# FACTORING POLYNOMIALS BY TYPE

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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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