Algebra Honors (Periods 6 & 7)

Differences of Two Squares 5-5 

(a + b)(a-b) = a² - b²
(a+b) is the sum of 2 numbers
(a-b) is the difference of 2 numbers
= ( first#)² - ( 2nd #) ²

( y -7)( y + 7) = y² - 49
We did the box method to prove this.

(4s + 5t) (4s - 5t)
16s² - 25t²

(7p + 5q)(7p-5q) = 49p² - 25q²
But then we looks at
(7p+5q)(7p+5q) that isn't the difference of two squares that is
49p² + 70pq + 25q²
So let's look at the difference of TWO Squares:

b² -36
So that is ( b + 6)(b -6)

m² - 25
(m + 5)(m -5)

64u² - 25v²
(8u + 5v)(8u -5v)

1 - 16a²

(1+4a)(1- 4a)

But what about 1- 16a⁴

( 1 + 4a²)(1 - 4a²) but we are NOT finished factoring because
(1 - 4a²) is still a difference of two squares so it becomes

( 1 + 4a²)(1 + 2a)(1 - 2a)

t⁵ - 20t³ + 64t

Factor out the GCF first

t(t⁴ - 20t² + 64)
t(t² -16)(t² -4)
YIKES... we have two Difference of Two Squares here...

t(t +4)(t-4)(t +2)(t -2)



81n² - 121
(9n +11)(9n -11)
3n⁵ - 48 n³


Factor out the GCF

3n³ (n² - 16)
3n³(n +4)(n-4)

50r⁸ - 32 r²

Factor out the GCF
2r²(25r⁶ - 16)

2r²(5r³ +4)(5³ -4)

u² - ( u -5) ²
think a² - b ² = (a + b)(a -b)

so
u² - ( u -5) ² =
[u + (u-5)][u - (u-5)]
(2u -5)(5)
= 5(2u-5)

t² - (t-1)²

[t +( t+1)]{t-(t-1)]
2t-1(+1)
=2t -1


What about x²ⁿ - y ⁶ where n is a positive integer

well that really equals
(xⁿ)² - (y³)²
so
(xⁿ + y³)(xⁿ - y³)

x²ⁿ - 25
(xⁿ + 5)(xⁿ - 5)

a⁴ⁿ - 81b⁴ⁿ
(a²ⁿ + 9b²ⁿ)(a²ⁿ - 9b²ⁿ)
= (a²ⁿ + 9b²ⁿ)(aⁿ + 3bⁿ)(aⁿ - 3bⁿ)


When multiplying to numbers such as (57)(63)
think
(60-3)(60 +3)
then the problem becomes so much easier

3600 - 9 = 3591 DONE!!!

(53)(47) = (50 +3)( 50-3)
2500 - 9 = 2491