Answer:
i used to know how to do this but i forgot sorry hope someone helps
Step-by-step explanation:
Good Luck!!
Answer: 2
Step-by-step explanation:
Answer:
![4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B2%7Dx%28%5Csqrt%5B3%5D%7By%7D%2B3xy%5Csqrt%5B3%5D%7By%7D%20%29)
Step-by-step explanation:
Let's start by breaking down each of the radicals:
![\sqrt[3]{16x^3y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B16x%5E3y%7D)
Since we're dealing with a cube root, we'd like to pull as many perfect cubes out of the terms inside the radical as we can. We already have one obvious cube in the form of
, and we can break 16 into the product 8 · 2. Since 8 is a cube root -- 2³, to be specific, we can reduce it down as we simplify the expression. Here our our steps then:
![\sqrt[3]{16x^3y}\\=\sqrt[3]{2\cdot8\cdot x^3\cdot y}\\=\sqrt[3]{2} \sqrt[3]{8} \sqrt[3]{x^3} \sqrt[3]{y} \\=\sqrt[3]{2} \cdot2x\cdot \sqrt[3]{y} \\=2x\sqrt[3]{2}\sqrt[3]{y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B16x%5E3y%7D%5C%5C%3D%5Csqrt%5B3%5D%7B2%5Ccdot8%5Ccdot%20x%5E3%5Ccdot%20y%7D%5C%5C%3D%5Csqrt%5B3%5D%7B2%7D%20%5Csqrt%5B3%5D%7B8%7D%20%5Csqrt%5B3%5D%7Bx%5E3%7D%20%5Csqrt%5B3%5D%7By%7D%20%5C%5C%3D%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot2x%5Ccdot%20%5Csqrt%5B3%5D%7By%7D%20%5C%5C%3D2x%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7By%7D)
We can apply this same technique of "extracting cubes" to the second term:
![\sqrt[3]{54x^6y^5} \\=\sqrt[3]{2\cdot27\cdot (x^2)^3\cdot y^3\cdot y^2} \\=\sqrt[3]{2}\sqrt[3]{27} \sqrt[3]{(x^2)^3} \sqrt[3]{y^3} \sqrt[3]{y^2}\\=\sqrt[3]{2}\cdot 3\cdot x^2\cdot y \cdot \sqrt[3]{y^2} \\=3x^2y\sqrt[3]{2} \sqrt[3]{y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B54x%5E6y%5E5%7D%20%5C%5C%3D%5Csqrt%5B3%5D%7B2%5Ccdot27%5Ccdot%20%28x%5E2%29%5E3%5Ccdot%20y%5E3%5Ccdot%20y%5E2%7D%20%5C%5C%3D%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7B27%7D%20%5Csqrt%5B3%5D%7B%28x%5E2%29%5E3%7D%20%5Csqrt%5B3%5D%7By%5E3%7D%20%5Csqrt%5B3%5D%7By%5E2%7D%5C%5C%3D%5Csqrt%5B3%5D%7B2%7D%5Ccdot%203%5Ccdot%20x%5E2%5Ccdot%20y%20%5Ccdot%20%5Csqrt%5B3%5D%7By%5E2%7D%20%5C%5C%3D3x%5E2y%5Csqrt%5B3%5D%7B2%7D%20%5Csqrt%5B3%5D%7By%7D)
Replacing those two expressions in the parentheses leaves us with this monster:
![2(2x\sqrt[3]{2}\sqrt[3]{y})+4(3x^2y\sqrt[3]{2} \sqrt[3]{y})](https://tex.z-dn.net/?f=2%282x%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7By%7D%29%2B4%283x%5E2y%5Csqrt%5B3%5D%7B2%7D%20%5Csqrt%5B3%5D%7By%7D%29)
What can we do with this? It seems the only sensible thing is to look for terms to factor out, so let's do that. Both terms have the following factors in common:
![4, \sqrt[3]{2} , x](https://tex.z-dn.net/?f=4%2C%20%5Csqrt%5B3%5D%7B2%7D%20%2C%20x)
We can factor those out to give us a final, simplified expression:
![4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B2%7Dx%28%5Csqrt%5B3%5D%7By%7D%2B3xy%5Csqrt%5B3%5D%7By%7D%20%29)
Not that this is the same sum as we had at the beginning; we've just extracted all of the cube roots that we could in order to rewrite it in a slightly cleaner form.
5x + 7
5x represents the weight of the 5 striped fish.
7 represents the fish that weighs 7 pounds.
Answer:
Step-by-step explanation:
This is a third degree polynomial because we are given three roots to multiply together to get it. Even though we only see "2 + i" the conjugate rule tells us that 2 - i MUST also be a root. Thus, the 3 roots are x = -4, x = 2 + i, x = 2 - i.
Setting those up as factors looks like this (keep in mind that the standard form for the imaginary unit in factor form is ALWAYS "x -"):
If x = -4, then the factor is (x + 4)
If x = 2 + i, then the factor is (x - (2 + i)) which simplifies to (x - 2 - i)
If x = 2 - i, then the factor is (x - (2 - i)) which simplifies to (x - 2 + i)
Now we can FOIL all three of those together, starting with the 2 imaginary factors first (it's just easier that way!):
(x - 2 - i)(x - 2 + i) = 
Combining like terms and canceling out the things that cancel out leaves us with:

Remembr that
, so we can rewrite that as
and

That's the product of the 2 imaginary factors. Now we need to FOIL in the real factor:

That product is

which simplifies down to

And there you go!