GCF and Factoring Trinomials   Factoring Out the Greatest Common Factor (GCF)   Consider the example         x (x - 3)  =  x2 - 3x  We can do the process in reverse: Look at          x2 - 3x  and notice that both have a common factor of x.   We can pull out the x terms (use the distributive property in reverse).         x2 - 3x  =  x (x - 3).   Example:   Consider the expression          2x3 - 4x2   Find the GCF Pull it out.   Solution: The GCF is 2x2 We can write          2x3 - 4x2  =  2x2 (x - 2)   Exercises: Find the GCF Pull it out. x2 - 5x                               5xy + 25y                          3x(2 - x) + 4(2 - x)             6x2y3 - 8x2y2  + 10x4y4     Factoring by Grouping Consider the expression         3x2 + 6x - 4x - 8 We can factor this by grouping two at a time:         (3x2 + 6x) - (4x + 8) We now pull out the GCF of each:         3x (x + 2) - 4 (x + 2) Pull out the GCF again:         (3x - 4) (x + 2)   Exercises Factor the following by grouping: 3x2 - 6x + x - 2        x3 - 3x2 + x - 3        Factoring Trinomials With constant Leading Coefficient Consider the trinomial:         x2 + 5x + 4 We want to factor this trinomial, that is do FOIL in reverse.  Since the leading coefficient is 1 we can write:         x2 + 5x + 4 = (x + a) (x + b)   Where a and b are unknown numbers. FOIL the right expression out to get:         x2 + 5x + 4          =  x2 + bx + ax + ab          =  x2 + (a + b) x + ab Hence we are searching for two numbers such that their product is 4 and their sum is 5.   Note that the only pairs of number whose product is 4 are         (2,2) and (1,4). Of these two pairs only (1,4) add up to 5. Hence         x2 + 5x + 4  =  (x + 1) (x + 4)   Two factor a trinomial with leading coefficient 1 we ask ourselves the following questions: What two numbers multiply to the last coefficient and add to the middle coefficient?   Example:  Factor         x2 - 3x - 10 The pairs that multiply to -10 are         (1,-10)     and     (-1,10)    and     (5,-2)   and    (2,-5) Only the last pair adds to -3 hence         x2 - 3x - 10  =  (x + 2) (x - 5)   Exercises:  Factor if possible x2 + 8x + 15        x2 + 6x + 8          x2 - 4x - 5            x2 + 3x - 18         x2 - 7x +12          x2 - 4x - 9            The AC Method What can we do when the leading coefficient is not 1? We use an extension of factoring by grouping called the AC method.   Step by Step method for factoring    Ax2 + Bx + C Step 1.  Multiply together AC and list the factors of AC. Step 2.  Find a pair that adds to B.  If you cannot find such a pair then the trinomial does not factor. Step 3.  Rewrite the middle term as a sum of terms whose coefficients are the chosen pair. Step 4.  Factor by grouping.   Remember you should always first pull out the GCF   Example:   Factor          2x2 + 5x - 25 AC = (2)(-25) = -50  the pairs are         (1,-50)     (-1,50)     (2,-25)     (-2,25)      (5,-10)   and   (-5,10). We see that          -5 + 10  =  5   hence we choose the pair (-5,10) We write           2x2 - 5x + 10x - 25 (2x2 - 5x) + (10x - 25)  =  x (2x - 5) + 5 (2x - 5)  =  (x + 5) (2x - 5)   Exercises:   Factor the following 15x2 - x - 6                 7x2 - 20x + 12            30x3 + 25x2 + 5x        4x4 + 8x2 + 3             For an alternate method for the AC method click here  e-mail Questions and Suggestions