Factoring problem... x^4- x^2-12=0?

Question

factor this......

x^4 - x^2 -12= 0

Answer 1

xˆ4 - xˆ2 - 12 = 0

xˆ4 - xˆ2 = 12

xˆ2(xˆ2 - 1) = 12

Answer 2

(x²+3)(x-2)(x+2)

Answer 3

Question Number 1 :

For this equation x^4 -x^2 - 12 = 0 , answer the following questions :

A. Use factorization to find the roots of the equation !

B. Use completing the square to find the roots of the equation !

Answer Number 1 :

The equation x^4 -x^2 - 12 = 0 is already in a*x^4+b*x^2+c=0 form.

In that form, we can easily derive that the value of a = 1, b = -1, c = -12.

1A. Use factorization to find the roots of the equation !

  x^4 -x^2 - 12 = 0

  ( x^2 - 4 ) * ( x^2 + 3 ) = 0

  It imply that x1^2 = 4 and x2^2 = -3

  To get the root, remember x1 = sqrt( 4 ) , x2 = sqrt( -3 ) ,  x3 = -sqrt( 4 ) and x4 = -sqrt( -3 )

  The answers are x1 = 2 , x2 = 1.73205080756888*i ,  x3 = -2 and x4 = -1.73205080756888*i ,

1B. Use completing the square to find the roots of the equation !

  x^4 -x^2 - 12 = 0 ,divide both side with 1

  So we get x^4 -x^2 - 12 = 0 ,

  And the coefficient of x is -1

  We have to use the fact that ( x^2 + q )^2 = x^4 + 2*q*x^2 + q^2 , and assume that q = -1/2 = -0.5

  So we have make the equation into x^4 -x^2 + 0.25 - 12.25 = 0

  So we will get ( x^2 - 0.5 )^2  - 12.25 = 0

  Which can be turned into (( x^2 - 0.5 ) - 3.5 ) * (( x^2 - 0.5 ) + 3.5 ) = 0

  By using the associative law we get ( x^2 - 0.5 - 3.5 ) * ( x^2 - 0.5 + 3.5 ) = 0

  Just add up the constants in each brackets, and we get ( x^2 - 4 ) * ( x^2 + 3 ) = 0

  By doing so we get x1^2 = 4 and x2^2 = -3

  So x1 = sqrt( 4 ) , x2 = sqrt( -3 ) ,  x3 = -sqrt( 4 ) and x4 = -sqrt( -3 )

  So we have the answers x1 = 2 , x2 = 1.73205080756888*i ,  x3 = -2 and x4 = -1.73205080756888*i ,

Answer 4

(x^2+3)(x^2-4)  sq minus sq can be further factored

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

Answer 5

(x^2-4)(x^2+3)

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