Factoring Polynomials III

Free Math Help Online presents Factoring Polynomials Part III.  In my first post, Factoring Polynomials, I covered factoring out a common factor from all terms, factoring quadratics, and factoring using the basics of the special rules for perfect squares, difference of squares, and sum and difference of cubes.  In Factoring Polynomials II I covered factoring by grouping, and recognizing special forms.  This post will cover factoring negative and fractional exponents, and putting it all together.  My next post will cover Solving Equations.


If you get lost in any of this, please go back to Factoring Polynomials and Factoring Polynomials II first and make sure you understand these before moving on.

1. Negative and Fractional Exponents: Once we have either negative or fractional exponents, we are actually no longer dealing with polynomials, but these can still often be factored using the same methods as polynomials.  You may or may not see questions like these in a high school algebra class, but you will in a pre-calculus class because expressions like this arise commonly in calculus problems.  First, lets consider how to factor out a common factor:
   7x^-5 - 3x^-3 + 4x^-2
   
   The common thing to do here is to factor out the most negative exponent, in this case, x^-5.  To do this, use the rules of exponents to rewrite each term with a factor of x^-5
   7x^-5 - 3x^-3 + 4x^-2
   7x^-5 - 3x^-5x² + 4x^-5x³
   x^-5(7 - 3x² + 4x³)
   
   Factoring fractional common factors is done similarly by using the rules of exponents to rewrite each term with the fractional exponent factor:
   8x^13/3 + x^7/3 - 5x^4/3
   8x^4/3x³ + x^4/3x - 5x^4/3
   x^4/3(8x³ + x - 5)
   
   You can also factor out negative fractional exponents:
   x^1/2 + x^-1/2
   x^-1/2x + x^-1/2
   x^-1/2(x + 1)
   
   Expressions with negative and fractional exponents can also often be written in quadratic or special forms:
   x^-2/3 - 8x^-1/3 - 20
   (x^-1/3 - 10)(x^-1/3 +2)
   
   x^-2 - y^-4
   (x^-1 + y^-2)(x^-1 - y^-2)
   
2. Putting It All Together: Sometimes you must put two or more of the factoring techniques together to completely factor an expression:
   3x⁶ - 192y¹² (factor out 3)
   3(x⁶ - 64y¹²) (use difference of squares***)
   3(x³ + 4y⁶)(x³ - 4y⁶) (use sum and difference of cubes)
   3(x + 2y²)(x² - 2xy² + 4y⁴)(x - 2y²)(x² + 2xy² + 4y⁴)
   
   ***Note: It is possible to use either use difference of squares or use difference of cubes on x⁶ - 64y¹².  When this is the case, always use difference of squares.  This is because the difference of cubes formula will leave two factors that both do factor, but the second one is not factorable by any of the methods I have shown in any of these posts, it is much harder.  To see this, lets factor x⁶ - 64y¹² by difference of cubes first:
   (x² - 4y⁴)(x⁴ + 4x²y⁴ + 16y⁸) (use difference of squares)
   (x + 2y²)(x - 2y²)(x⁴ + 4x²y⁴ + 16y⁸)
   
   Now x⁴ + 4x²y⁴ + 16y⁸ does factor to (x² - 2xy² + 4y⁴)(x² + 2xy² + 4y⁴) but you cannot figure that out by any of the methods I have shown.  You can verify it is true by multiplying the factors.  For this reason, always use difference of squares rather than difference of cubes when there is a choice.
   
   Here is another expression to factor I recently saw:
   x² - 12x + 36 - 49y² (group the first three terms)
   (x² - 12x + 36) - 49y² (use perfect square)
   (x - 6)² - 49y² (use difference of squares)
   (x - 6 + y)(x - 6 - y)
   
   Finally, here is an expression I just made up:
   (x + 12)(x - 1)(x - 3)^-1 + (x - 3)
   (x + 12)(x - 1)(x - 3)^-1 + (x - 3)^-1(x - 3)²
   (x - 3)^-1((x + 12)(x - 1) + (x - 3)²)
   (x - 3)^-1(x² + 11x - 12 + x² - 6x + 9)
   (x - 3)^-1(2x² + 5x - 3)
   (x - 3)^-1(2x - 1)(x + 3)

Well, I hope you enjoyed reading this as much as I enjoyed writing it, or at least I hope it has helped you.  My next post covers Solving Equations, with more to come soon.