Results 1 to 7 of 7

Math Help - Algebra with cubics

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    8

    Algebra with cubics

    Ok im studying for a maths exam and there is one equation i cant do..

    If a factor of P(x)=-7+ax+5x^2+15x^3+bx^4 is (x^2-1) find the values of a and b?

    here is the answer i cant work out how to do it any help is appreciated thanks

    answer: a= -15 b= 2

    thanks Tarik
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    x^2 - 1 = (x - 1)(x + 1).

    So that means x - 1 and x + 1 are factors.


    From this, you have P(1) = 0 and P(-1) = 0 by the factor theorem.

    Substitute and solve the two resulting equations simultaneously.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2010
    Posts
    8
    ok another similiar question popped up and i tried to apply the same logic and bam i failed again :S

    if (x+3) is a factor of f(x)= -x^3+bx^2+ax-18 and g(x)=ax^2+bx-75, find values of a and b

    I begun with

    f(-3)= -(-3)^3 + b(-3)^2+a(-3)-18
    0=9+9b-3a
    therefore a =3b+3

    not sure what from here
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    x + 3 is also a factor of g(x).

    So evaluate g(-3), set it = 0 and get another equation in terms of a and b.

    Then solve them simultaneously for a, b.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2010
    Posts
    8
    thanks alot Prove it.. ok im going to start to become a nuisance here but i am now stuck on another question.. and btw im away so i have no teachers to help me before my exam soon


    the graph of y= (a)/(x-b) + c has a vertical asymptote at x=2 and a horizontal asymptote at y=-1


    a.) find values of b and c
    which will be b = 2 and c=1


    thus new equation is y= (a)/(x-2) + 1


    This graph then udergoes following transformations
    . reflection in the x axis
    . dilation by a factor of 3 from the x axis
    .horizontal shif of 2 units right

    b.) if the intersection of the two graphs is at (m,2), find the value of m

    c.)hence find the equation of the transformed graph

    now im very lost here...

    p.s is there any good way of doing fractions?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    Ok, first it would really help if you'd learn some LaTeX, makes reading the questions a lot easier.

    I assume that you have y = \frac{a}{x - b} + c with vertical asymptote at x = 2 and horizontal asymptote at y = -1.

    Your b value is correct, your c value is not. It should be -1.

    This is because it is a standard vertical translation of the hyperbola y = \frac{1}{x}. This has horizontal asymptote at y = 0. By subtracting -1, it translates everything down by 1 unit. So the c value is -1.

    So your equation is y = \frac{a}{x - 2} - 1.


    I'm sure you must have studied transformations, what you have had to do to translate, dilate and reflect your graph (they will be new constants appearing somewhere in the equation). Reread your notes to familiarise yourself with where these constants go to do these particular transformations.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2010
    Posts
    8
    oh whoops thanks, the only part im confused with is how to work out the A variable.
    And how do i do latex?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cubics!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 12th 2010, 01:16 PM
  2. Cubics
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 12th 2010, 01:16 PM
  3. cubics
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 2nd 2010, 08:25 PM
  4. cubics
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 9th 2009, 04:45 AM
  5. Cubics fun
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 13th 2008, 06:50 AM

Search Tags


/mathhelpforum @mathhelpforum