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FACTORING

FACTORING. Factoring the Common Factor.

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FACTORING

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  1. FACTORING

  2. Factoring the Common Factor • When factoring the common factor, look for a number or variable that divides into both (or all) terms.  If there is more than one common factor, be sure to get the largest common factor you can find.  First write down the common factor.  Then, open parentheses, and put down all the other factors that are left over.

  3. The first step in any factoring problem is to factor the common factor. Factoring the common factor is simply using the distributive property in reverse.  Study the examples below.           EXAMPLES of  EXAMPLES of DISTRIBUTIVE PROPERTY                  FACTORING:                    6(x + 7) = 6x + 42                                                 6x + 42 = 6(x + 7)                  7(2x + 3) = 14x + 21                                               14x + 21 = 7(2x + 3)                    9(x − 4) = 9x − 36                                                 9x − 36 = 9(x − 4)                12(2x + 1) = 24x + 12                                            24x + 12 = 12(2x + 1)           5(3x − 2y + 4) = 15x − 10y + 20                          15x − 10y + 20 = 5(3x − 2y + 4)                   5x(x + 4) = 5x2  + 20x                                    5x2  + 20x = 5x(x + 4)  

  4. EXAMPLE 1.     12x + 36 SOLUTION:  There are several numbers that divide evenly into both 12 and 36:  1,2,3,4,6, and 12.  Take the largest common factor, which is 12.  Write down the 12, then open parentheses:  12 (    +    ).  In the parentheses you put the remaining factors, x and 3, like this:                         12x + 36 = 12 (     +     )                                        = 12 ( x   +  3 )

  5. RULES • In order to factor a common factor, you must have an identical factor common to all terms.      Be sure to count terms first. • When factoring powers, take out the lowest exponent (power) of the factor.  Then subtract exponents.

  6. FACTORING BY GROUPING EXAMPLE 1.             Factor  x3 + 2x2 + 8x + 16 Solution:         There are no common factors to all four terms.  It is not a trinomial, and nothing discussed so far works to factor this.  So, try grouping the first two terms together, and the last two terms together, and factor out the common factor within each grouping as follows:                                                 (x3 + 2x2) + (8x + 16) =  x2 (x + 2)+ 8(x + 2)                          Notice that there is a common factor of (x+2) that can be factored out:                                                                                     =  (x + 2)( x2 + 8)  EXAMPLE 2.            Factor  xy− 4y+ 3x− 12 Solution:         Again, there are no common factors, and this is not a trinomial.  Group the first two and the last two terms together, and factor out the common factor from each grouping:                                                  (xy− 4y)+ (3x− 12) =  y(x− 4) + 3(x− 4)                  Now, take out the common factor, which is (x− 4):                                                                                     =  (x− 4)(y + 3)

  7. Factoring Trinomials • RULES:          • 1.   When the sign of the LAST is positive, the signs are the SAME. You find middle term by ADDING the O and I terms. • 2.  When the sign of the LAST is negative, the signs are OPPOSITE. You find middle term by SUBTRACTING the O and I terms.

  8. EXAMPLE 1. Factor      x2 + 7x + 10. Solution:                       x2 + 7x + 10        (         )(         )      Product of two binomials; (x       )(x       )      F term is x2, which is x times x;                                     (x       )(x       )      L term is +10.  You must find two numbers whose product is +10.  Probably 2 times 5, or it could be 1 times 10.  Try 2 times 5.  Since the last sign is positive, it will be positive times positive, or negative times negative.  Also, the middle terms (O and I) must be added together.    (x + 2)(x + 5)     OI term is 7x.   This means the outer times outer and the inner times  inner terms must add up to 7x.     [NOTICE:  The order does not matter!  If you wrote (x + 5)(x + 2), this is exactly equivalent to (x + 2)(x + 5). ]

  9. If the "L" term is negative, this means that the signs are opposite, and you must subtract the numbers to get the middle term:                                     F    OI   L                                                        F    OI   L     EXAMPLE 3.             x2 + 2x − 8                  EXAMPLE 4.             x2 − 2x − 8 Solutions:        Because the L term is negative in each of these examples, you must have opposite signs, probably −2 times 4 or −4 times 2. x2 + 2x − 8                                                 x2 − 2x − 8                                              (x +   )(x −    )                                       (x +   )(x −    )    Answers:                  (x + 4)(x − 2)                                       (x + 2)(x − 4) [Also correct:  (x − 2)(x + 4)                                                  (x − 4)(x + 2)  ]

  10. In the factoring of trinomials of the previous sections, it may have been assumed that the coefficient of x2 is 1 (or that the coefficient is a common factor of the entire trinomial).  Notice that as you work through this section, because of combinations of numbers, the trial and error process becomes more and more challenging.   However, let’s begin with some examples that do not involve too many combinations of numbers. • EXAMPLE 1.    5x2 + 6x + 1. • Solution:  Remember that this should be factored by F L OI. The F term is 5x2,                                L is 1,      5x2 + 6x + 1                               5x2 + 6x + 1    (5x      )(x       )                          (      +  1)(     +  1) and OI adds up to 6x, as follows: 5x2 + 6x + 1 Final answer:    (5x + 1)(x + 1)

  11. Of course, with larger numbers, with many more combinations of numbers this can become a very lengthy process of trial and error.  There are some systematic methods of factoring these trinomials, which usually turn out to be somewhat complicated.  In problems that are not too difficult, the trial and error method will be fairly simple and more than adequate for now.  EXAMPLE 6.     5x2 + 8x + 3(Trial and Error!!!) Solution:         The F term is 5x2,                                     L is 3, 5x2 + 8x + 3                                     5x2 + 8x + 3 (5x      )(x       )               (     +  3)(     +  1)  or (     + 1)(      + 3) The OI term must add up to 8x.     5x2 + 8x + 3 (5x + 3)(x + 1)  or   (5x + 1)(x + 3)     (The first is correct, the second is NOT!) Final answer:              (5x + 3)((x + 1)

  12. SUM AND DIFFERENCE OF CUBES  In this section, formulas and procedures will allow you to factor expressions in the form x3− y3 and also x 3 + y 3.  Recall from previous sections that  x2− y2 = (x− y)(x + y) and that x2 + y2 cannot be factored.  Begin with the multiplication problems:            (x− y)( x2 + xy + y2 )  =  x3 + x2y + xy2 − x2y −xy2 − y3                                                 =      x3          −      y3            (x + y)( x2−xy + y2 ) =  x3−x2y + xy2     + x2y −xy2 + y3                                               =       x3        +      y3 . This derives the formulas, known as the "sum and difference of cubes formulas."

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