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)
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)
For the second example, how did you get to 2x/5 and 4/5?
To completely factor an expression, we need to identify any common factors and then apply various factorization techniques. The instructions provided below will help you factor each of the given expressions.
1. 250x^3 - 128:
First, notice that both terms have a common factor of 2. So, we can factor out 2 from both terms:
2(125x^3 - 64). Now, this expression can be factorized further using the difference of cubes formula: a^3 - b^3 = (a - b)(a^2 + ab + b^2).
In this case, a = 5x and b = 4. Applying the formula, we have:
2[(5x)^3 - 4^3] = 2(5x - 4)(25x^2 + 20x + 16).
2. x^6 - 1:
This expression is in the form of a difference of squares: a^2 - b^2 = (a + b)(a - b). Here, a = x^3 and b = 1.
Using the difference of squares formula, we can factorize the expression as:
(x^3 + 1)(x^3 - 1).
3. x^6 + 1:
This expression doesn't have any readily apparent factorization. However, we can use the sum of cubes formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2).
Here, a = x^2 and b = 1. Applying the formula, we get:
(x^2 + 1)(x^4 - x^2 + 1).
4. x^2 - 8xy + 16y^2 - 25:
This expression can be factorized using the difference of squares: a^2 - b^2 = (a + b)(a - b). In this case, we see that the expression is a perfect square trinomial (x - 4y)^2 - 5^2.
Using the difference of squares formula, we can rewrite it as:
[(x - 4y) + 5][(x - 4y) - 5] = (x - 4y + 5)(x - 4y - 5).
5. 16x^4a - y^8a:
This expression does not have a common factor, and we cannot factor it further using any specific formula. Therefore, it remains in factored form as: 16x^4a - y^8a.
6. (8x^3/125) + (64/y^3):
To factorize this expression, we need to first find the least common denominator (LCD) of the two fractions, which is 125y^3. Then, we can rewrite the expression with the LCD:
(8x^3 * (y^3/y^3))/125 + (64 * (125/x^3))/(y^3 * (125/x^3)).
Simplifying the expression gives us:
(8x^3y^3 + 64(125/x^3))/(125y^3).
Now, let's factor out the common factor of 8 from the numerator:
[8(x^3y^3 + 8(125/x^3))]/(125y^3).
Finally, we can further simplify the expression like this:
8(x^3y^3 + 1000/y^3)/(125y^3).
These instructions provide a general approach to factorizing different types of expressions. It's important to familiarize yourself with various factorization techniques and keep practicing to improve your skills.