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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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