1. ## Find the product

can someone help me with the following: find the product of -3a^3b(2a^0b^4 - 4a^2b^3) Thanks much!

2. Originally Posted by jay1
can someone help me with the following: find the product of -3a^3b(2a^0b^4 - 4a^2b^3) Thanks much!
what have you tried? use the distributive law. take the $-3a^3b$ and multiply each term in the brackets. remember $x^a \cdot x^b = x^{a + b}$

3. hi Jay1,

$-3a^3b(2a^0b^4-4a^2b^3)=-3a^3b(2b^4-4a^2b^3)$

since any value to the power of zero is 1.

$=-6a^3b^{1+4}+12a^{3+2}b^{1+3}=6a^3b^4(2a^2-b)$

4. ## OOps

I typed this wrong it should read: -3a^3b(2a^0b^4 - 4a^2b^3) s the answer -6b^4 + 12a^6b^3?

5. That one is the same as you typed originally,
is the equation i wrote what you should have to start with?

6. yes it is. I'm not arriving at the same answer however. ???

7. Ok Jay1,

then we can show you why it's not working out.

8. I can see that you are unsure when you add the powers (exponents)
and when you multiply them.
Can you see this?

9. $-3a^2b(2a^0b^4)=-3(2)a^2a^0b^1b^4$

$b^1b^4=b[(b(b)b(b)]=b^5=b^{4+1}$

$(b^2)^3=[b(b)][b(b)][b(b)]=b^6=b^{2(3)}$

This illustrates the difference between addition and multiplication of powers

10. Okay I think that I have it. Thank you!

11. So the product of (x-2y)^2 is the answer x^2 - 4xy+4y2??? I also need help with solving the equation 3z^3 - 300z = 0. Is this already prime?

12. Originally Posted by jay1
So the product of (x-2y)^2 is the answer x^2 - 4xy+4y^2???
yes

I also need help with solving the equation 3z^3 - 300z = 0. Is this already prime?
no it isn't. you can factor 3z from the left side.

and from now on, post new problems in new threads.

13. Yes, for the first one, you just multiply it out term by term.

There are common terms in the 2nd one, though.

$3z^3-300z=0$

$3z(z^2-100)=0$

z could be zero or $z^2$ could be 100

Remember 100 has a positive and negative square root.