Algebra Honors (Period 6 & 7)

Using Several Methods of Factoring Section 5-11

1) Always factor the GCF first
2) look for he difference of 2 SQ's
3) Look for a Perfect SQ trinomial
4) If trinomial is NOT SQ look for a pair of factors
5) If 4 or more terms-- look for a way to group the terms into pairs or into a group of 3 terms that is a perfect SQ trinomial
6) Make sure each binomial or trinomial factor is PRIME
7) check your work
-4n⁴ + 40 n³ -100n²
GCF
-4n²(n² -10n + 25)
-4n²(n-5)²

What about
5a³b² + 3a⁴b - 2a²b³
Again factor the GCF
a²b(5ab + 3a²-2b²)
reorder this and use either XBOX or factor pairs to solve

a²b(3a²+5ab-2²)


a²b(3a²+6ab-ab-2b²)


a²b[(3a²+6ab)+(-ab-2b²)]

a²b[3a(a+2b)-b(a + 2b)]
a²b(a+2b)(3a-b)


a²bc -4bc + a² -4b
b(a²c-4c+a²-4)
b(a²c+a²-4c-4)
b[(a²c+a²) -(4c+4)]
b[a²(c + 1) -4(c +1)]
b(c+1)(a²-4)
but we aren't finished...
b(c+1)(a+2)(a-2)



6c²+18cd+12d²
6(c²+3cd+2d²)
6(c+2d)(c+d)

3xy²-27x³

3x(y²-9x²)
3x(y+3x)(y-3x)


-n⁴-3n²-2n³

-n²(n²+3+2n)
-n²(n²+2n+3)

Its factored completely!!


16x²+16y -y²-64
16x²-y²+16y-64
16x²-y²+16y-64
16²-(y²-16y+64)
16x²-(y-8)²
becomes the difference of two squares
(4x+y-8)(4x-y+8)

x¹⁶ -1
(x⁸+1)(x⁸-1) =
(x⁸+1)(x⁴+1)(x⁴-1)=
(x⁸+1)(x⁴+1)(x²+1)(x² -1) =
(x⁸+1)(x⁴+1)(x²+1)(x +1)(x-1)


2(a +2)² + 5(a +2) - 3
Think of this as letting a+ 2 = x

2x² +5x -3
Factoring that is easy
(2x-1)(x +3)
so substitute in a + 2 for each x
[2(a+2) -1]{a+2 +3]


2a + 4 -1)(a +5)
(2a +3)(a +5)