how do you factor x^4-5x^2-6?

Comments

• anonymous

  May 2021

  = (x^2 + 1)(x^2 - 6) = (x^2 + 1)(x - sqrt6)(x + sqrt6)
• mred11

  May 2021

  x ^ 4 - 5 x ^ 2 - 6

  y = x ^ 2

  y ^ 2 - 5 y - 6

  ( y + 1 )( y - 6 )

  ( x ^ 2 + 1)( x ^ 2 - 6 )

  Solve each of the two factors using quadratic formula,

  For ( x ^ 2 + 1 ),

  x = ( 0 +- ( ( 0 ^ 2 ) - 4( 1 )( 1 ) ) ^ 1/2 ) / ( 2( 1 ) )

  = +- ( -4 ^ 1/2 ) / 2

  = +- 2i / 2

  = +- i

  So ( x ^ 2 + 1 ) = ( x + i )( x - i )

  For ( x ^ 2 - 6 ),

  x = ( 0 +- ( ( 0 ^ 2 ) - 4 ( 1 )( -6 ) ) ^ 1/2 ) / ( 2 ( 1 ) )

  = +- ( 24 ^ 1/2 ) / 2

  = +- 2( 6 ^ 1/2 ) / 2

  = +- ( 6 ^ 1/2 )

  So ( x ^ 2 - 6 ) = ( x + ( 6 ^ 1/2 ) )( x - ( 6 ^ 1/2 ) )

  Therefore,

  x ^ 4 - 5 x ^ 2 - 6 =

  ( x + i )( x - i )( x + ( 6 ^ 1/2 ) )( x - ( 6 ^ 1/2 ) )
• alwbsok

  May 2021

  This is a quadratic in x^2 (rather than in x, like you may be used to). We factorise the same way:

  x^4 - 5x^2 - 6

  = (x^2 - 6)(x^2 + 1)

  Now, we have some differences of two squares:

  = (x^2 - (sqrt(6))^2)(x^2 - i^2)

  = (x - sqrt(6))(x + sqrt(6))(x - i)(x + i)
• syed-muhammad-hussain

  May 2021

  How to factorize x^4 + 5x^2 +2
• cheese

  May 2021

  an easy way to do this is to let a=x^2

  a^2-5a-6=0

  (a+1)(a-6)=0

  sub a=x^2 back in.

  x^2=-1 (not possible for real) x^2=6

  x=+or-sqrt(6)

  for complex... x=i and x=+or-sqrt(6)
• kusmosis

  May 2021

  Well, to simplify this, there is a very easy method.

  First, we'll let r = x^2

  So, we can simplify the first equation to:

  r^2 - 5r - 6

  By factoring that we get:

  (r - 3)(r - 2)

  then just substitute x^2 for r back in.

  (x^2 - 3)(x^2 - 2)

  If you wanted the roots for this equation, you would just set each of those to equal zero.

  x^2 = 3

  x = +- 3

  x^2 = 2

  x = +- 2
• steve

  May 2021

  let u = x^2, then treat it like a quadratic. It's easy!