Special factorization
I know someone has helped me with this before, but I'm really stuck on how to attack these two problems
[(8x^3)/125]+[64/(y^3)]
X^2-8xy+16y^2-25
I am not sure what you want but
8 = 2^3
125 = 5^3
64 = 2^6 = 4^3
(2x/5)^3 + (4/y)^3
but a^3 + b^3 = (a+b)(a^2-ab+b^2)
(2x/5+4/y)(4x^2/25-8x/y+16/y^2)
maybe complete the square?
(x^2-8xy+16y^2) -16y^2 +16y^2-25
(x-4y)^2 - 5^2
(a-b)^2 = (a+b)(a-b)
(x-4y+5)(x-4y+5)
Is it not 8x/5y not 8x/y. I'm only basing this off of what you gave me
Yes, you are right
and in the second one a sign error
(x-4y+5)(x-4y-5)
Thank you sir!!! And thanks for being awake(unless you're not from the U.S)
You are welcome and I am from a fishing town north of Boston where we rise early.
Pretty darn early, and I'm 30 mins away from philly
To simplify the expressions and find the special factorizations for the given problems, follow these steps:
Problem 1: Simplify [(8x^3)/125] + [64/(y^3)]
Step 1: Simplify each term individually by canceling out the common factors:
[(8x^3)/125] can be simplified to (2x/5)^3
[64/(y^3)] can be simplified to (4/y)^3
Step 2: Now that each term is written in a cube form, rewrite the expression:
[(8x^3)/125] + [64/(y^3)] = (2x/5)^3 + (4/y)^3
Step 3: Apply the special factorization formula for the sum of two cubes:
A^3 + B^3 = (A + B)(A^2 - AB + B^2)
In our case, A = 2x/5 and B = 4/y:
(2x/5)^3 + (4/y)^3 = ((2x/5) + (4/y))((2x/5)^2 - (2x/5)(4/y) + (4/y)^2)
Therefore, the special factorization for the expression [(8x^3)/125] + [64/(y^3)] is:
((2x/5) + (4/y))((2x/5)^2 - (2x/5)(4/y) + (4/y)^2)
Problem 2: Factorize x^2 - 8xy + 16y^2 - 25
Step 1: Recognize that the given expression is a quadratic expression.
Step 2: Try to factor the expression by splitting the middle term:
x^2 - 8xy + 16y^2 - 25 = (x^2 - 4xy - 4xy + 16y^2) - 25
Step 3: Group the terms and factor by grouping:
(x^2 - 4xy) + (-4xy + 16y^2) - 25 = x(x - 4y) - 4y(x - 4y) - 25
Step 4: Simplify further:
x(x - 4y) - 4y(x - 4y) - 25 = (x - 4y)(x - 4y) - 25
Step 5: Notice that we have (x - 4y) repeated, which can be rewritten as (x - 4y)^2:
(x - 4y)(x - 4y) - 25 = (x - 4y)^2 - 25
Step 6: Apply the special factorization formula for the difference of two squares:
A^2 - B^2 = (A - B)(A + B)
In our case, A = (x - 4y) and B = 5:
(x - 4y)^2 - 25 = [(x - 4y) - 5][(x - 4y) + 5]
Therefore, the special factorization for the expression x^2 - 8xy + 16y^2 - 25 is:
[(x - 4y) - 5][(x - 4y) + 5]