FACTORING POLYNOMIALS BY TYPE



TYPE

    I	         GCF (greatest common factor)….. no matter how complicated, always do this first!
	         ab + ac = a(b + c)
	         ab + a = a(b + 1)

    II	         Difference of two perfect squares:
	         a2 – b2 = (a – b)(a + b)
	         36x2 – 49 = (6x – 7)(6x + 7)    use FOIL to check:  36x2 + 42x – 42x – 49

    III	         Perfect square trinomials:  (note – on all types of trinomials, the middle term is always     

                         key when you multiply the outside terms and add them to the product of the inside terms;                                  always test the middle term, and if it checks out, you most likely have it properly  

                        factored!)                       a2 + 2ab + b2 = (a + b)2                          The marks of a Type III: 1st and last terms are perfect squares, middle term is an even

                        number, and the first and last terms are always positive!                          16x2 + 24x + 9 = (4x + 3)2                          To check: since the middle term equals 2ab when you FOIL it, multiply 4x (first term of

                        the factor) times 3 (last term of the factor) and then times 2 = 24x; the middle term

                         checks! a2 – 2ab + b2 = (a – b)2

                         25a2 – 40ab + 16b2 = (5a – 4b)2                          To check: multiply 5a times -4b, then times 2 = -40ab, the same as the middle term,

                         which checks. The sign of the middle term of the perfect square trinomial always

                         determines the sign of the factor.

  IV Imperfect square trinomial:                           x2 + bx + c (find the factors of c such that their sum equals b)                           For help with signs see chart:                           b        c            factored signs example                           +        +        =     ( + )( + ) x2+ 9x + 20 = (x + 4)(x + 5)                           –        –         =    ( – largest number)( + ) x2 – x – 20 = (x – 5)(x + 4)                           +        –         =    ( + largest number)( – ) x2 + x – 20 = (x + 5)(x – 4)                           –        +        =     ( – )( – ) x2 – 9x + 20 = (x – 5)(x – 4)

                         Again, always check the middle term by adding the products of the inside terms and the

                         outside terms: example 1: 4x + 5x = 9x; example 2: -5x + 4x = -x;

                         example 3: 5x – 4x = x; and example 4: -5x – 4x = -9x

    V Imperfect square trinomial with first term coefficient:                           ax2 + bx + c (like a Type IV, but you must factor both a and c, such that their sum using

                         FOIL equals b)                           3x2 + 17x + 24 = (3x + 8)(x + 3)                           4x2 – 4x – 15 = (2x – 5)(2x + 3)                           With Type V’s you must check all the factors of the first and last terms and all their

                          combinations of sums such that when you add their products you again get the middle

                          term.

    VI 4-term polynomial: Factor using the following strategies:                           1) separate it into 2 binomials such that when you take out the GCF of each, you have

                          a common binomial factor, e.g. ax + b + a + bx                           First, rearrange it into pairs with like factors: ax + a + bx + b, then factor out "a" in the

                         first two terms and "b" in the last two terms: a(x + 1) + b(x + 1); now you have (x + 1)

                   as a common factor and that must be taken out from these two new terms: (x + 1)(a + b)                           2) separate the problem into the difference of a perfect square trinomial (Type III) and a

                          perfect square: a2 + 2ab + b2 – x2                           Now factor the trinomial: (a + b)2 – x2, and now you can see that you have a

                          difference of 2 perfect squares, which factors into [(a + b) – x][(a + b) + x]                           another example: x2 – a2– 2ab – b2 = x2– (a2 + 2ab + b2 ) = x2(a + b)2 , then                            [x – (a + b)][x + (a + b)], or (x – a – b)(x + a + b)

    VII Perfect cube binomial: a3 + b3 = (a + b)(a2 – ab + b2 ) a3 – b3 = (a – b)(a2 + ab + b2 )


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