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ALGEBRA I - SECTION 8-7 (Factoring Special Cases)

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1 ALGEBRA I - SECTION 8-7 (Factoring Special Cases)
@ SECTION 8-7 : FACTORING SPECIAL CASES

2 (x + 3)(x + 3) (x – 2)(x – 2) (x + 5)2 (x – 6)2 (x + 7)(x – 7)
FOIL the problem associated with your group number. (x + 3)(x + 3) (x – 2)(x – 2) (x + 5)2 (x – 6)2 (x + 7)(x – 7) (x + 10)(x – 10) (x – 4)(x + 4) (x – 11)(x + 11) x2 + 6x + 9 x2 – 4x + 4 x2 + 10x + 25 x2 – 12x + 36 x2 – 49 x2 – 100 x2 – 16 x

3 SQUARE of a SUM a2 + 2ab + b2 = (a + b)2 SQUARE of a DIFFERENCE a2 – 2ab + b2 = (a – b)2 DIFFERENCE of SQUARES a2 – b2 = (a + b)(a – b)

4 To determine if a trinomial if a square :
NOTE : The 1st and 3rd terms must be positive squares. Drop the sign of the middle term to the middle of the binomial. x2 + 8x + 16 ( )2 x + 4 Take the square root of the first term Take the square root of the third term Multiply the two terms together. Double the result. If it is the middle term, voila, you have a square.

5 Factor completely. 9) x2 – 10x + 25 11) x2 + 2x + 1 10) 4x2 – 28x + 49
ANSWER : (x – 5)2 ANSWER : (x + 1)2 10) 4x2 – 28x + 49 12) 9m2 + 12m + 4 ANSWER : (2x – 7)2 ANSWER : (3m + 2)2

6 But, can it still be factored?
13) x2 – 5x - 24 But, can it still be factored? ANSWER : (x – 8)(x + 3) 14) 25x2 + 35x + 9 ANSWER : prime 15) 16z2 – 24z + 9 ANSWER : (4z – 3)2

7 Are these binomials the Difference of Squares? If so, factor.
18) x2 - 64 ANSWER : (a + 5)(a – 5) ANSWER : (x + 8)(x – 8) 17) y2 - 42 19) 9y2 - 49 ANSWER : not the difference of squares ANSWER : (3y + 7)(3y – 7)

8 20) 121 – p2 22) j2 + 25 21) 25p3 – 100p ANSWER : (11 + p)(11 – p) or
ANSWER : you can’t factor the SUM of squares….yet. 21) 25p3 – 100p ANSWER : 25p(p + 2)(p – 2)

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