

      x2 + x + 6 = –1(x2x – 6) = –1(x + 2)(x – 3)

There is one other type of quadratic that looks kind of different, but it works in exactly the same way:

  • Factor 6x2 + xy – 12y2.
  • This may look bad, what with the y2 at the end, but it factors just like all the ones above. Remember, from the "simple" case, that we knew that the factors had to be of the form:

      (x + something)(x + something else)

    ...because we knew we'd multiplied factors that looked like this in order to get the quadratic in the first place. This was how we knew that we needed x's in the fronts of our parentheses. In the same way, we know that we must have multiplied factors of the form:

      (an x term + a y term)(another x term + another y term)

    ...to get that y2 term at the end of our quadratic. So we'll need to put y's at the ends of our parentheses. But, other than this, the process will work as usual. First, I need to find factors of (6)(–12) = –72 that add to +1; I"ll use +9 and –8.  Then "box" gives me:

      6x^2 + xy - 12y^2 = (3x - 4y)(2x + 3y)

    So 6x2 + xy – 12y2 factors as (2x + 3y)(3x – 4y).

Quadratic factoring can pop up in even more exotic forms than the last example above....

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Cite this article as:

Stapel, Elizabeth. "Factoring Quadratics: The Hard Case (Examples)." Purplemath. Available from
    http://www.purplemath.com/modules/factquad3.htm. Accessed
 

 

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