These are all difference of squares or sum/difference of cubes.
You need to be able to recognize such values, and apply the formulas:
a^2-b^2 = (a-b)(a+b)
a^3±b^3 = (a±b)(a^2∓ab+b^2)
For example,
16x^4a-y^8a = a(16x^4-y^8)
= a(4x^2+y^4)(4x^2-y^4)
= a(4x^2+y^4)(2x+y^2)(2x-y^2)
((8x^3)/125)+(64/y^3)
If you let a = 2x/5 and b = 4/5, then you have a^3+b^3, so
= (2x/5 + 4/y)(4x^2/25 - 8x/5y + 16/y^2)
You can probably tackle the others now, right?
Note that something might be both a difference of squares and a difference of cubes:
a^6-b^6 = (a^2-b^2)(a^4+a^2b^2+b^4)
a^6-b^6 = (a^3-b^3)(a^3+b^3)
= (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)
The tricky part is showing these are really the same:
a^4+a^2b^2+b^4
= a^4+2a^2b^2+b^4 - a^2b^2
= (a^2+b^2)^2 - (ab)^2
= (a^2+ab+b^2)(a^2-ab+b^2)
Special factorization
Completely factor the following expressions
250x^3-128
X^6-1
X^6+1
x^2-8xy+16y^2-25
16x^4a-y^8a
((8x^3)/125)+(64/y^3)
I was given a set on how to do this skill, but it didn't really cover on how to attack these kind of expressions. I was wondering if someone could help walk me through this, or teach me a set of instructions that I could use to apply to each one when I am encountering one (if there is one)
2 answers
For the second example, how did you get to 2x/5 and 4/5?
3 answers