
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 patternrecognition once you catch on to how to do them.
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 x^{2}", rather than "in x". In fact, I can rewrite the expression as:
(x^{2})^{2} – 2(x^{2}) – 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 x^{2} at the beginning of each factor: (x^{2} )(x^{2} ) Factors of –8 that add to –2 are –4 and +2 , so I get: (x^{2} – 4)(x^{2} + 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, x^{2} – 4, that I can factor: x^{4} – 2x^{2} – 8 = (x^{2} – 4)(x^{2} + 2) = (x – 2)(x + 2)(x^{2} + 2) Some books will encourage you to switch variables, plugging in a "y" for the "x^{2}", so you'd get: y^{2} – 2y – 8 This is clearly a quadratic, which you can factor as usual: y^{2} – 2y – 8 = (y – 4)(y + 2) Then you'd plug the "x^{2}" back in for the "y" to get: x^{4} – 2x^{2} – 8 It is not necessary to do that substitution of "y" for "x^{2}"; 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.
The power on the middle term is 3, which is half of the power on the leading term, so this is a quadratic in x^{3}. I will factor as usual: Copyright © Elizabeth Stapel 19992011 All Rights Reserved x^{6} + 6x^{3} + 5 Since the second of these factors is a sum of cubes, I can factor further: x^{3} + 1 = (x)^{3} + (1)^{3} = (x + 1)(x^{2} – x + 1) Then the complete answer is: x^{6} + 6x^{3} + 5 = (x^{3} + 5)(x + 1)(x^{2} – 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.
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 x^{1/3}. I will factor as usual: x^{2/3} – x^{1/3} – 6
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 x^{1/2}. I will factor as usual: x + 5x^{1/2} + 4
This is just a difference of squares. I will factor as usual: 4x^{4} – 25
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 Since neither of these quadratics factors, I am done. (x – 3)^{4} + 2(x – 3)^{2} – 8 = (x^{2} – 6x + 13)(x^{2} – 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 fourthdegree polynomial: x^{4} – 12x^{3} + 56x^{2} – 120x + 91. << Previous Top  1  2  3  4  Return to Index


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