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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 aren't that hard, once you catch on to how to do them. I'll show some examples.

• Factor x⁴ – 2x² – 8.

At first glance, this does not appear to be a quadratic. But it is -- sort of. Take another look at those exponents. 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, you have the powers 2 and 1, where 1 is half of 2. So this polynomial follows the pattern, and is therefore a quadratic "in x²", rather than "in x". That is, you could rewrite this as:

(x²)² – 2(x²) – 8

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

y² – 2y – 8

This is clearly a quadratic! So factor as usual:

y² – 2y – 8 = (y – 4)(y + 2)

Now plug the "x²" back in for the "y":

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

At this point, you just need to check to see if there is any further factoring possible. In this case, you still have a difference of squares that can factor:

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

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

What follows are some fairly typical examples. Note that there are always three terms (or two, if it's a difference of squares), and the power on the middle term is 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. Factor as usual:   Copyright © Elizabeth Stapel 2006-2008 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 problem.

• Factor x^2/3 – x^1/3 – 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 x^1/3. Factor as usual:

x^2/3 – x^1/3 – 6
    = (x^1/3)² – 1(x^1/3) – 6
    = (x^1/3 – 3)(x^1/3 + 2)

• Factor x + 5x^1/2 + 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 x^1/2. Factor as usual:

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

• Factor 4x⁴ – 25.

This is just a difference of squares. 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. 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, you are done.

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

You can check this last one by multiplying the two factors, and then multiplying out the original expression and verifying that they both simplify to x⁴ – 12x³ + 56x² – 120x + 91.

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