Chabot Mathematics. §5.5 Factor Special Forms. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] MTH 55. 5.4. Review §. Any QUESTIONS About §5.4 → Factoring TriNomials Any QUESTIONS About HomeWork §5.4 → HW-14. §5.5 Factoring Special Forms.
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Chabot Mathematics
§5.5 FactorSpecial Forms
Bruce Mayer, PE
Licensed Electrical & Mechanical [email protected]
MTH 55
5.4
A2 + 2AB + B2 = (A + B)2
A2− 2AB + B2 = (A−B)2
1. Two terms, A2 and B2, must be squares, such as: 9, x2, 100y2, 25w2
2. Neither A2 or B2 is being SUBTRACTED.
3. The remaining term is either 2 A B or −2 A B
a) x2 + 8x + 16 b) t2− 9t− 36
c) 25x2 + 4 – 20x
It helps to write it in descending order.
25x2– 20x + 4
A2 + 2AB + B2 = (A + B)2
A2− 2AB + B2 = (A−B)2
b) 25x2− 20x + 4
x2 + 8x + 16 = x2 + 2 x 4 + 42 = (x + 4)2
A2 + 2 A B + B2 = (A + B)2
b) 25x2− 20x + 4
25x2– 20x + 4 = (5x)2–2 5x 2 + 22 = (5x– 2)2
A2– 2 A B + B2 = (A – B)2
16a2− 24ab + 9b2= (4a)2− 2(4a)(3b) + (3b)2
= (4a− 3b)2 = (4a− 3b)(4a− 3b)
(4a− 3b)(4a− 3b) = 16a2− 24ab + 9b2
12a3− 108a2 + 243a = 3a(4a2− 36a + 81)
= 3a[(2a)2− 2(2a)(9) + 92]
= 3a(2a− 9)2
a) 16x2− 25b) 36 −y5c) −x12 + 49
The 2nd expression is a sq: 25 = 52
A2−B2 = (A + B)(A−B)
b) y2− 16w2 = y2− (4w)2 = (y + 4w)(y− 4w)
A2−B2 = (A + B) (A−B)
c) 25 − 36a12 = 52− (6a6)2 = (5 + 6a6)(5 − 6a6)
d) 98x2− 8x8
Alwayslook for a common factor. This time there is one, 2x2:
98x2− 8x8 = 2x2(49 − 4x6)
= 2x2[(72− (2x3)2]
= 2x2(7 + 2x3)(7 − 2x3)
rearranging
m2 + 14m + 49 − 4b4
GROUPING
(m2 + 14m + 49) − 4b4
(m2 + 14m + 49) − 4b4
Perfect Sqs
(m + 7)2− (2b2)2
(m + 7)2− (2b2)2
Diff-of-Sqs → (A − B)(A + B)
([m+7] − 2b2)([m + 7] + 2b2)
Simplify → ReArrange
(−2b2 + m + 7)(2b2 + m + 7)
SAME Sign
OPP Sign
SAME Sign
OPP Sign
Recognize Pattern as Sum of CUBES
Determine Values that were CUBED
Map Values to Formula
Substitute into Formula
Simplify and CleanUp
Recognize Pattern as Difference of CUBES
Determine CUBED Values
Simplify by Properties of Exponents
Map Values to Formula
Sub into Formula
Simplify & CleanUp
Use Distributive property
Use Comm & Assoc. properties, and Adding-to-Zero
5x4− 3125 = 5(x4− 625)
= 5[(x2)2− 252]
= 5(x2− 25)(x2 + 25)
= 5(x− 5)(x + 5)(x2 + 25)
Sum ofTwoCubes