Results 1 to 4 of 4

Math Help - Factoring Question

  1. #1
    Member
    Joined
    Jun 2012
    From
    Australia
    Posts
    86

    Factoring Question

    Hello,

    I can't figure out how I should factor this expression.

    x^4 - 18x^2 + 1

    I know the answer is (x^2 - 4x - 1)(x^2 + 4x - 1)

    But I don't know how it gets to this point. I've tried substituting x squared with y to get this...

    y^2 - 18y + 1

    but this can't really be factored much more, so I don't know what I should be doing..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Factoring Question

    Quote Originally Posted by astuart View Post
    Hello,

    I can't figure out how I should factor this expression.

    x^4 - 18x^2 + 1

    I know the answer is (x^2 - 4x - 1)(x^2 + 4x - 1)

    But I don't know how it gets to this point. I've tried substituting x squared with y to get this...

    y^2 - 18y + 1

    but this can't really be factored much more, so I don't know what I should be doing..
    You could complete the square...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539

    Re: Factoring Question

    Hello, astuart!

    This require complete-the-square,
    . . but in a different manner.


    \text{Factor: }\:x^4 - 18x^2 + 1

    We have: . x^4 + 1 - 18x^2

    Subtract and add 2x^2\!:\;x^4\; {\color{red}-\; 2x^2} + 1 - 18x^2 \;{\color{red}+\; 2x^2}

    . . . . . . . . . . . . . . =\;(x^4 - 2x^2 + 1) - 16x^2

    . . . . . . . . . . . . . . =\;(x^2-1)^2 - (4x)^2

    . . . . . . . . . . . . . . =\;\left([x^2-1)-4x\right]\left([x^2-1]+4x\right)

    . . . . . . . . . . . . . . =\;(x^2-4x-1)(x^2+4x-1)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2012
    From
    Australia
    Posts
    86

    Re: Factoring Question

    Quote Originally Posted by Soroban View Post
    Hello, astuart!

    This require complete-the-square,
    . . but in a different manner.



    We have: . x^4 + 1 - 18x^2

    Subtract and add 2x^2\!:\;x^4\; {\color{red}-\; 2x^2} + 1 - 18x^2 \;{\color{red}+\; 2x^2}

    . . . . . . . . . . . . . . =\;(x^4 - 2x^2 + 1) - 16x^2

    . . . . . . . . . . . . . . =\;(x^2-1)^2 - (4x)^2

    . . . . . . . . . . . . . . =\;\left([x^2-1)-4x\right]\left([x^2-1]+4x\right)

    . . . . . . . . . . . . . . =\;(x^2-4x-1)(x^2+4x-1)
    Ah, so by subtracting 2x^2 from the equation, it results in 16x^2, which is a perfect square.
    I was confused by the subtraction of the 2x^2, I didn't know what the reason for it was..

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Factoring Question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 3rd 2012, 03:08 PM
  2. Need help factoring a question.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 20th 2011, 10:06 PM
  3. Question on Factoring
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 30th 2008, 10:55 AM
  4. Factoring question
    Posted in the Algebra Forum
    Replies: 9
    Last Post: July 8th 2007, 07:32 PM
  5. Factoring question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 27th 2007, 09:58 AM

Search Tags


/mathhelpforum @mathhelpforum