Question

4x^4-x^2-12x-36 factorise the following

Answer 1

Hi ,

Let p( x ) = 4x⁴ - x² - 12x - 36

= 4x⁴ - ( x² + 12x + 36 )

= 4x⁴ - [ x² + 2 × x × 6 + 6² ]

= ( 2x² )² - ( x + 6 )²

= [ 2x² + ( x + 6 ) ][ 2x² - ( x + 6 ) ]

= ( 2x² + x + 6 )( 2x² - x - 6 )

= ( 2x² + x + 6 )[ 2x² - 4x + 3x - 6 ]

= ( 2x² + x + 6 ) [ 2x( x - 2 ) + 3( x - 2 )]

= ( 2x² + x + 6 )( x - 2 )( 2x + 3 )

I hope this helps you.

: )

Answer 2

ANSWER :

[$2x^{2}$+x+6] [x-2] [2x+3]

STEP BY STEP EXPLANATION:

$4x^{4} -x^{2} -12x-36$

  = $4x^{4} -[x^{2} +12x+36]$

  =$4x^{4} -[x^{2} +6x+6x+36]$

  =$4x^{4} -[x(x+6)+6(x+6)]$

  =$4x^{4} -[x+6][x+6]$

  =$4x^{4} -[x+6]^{2}$

  =$(2x^{2} )^{2} - [x+6]^{2}$

  =$[2x^{2} +[x+6]][2x^{2} -[x+6]$

 =$[2x^{2} +x+6] [2x^{2} -x-6]$

 =$[2x^{2} +x+6] [2x^{2} -4x+3x-6]$

 =$[2x^{2} +x+6] [2x(x-2)+3(x-2)]$

 =$[2x^{2} +x+6] [ (x-2) (2x+3)]$

ANSWER:

=$[2x^{2} +x+6] [ x-2] [2x-3]$

 

HOPE IT HELPS YOU ........

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