Algebra and Trig question. Please show work. Serious replies only!!?

Question

Simplify: 3 squareroot of 432 a ^5 b^ 7

a).2 a^3b^3 ^3 squareroot 54 a^2 b
b). 27 a^3 b^3 ^3 squareroot 16 a ^2 b^4
c). 3ab^2 squareroot 16 a^2b
d). 6 ab^2 ^3squareroot 2 a^2b

Answer 1

Where is the trig. If you have a sqrt, you can "pull out", or remove from the sqrt( ), even powers since sqrt(a^2n)= a^n. If you have an odd power of 3 or more, you can pull out the next lowest power, which is even. You can also pull out numbers whose sqrt is an integer.
Problem 1: 432 is a multiple of 144.
So you can remove an a^4 and a b^6 to get
36 a^2 b^3 sqrt(3 ab)

The others are pretty much the same in terms of what you do, so you can do it yourself.

Answer 2

C
3 squareroot of 432 a ^5 b^ 7
432 is not a perfect squareroot; so you must look for the largest numbers that will multiple to 432, and one of the numbers MUST be a perfect squareroot. So, 27 and 16.
16 is not a perfect square, but 27 is (the square root is 3).
Split a ^5 into a^3 and a^2. 3 will divide evenly into a^3; a^2 is the left over.
And Split b^7 into b^6 and b. 3 will divide evenly into b^6; b is the left over.
c). 3ab^2 squareroot 16 a^2b is the answer.

Answer 3

sqt of 432 = 20.7846
3 x sqt of 432 = 62.3538
sqt a^5 = a^2.5
sqrt b^7 = b^3.5

ans: 62.3538 a^2.5 b^3.5