Factor over the reals completely
32a^4b-18b^3
If you factor out a two b
2b(16a^4-9b^2) I see a difference of two squares...
2b( [4a^2]^2 -[3b]^2 )
To factor the expression 32a^4b - 18b^3 completely over the reals, we need to first look for any common factors among the terms. In this case, we notice that both terms have a factor of 2 and a factor of b. Therefore, we can factor out these common factors:
32a^4b - 18b^3 = 2b(16a^4 - 9b^2)
Now, we are left with the expression 16a^4 - 9b^2. This expression is a difference of squares, which can be factored as follows:
16a^4 - 9b^2 = (4a^2)^2 - (3b)^2
Using the difference of squares formula, (a^2 - b^2) = (a + b)(a - b), we can rewrite the expression as:
(4a^2)^2 - (3b)^2 = (4a^2 + 3b)(4a^2 - 3b)
Putting it all together, the completely factored expression is:
32a^4b - 18b^3 = 2b(4a^2 + 3b)(4a^2 - 3b)