Factoring Polynomials II

Free Math Help Online presents Factoring Polynomials Part II.  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 this post I will cover factoring by grouping, and recognizing special forms.  In Factoring Polynomials III I will cover factoring negative and fractional exponents, and putting it all together.  My next post will cover Solving Polynomial Equations.


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

1. Factoring by Grouping: Consider the following expression:
   6x² - 15x + 14x - 35
   
   If asked to factor this, the most common approach would be to first add like terms and then factor the quadratic like we did in Factoring Polynomials.  Doing this we could find that
   6x² - 15x + 14x - 35 = (3x + 7)(2x - 5)
   
   Being very conscientious, we would then check our work by multiplying with the FOIL method
   (3x + 7)(2x - 5) = 3x∙2x + 3x∙(-5) + 7∙2x + 7∙(-5)
   (3x + 7)(2x - 5) = 6x² - 15x + 14x - 35
   
   Notice that before combining the Inside and Outside terms, this is exactly what we started with.  To factoring by grouping you first group the terms in a convenient way so that each group is factorable, then factor each group, and finally look to see if you have created a new whole expression that can be factored itself.  For this example:
   6x² - 15x + 14x - 35
   (6x² - 15x) + (14x - 35)
   (3x∙2x + 3x∙(-5)) + (7∙2x + 7∙(-5))
   3x(2x - 5) + 7(2x - 5)
   (2x - 5)(3x + 7)
   
   Notice in the last step that the binomial (2x - 5) is now a factor in each term of the whole expression so it can be factored out leaving the answer (2x - 5)(3x + 7).  This is the same answer I got before, just with the factors written in reverse order.  This is, of course, perfectly OK.  Also, given that the problem was written with the - 15x + 14x rather than - x, factoring by grouping is much faster than the trial and error approach used before.
   
   Now, how can we figure out how to rewrite a quadratic binomial into an expression that can be factored by grouping? The answer, called the AC method, is very much like factoring a quadratic trinomial with leading coefficient of 1. Generically speaking, quadratics can be written in the form ax² + bx + c. When a is 1 all we had to do was find factors of c that add to b. When a is not 1 all we have to do is find factors of ac that add to b. Lets look again at 6x² - 15x + 14x - 35 = 6x² - x - 35. What are the factors of 6(-35) = -210 that add to -1? The factors of 210 are
   1, 210
   2, 105
   3, 70
   5, 42
   6, 35
   7, 30
   10, 21
   14, 15
   
   Since we actually need factors of -210 then we must consider each pair above with one positive and one negative. It is now easy to see that the factors we want are 14 and -15. The only question that remains is the order in which to put them. Fortunately it does not matter, but I recommend putting a positive number last if possible to reduce the likelihood of making a mistake factoring out a negative. I will do it the opposite way here, but pay attention to how the negative is handled.
   6x² + 14x - 15x - 35
   6x² + 14x + (-15x - 35)
   2x(3x + 7) - 5(3x + 7)
   (3x + 7)(2x - 5)
   
   Unfortunately, most cubic expressions are not easy factor by grouping, but in a High School Algebra course you are very likely to see cubic expressions that can be factored by grouping such as the following:
   x³ + 2x² - 7x - 14
   
   Again, there is one part that is tricky to most students, so pay close attention to how negative values are handled.  Notice that -7x - 14 is NOT the same thing as -(7x - 14) because when you distribute the negative in the second expression you change the sign of both terms on the inside: -(7x - 14) = -7x + 14 ≠ -7x - 14.  When we group here we will group the entire (-7x - 14) and then factor out the -7.
   x³ + 2x² - 7x - 14
   (x³ + 2x²) + (-7x - 14)
   x²(x + 2) - 7(x + 2)
   (x + 2)(x² - 7)
   
   Technically, the (x² - 7) can be factored further, but not with rational coefficients.  I will discuss this more in Factoring Polynomials III
   
   
2. Recognizing Special Forms: Here we will be using the quadratic form and special forms introduced in Factoring Polynomials.  Consider this expression:
   5x⁴ + 34x² + 24
   
   This is not a quadratic expression, it is a 4^th degree polynomial.  With a convenient substitution, however, it can be written in quadratic form.  Let t = x².
   
   5x⁴ + 34x² + 24
   5(x²)² + 34(x²)+ 24
   5t² + 34t+ 24
   (5t + 4)(t + 6)
   (5x² + 4)(x² + 6)
   
   Once you see how it is done, you probably won’t want to actually write the three step in the middle.  Instead it will look more like:
   5x⁴ + 34x² + 24
   (5x² + 4)(x² + 6)
   
   There is still trial and error going on to find the factors of 24 that work with the factors of 5 to give the 34, I just am not showing that part here.  Now lets try some that are of the other special forms:
   16x¹⁰ - 121z⁴ (Difference of Squares form, let t = 4x⁵ and u = 11z²)
   (4x⁵)² - (11z²)²
   t² - u²
   (t + u)(t - u)
   (4x⁵ + 11z²)(4x⁵ - 11z²)
   
   a¹² + 6a⁶b² + 9b⁴ (Perfect Square form, let t = a⁶ and u = 3b²)
   (a⁶)² + 2(a⁶)(3b²) + (3b²)²
   t² + 2tu + u²
   (t + u)²
   (a⁶ + 3b²)²
   
   27w⁹ - 64y¹² (Difference of Cubes form, let t = 3w³ and u = 4y⁴)
   (3w³)³ - (4y⁴)³
   t³ - u³
   (t - u)(t² + tu + u²)
   (3w³ - 4y⁴)((3w³)² + (3w³)(4y⁴) + (4y⁴)²)
   (3w³ - 4y⁴)(9w⁶ + 12w³y⁴ + 16y⁸)
   
   The more you practice these kinds of problems, the fewer of the substitution steps you will need to show.

If you need more factoring help, go on to the next post, Factoring Polynomials III as soon as it is available.