Results 1 to 6 of 6

Math Help - How do I simplify from my answer to the tutor's answer?

  1. #1
    Newbie
    Joined
    Oct 2011
    Posts
    8

    How do I simplify from my answer to the tutor's answer?

    Hi

    There's an algebra problem I've been working on - I've got as far as I can. My tutor says my answer is correct but can be simplified down to his answer - but I can't see how. If anyone can explain the steps on how to simplify to the answer I'd be really grateful!

    My answer:

    (x+3)x^2(2) - (2x+7)[(x+3)2x + x^2]
    /
    [(x+3)x^2]^2


    My tutor's simplified answer:

    -6x^3 -12x^2 - 13x -42
    /
    (x+3)^2 x^3
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member Youkla's Avatar
    Joined
    Oct 2011
    Posts
    26

    Re: How do I simplify from my answer to the tutor's answer?

    Unless you wrote it down incorrectly or I translated your heiroglyphics incorrectly, I got a different answer:

    \frac{2x^2(x+3) - (2x+7)[2x(x+3) + x^2]}{[x^2(x+3)]^2}

    Distribute and get:

    \frac{2x^3 + 6x^2 - (2x+7)(3x^2 + 6x)}{[x^2(x+3)]^2}

    Distribute some more and get:

    \frac{2x^3 + 6x^2 - 6x^3 - 12x^2 - 21x^2 - 42x}{[x^2(x+3)]^2}

    Combine like terms and get:

    \frac{-4x^3 - 27x^2 - 42x}{x^4(x+3)^2}

    Factor an x out of the numerator and cancel with an x in the denominator and get:

    \frac{-4x^2 - 27x - 42}{x^3(x+3)^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398

    Re: How do I simplify from my answer to the tutor's answer?

    Quote Originally Posted by econolondon View Post
    Hi

    There's an algebra problem I've been working on - I've got as far as I can. My tutor says my answer is correct but can be simplified down to his answer - but I can't see how. If anyone can explain the steps on how to simplify to the answer I'd be really grateful!

    My answer:

    (x+3)x^2(2) - (2x+7)[(x+3)2x + x^2]
    /
    [(x+3)x^2]^2

    {(x+3)x^2(2) - (2x+7)[(x+3)2x + x^2]} / [(x+3)x^2]^2

    My tutor's simplified answer:

    -6x^3 -12x^2 - 13x -42
    /
    (x+3)^2 x^3

    (-6x^3 -12x^2 - 13x -42)/[(x+3)^2 x^3]
    In LaTeX , your answer is: \frac{(x+3)x^2(2) - (2x+7)[(x+3)2x + x^2]}{((x+3)x^2)^2}

    Your tutor's answer is: \frac{-6x^3 -12x^2 - 13x -42}{(x+3)^2 x^3}

    Factor x out of the numerator. Cancel with a factor of x in the denominator.

    Expand (multiply out) all of the terms in the numerator, then combine like terms .

    There should not be an x^3 term in the numerator after cancelling. The tutor's answer is in error.

    Youkla's answer is correct.
    Last edited by SammyS; October 21st 2011 at 11:15 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2013
    From
    Iowa City, IA
    Posts
    13

    Re: How do I simplify from my answer to the tutor's answer?

    find a better math tutor Search: math - Tutor Universe
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,100
    Thanks
    67

    Re: How do I simplify from my answer to the tutor's answer?

    WHY are you reactivating old threads, Mr Spam?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2013
    From
    Iowa City, IA
    Posts
    13

    Re: How do I simplify from my answer to the tutor's answer?

    my apologies. I didn't realize this pushes the threads up. I'm new to this place. I will stop replying to old threads.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: August 31st 2010, 05:08 PM
  2. Replies: 1
    Last Post: February 2nd 2010, 03:04 AM
  3. Find the difference quotient and simplify answer
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: November 9th 2009, 10:14 PM
  4. How to simplify this answer?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 1st 2009, 08:14 AM

Search Tags


/mathhelpforum @mathhelpforum