Results 1 to 6 of 6

Math Help - More questions to solve!

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    49

    More questions to solve!

    I've done my homework this time and I think I have answers to some of my questions!
    The thing is I don't know if I'm right.......

    Factor the following polynomials.

    1
    x^2+x-6

    I used brute force and got the answer (x+3)(x-2), I do not remember how to show working for this kind of question though.

    2
    6x^2+13x-5

    Same deal here....Brute force, no idea how to show working.....


    The area of a right triangle is 250 in^2. find the lengths of its legs if one leg is 5 inches longer than the other.(use the five step method)

    I know the lengths: 20 in and 25 in, but as above I use brute force, so--yeah, I'm a little clueless as to how to show my work.


    x^3+5x^2=6x

    I think that the answer could be 0,1 or -6, I found -6 using a math program so I don't quite know how I would show that--are they all correct, or just 2 of them? Could I just put {-6,0,1} and call that my answer?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jul 2009
    Posts
    16
    1,2:

    You have to find the roots of the quadratic equation, because

    6\left(x-x_{1}\right)\left(x-x_{2}\right)=6x^{2}+13x-5,

    where x_{1},x_{2} are roots of the equation.

    3:

    It is known that the area of a right triangle is \frac{ab}{2}, because that is a half rectangle, where a,b are the legs of the triangle.

    \frac{x(x+5)}{2}=250
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    Another method to use when trying to factor a quadratic ax^2+bx+c, is to multiply the coefficient of the x^2 term and the constant term (ac), then look for a pair of factors for this whose sum is the coefficient of the linear term (b). So for problem 1, ac=1(-6)=-6; so we look for two numbers whose product is -6 and whose sum is 1. Since ac=-6 is negative, one factor will be negative, and the other positive; so our problem is equivalent to finding factors of 6 whose difference is 1. Since the pairs of factors that multiply to 6 are 1 & 6 or 2 & 3, we see that 3-2=1, and so 3*(-2)=-6, and 3+(-2)=1. These two factors form the two x terms produced in FOIL that add to become the x term in the quadratic, so
    x^2+x-6=x^2+3x-2x-6=(x^2+3)+(-2x-6)=x(x+3)-2(x+3)=(x-2)(x+3).

    Similarly for problem 2, ac=6*(-5)=-30, so we need a pair of numbers that multiply to -30 and add to 13; looking at pairs of factors of 30 and their differences:
    30 & 1; difference=29
    15 & 2; difference=13*
    so we have 15*(-2)=-30, 15+(-2)=13,
    so
    6x^2+13x-5=6x^2+15x-2x-5
    \;\;\;=(6x^2+15x)+(-2x-5)
    \;\;\;=3x(2x+5x)-(2x+5)
    \;\;\;=(3x-1)(2x+5).

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2010
    Posts
    49
    ROFLMAO!
    I forgot to add the answer I found to my second question, I see that I'm a little off with it, but I'll still post it:

    2
    6x^2+13x-5


    (2x-5)(3x+1)
    Same deal here....Brute force, no idea how to show working.....

    Thank you guys, I think I understand how to work those questions!
    WOOOO!!

    One small thing:
    What do I do here?

    x^3+5x^2=6x

    I think that the answer could be 0,1 or -6, I found -6 using a math program so I don't quite know how I would show that--are they all correct, or just 2 of them? Could I just put {-6,0,1} and call that my answer?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    x^3+5x^2=6x
    x^3+5x^2-6x=0
    x(x^2+5x-6)=0
    For this to be true,
    Either x=0 (which gives us one solution)
    or x^2+5x-6=0
    Because when timsed, the result must equal 0
    By factorizing that quadratic, can you finish from here?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,116
    Thanks
    68
    Quote Originally Posted by quikwerk View Post
    What do I do here?
    x^3+5x^2=6x
    Set = 0 : x^3 + 5x^2 - 6x = 0
    Common factor x: x(x^2 + 5x - 6) = 0
    Factor: x(x - 1)(x + 6) = 0
    Sooooo: x = 0 or x = 1 or x = -6
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2 Questions I cannot Solve
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 13th 2010, 04:51 PM
  2. Replies: 4
    Last Post: July 19th 2008, 08:18 PM
  3. Replies: 6
    Last Post: June 17th 2007, 12:31 PM
  4. please help me to solve that questions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 26th 2006, 01:23 PM
  5. please solve these questions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 20th 2006, 05:37 PM

Search Tags


/mathhelpforum @mathhelpforum