Factorize the following expressions completely.
6x2y3 +18xy + 3x2y2 + 9x
y2 – x2 + 8y + 16
64x6 – y6
9x4y5+18xy+9x
I think this is the answer to the first one.
Walker's answer is very very wrong!
You cannot add unlike terms.
this is done by grouping
6x2y3 +18xy + 3x2y2 + 9x
= 6x^2y^3 + 3x^2y^2 + 18xy + 9x
= 3x^2y^2(2y + 3) + 9x(2y + 3)
= (2y+3)(3x^2y^2 + 9x)
= x(2y+3)(3xy^2 + 9)
y2 – x2 + 8y + 16 = (y-4)^2 - x^2
= (y-4+x)(y-4-x)
64x^6 - y^6 is also the difference of two squares, and so can be written
(8x^3 -y)(8x^3 + y)
Additional factoring, which is possible, would require fractional powers of y, so I will stop there
I forgot about that. Thank You for correcting me :).
To factorize the given expressions completely, we need to break them down into their simplest possible form by finding common factors and using factoring techniques. Let's start with each expression:
1. 6x^2y^3 + 18xy + 3x^2y^2 + 9x
First, let's look for common factors among the terms:
Common factors: 3x
Now, factor out the common factor:
3x(2xy^3 + 6y + xy^2 + 3)
Next, let's pair up the terms and look for common factors again:
Common factors: y
y(2xy^2 + 6 + xy + 3)
Finally, we can factorize it completely:
3xy(y + 2)(x + 1)
2. y^2 – x^2 + 8y + 16
This expression has four terms. Let's look for differences of squares and try to factorize accordingly.
Rearranging the terms, we get:
(y^2 + 8y) – (x^2 – 16)
Now, factorize using the difference of squares:
y(y + 8) – (x – 4)(x + 4)
So, the completely factorized form is:
y(y + 8) – (x – 4)(x + 4)
3. 64x^6 – y^6
This expression is a difference of cubes. Let's factorize using that property.
Using the formula for difference of cubes, we get:
(4x^2)^3 – y^6
We can now factor it as:
(4x^2 – y^2)(16x^4 + 8x^2y^2 + y^4)
Therefore, the completely factorized form is:
(4x^2 – y^2)(16x^4 + 8x^2y^2 + y^4)
By following these factoring techniques and finding common factors among the terms, we were able to completely factorize the given expressions.