FACTOR DIFFICULT TRINOMIALS

TWO EASY WAYS TO  

FACTOR DIFFICULT TRINOMIALS

Factoring trinomials can sometimes seem like a gigantic task, especially when the numbers get large or there’s a coefficient other than 1 in front of the leading variable.  Thinking of the two numbers that multiply up to the constant and also add up to the middle term is actually ‘double think.’  It requires keeping two operations in mind at the same time….ugh.
Consider the equation y = 24x^2 + 2x - 35.  First, find all the factors of 24.  Then find all the factors of 35.  Then consider each factor pair of 24 in combination with each factor pair of 35 to see if any of the products add up to 2.  My brain is a jumble of numbers at this point… 1 times 24 plus or minus 1 times 35, 2 times 12 plus or minus 1 times 35, 3 times 8 plus or minus 1 times 35, and on and on and on.  This is too much work, and as a mathematician, I’m always looking for the “elegant solution.”  For the more complicated factoring problems, I prefer systems that reduce calculations to one-at-a-time (and self-correcting would be a bonus).

Here are two procedures that exploit methods we’ve already learned for simple problems and avoid some of the typical mistakes.

                      BOTTOMS UP
 

Given:                                                          y = 24x^2 + 2x - 35

Step 1: multiply the leading coefficient        (24)(35) = 840
        and the constant,
        and rewrite the problem                       y = x^2 + 2x - 840
————————————————————————————-
Step 1 turns an ugly trinomial into one that looks more like the first exercises practiced when learning to factor.  The next 3 steps should be very familiar.
————————————————————————————-
Step 2:  factor the product                                840  
                                                                      (1, 840)

                                                                      (10, 84)
                                                                      (20, 42)
                                                                      (24, 35)
 

Step 3:  select factors that SUM to              (28, 30)
          the middle term

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Step 4:  enter the factors into binomials        y = (x + 30)(x - 28)
        (Notice that the MIDDLE TERM
        determines the sign of the larger
        factor, while the CONSTANT tells
        if the factors are the same sign or
        different signs. A negative constant
        indicates the signs are DIFFERENT,
        while a positive constant says the
        signs are THE SAME.)

Step 5:  Divide the integers by the
        original leading coefficient                     y = (x + 30/24)(x - 28/24)

        and reduce.                                           y = (x + 5/4)(x - 7/6)

NOW, the step that gives the method
its name…..
Step 6: bring the “BOTTOMS UP”                y = (4x + 5)(6x - 7)

    (Notice that at Step 5, the problem is set up to solve.)

 
   
While this plan makes numbers easier to manipulate, the next alternative approach is also SELF CORRECTING.

               DE-FOIL

Follow Steps 1- 3 but don’t rewrite the equation.

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Step 1: multiply the leading coefficient        (24)(35) = 840
        and the constant.               

Step 2: factor the product                                  840  
                                                                       (1, 840)                                                

                                                                       (10, 84)
                                                                       (20, 42)
                                                                       (24, 35)
 

Step 3:  select factors that SUM to              (28, 30)
          the middle term

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Step 4:  Rewrite the equation into one
        with the FOUR TERMS that would

        result from FOIL:

    -- the first term comes as is
    -- the constant comes as is                       y = 24x^2…….………- 35

    -- instead of the initial middle
        term, enter the chosen
        factors                                                   y = 24x^2 - 28x + 30x - 35

     (If there are opposite signs, I like to put the negative one first so I don't lose it in the next step.)

Step 5:  Now, think about what would
        be done to factor any polynomial 

        with four terms........GROUP                 y = (24x^2 - 28x) + (30x - 35)

Step 6:  Factor out commons                       y = 4x(6x - 7) + 5(6x - 7)

Step 7:  Factor out commons                       y = (6x - 7)(4x + 5)

We do a lot of factoring in Algebra, but once we get to simplifying radical expressions it becomes mandatory to be at ease with the task.  One of these alternate methods could be a magic key to making the job less daunting.

The SUM of Two Squares -- BREAK THE RULES!

Here’s something nobody else in your Algebra II class (maybe even the teacher) will know.  (Do not try to be the wise-cracker until you're ready to PROVE that this works!!)

Since first learning to factor, we've been told that "the rules” are... you

    -  CAN factor the DIFFERENCE OF TWO SQUARES (referred to by some texts as ‘DOTS’)  into conjugates.....

    x^2 - 16 = (x - 4)(x + 4)

    - CAN NOT factor the SUM OF TWO SQUARES.  This second admonition is true when the instructions say "factor completely across the integers" because, as you are about to see, the answers are NOT integers.

BUT You CAN factor the sum of two squares if you've studied imaginary numbers and know a simple little TRICK.

        You will still use conjugates, but there needs to be an i in the second term of each binomial.....

    x^2 + 16 = (x - 4i)(x + 4i)

Why does it work?  Because when these two binomials are FOILed, the middle terms (O and I) cancel each other out.  AND since i^2 is simply -1, the multiplication of the last terms gives you the task of subtracting a negative number, which is simplified to ADDING!!

Try factoring these sums of perfect squares and FOIL the binomials to verify that you actually CAN “break the old rules” when you know enough math!!

PROBLEM A:     p2 + 64

PROBLEM B:     121 + y2 (hint: use the commutative property first and attach the i to the square root of 121)

PROBLEM C:     x2 + 36 y2

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 And this concludes the demonstration of how a math nerd can break all the "rules" and get away with it, mathematically speaking!!