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Factoring Quadratics: The Weird Case (page 4 of 4)

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


This is the case where it doesn't seem like you're factoring a quadratic, but you really are. You'll need to be clever with these, but they reduce to little more than pattern-recognition once you catch on to how to do them.

  • Factor x4 2x2 8.
  • At first glance, this does not appear to be a quadratic. But it is sort of. Taking a closer look at those exponents, I see that the power on the leading term is "4", and the power on the middle term is "2", which is half of 4. With a "regular" quadratic, I would have the powers 2 and 1, where 1 is half of 2. So this polynomial follows the quadratic pattern: it is a quadratic "in x2", rather than "in x". In fact, I can rewrite the expression as:

     

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      (x2)2 2(x2) 8

    My leading coefficient is just 1, so this is an "easy" quadratic. I'll draw my parentheses as usual, but I'll put an x2 at the beginning of each factor:

      (x2      )(x2      )

    Factors of 8 that add to 2 are 4 and +2 , so I get:

      (x2 4)(x2 + 2)

    I still need to check to see if there is any further factoring possible. In this case, I still have a difference of squares, x2 4, that I can factor:

      x4 2x2 8 = (x2 4)(x2 + 2) = (x 2)(x + 2)(x2 + 2)

Some books will encourage you to switch variables, plugging in a "y" for the "x2", so you'd get:

    y2 2y 8

This is clearly a quadratic, which you can factor as usual:

    y2 2y 8 = (y 4)(y + 2)

Then you'd plug the "x2" back in for the "y" to get:

    x4 2x2 8
        = (x2)2 2(x2) 8
        = y2 2y 8
        = (y 4)(y + 2)
        = (x2 4)(x2 + 2)

It is not necessary to do that substitution of "y" for "x2"; some students find it helpful, but many find it confusing. You should do whatever works better for you.


What follows are some fairly typical examples of "weird" quadratics. Note that there are always three terms in the expression (or two, if it's a difference of squares), and the power (or exponent) on the middle term is always half of the power on the leading term.

  • Factor x6 + 6x3 + 5.
  • The power on the middle term is 3, which is half of the power on the leading term, so this is a quadratic in x3. I will factor as usual:   Copyright Elizabeth Stapel 1999-2011 All Rights Reserved

      x6 + 6x3 + 5
          = (x3)2 + 6(x3) + 5
          = (x3 + 5)(x3 + 1)

    Since the second of these factors is a sum of cubes, I can factor further:

      x3 + 1 = (x)3 + (1)3 = (x + 1)(x2 x + 1)

    Then the complete answer is:

      x6 + 6x3 + 5 = (x3 + 5)(x + 1)(x2 x + 1)

In these two examples, after I'd factored the "quadratic", I still had to do some more factoring. This will not always be the case, but will generally be the case on tests when the instructor will be seeing if you're on top of your game. So keep in mind that, just because you've done one factoring step, this doesn't mean that you're done with the exercise.

  • Factor x2/3 x1/3 6.
  • The power on the middle term is 1/3, which is half of the power on the leading term, so this is a quadratic in x1/3. I will factor as usual:

      x2/3 x1/3 6
          = (x1/3)2 1(x1/3) 6
          = (x1/3 3)(x1/3 + 2)

  • Factor x + 5x1/2 + 4.

    The power on the middle term is 1/2, which is half of the power on the leading term, so this is a quadratic in x1/2. I will factor as usual:

      x + 5x1/2 + 4
          = (x1/2)2 + 5(x1/2) + 4
          = (x1/2 + 4)(x1/2 + 1)

  • Factor 4x4 25.
  • This is just a difference of squares. I will factor as usual:

      4x4 25
          = (2x2)2 (52)
          = (2x2 5)(2x2 + 5)

  • Factor (x 3)4  + 2(x 3)2 8.
  • The power on the middle term is 2, which is half of the power on the leading term, so this is a quadratic in (x 3)2. I will factor as usual:

      (x 3)4 + 2(x 3)2 8
          = ((x 3)2)2 + 2(x 3)2 8
          = y2 + 2y 8
          = (y + 4)(y 2)
          = ((x 3)2 + 4)((x 3)2 2)
          = ((x2 6x + 9) + 4)((x2 6x + 9) 2)
          = (x2 6x + 13)(x2 6x + 7)

    Since neither of these quadratics factors, I am done.

      (x 3)4 + 2(x 3)2 8 = (x2 6x + 13)(x2 6x + 7)

You can check the answer this last exercise by multiplying the two polynomial factors, and then multiplying out the original expression and verifying that they both simplify to the same fourth-degree polynomial: x4 12x3 + 56x2 120x + 91.

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Cite this article as:

Stapel, Elizabeth. "Factoring Quadratics: The Weird Case." Purplemath. Available from
    http://www.purplemath.com/modules/factquad4.htm. Accessed
 

 



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