Results 1 to 13 of 13

Math Help - Find the product

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    122

    Find the product

    can someone help me with the following: find the product of -3a^3b(2a^0b^4 - 4a^2b^3) Thanks much!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by jay1 View Post
    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}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    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)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2009
    Posts
    122

    OOps

    I typed this wrong it should read: -3a^3b(2a^0b^4 - 4a^2b^3) s the answer -6b^4 + 12a^6b^3?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    That one is the same as you typed originally,
    is the equation i wrote what you should have to start with?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2009
    Posts
    122
    yes it is. I'm not arriving at the same answer however. ???
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Ok Jay1,

    can you write your steps showing how you got your answer,
    then we can show you why it's not working out.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    I can see that you are unsure when you add the powers (exponents)
    and when you multiply them.
    Can you see this?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    -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
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Dec 2009
    Posts
    122
    Okay I think that I have it. Thank you!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Dec 2009
    Posts
    122
    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by jay1 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find the product
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 9th 2010, 08:38 PM
  2. Find the Product
    Posted in the Algebra Forum
    Replies: 6
    Last Post: April 21st 2009, 09:01 PM
  3. Find each product. please help :(
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 14th 2007, 03:59 PM
  4. please help :( Find the product.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 14th 2007, 02:39 PM
  5. Find the product of...
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: January 30th 2007, 06:42 PM

Search Tags


/mathhelpforum @mathhelpforum