How do you factor x^4+5x^2+9?

Question

Please explain. Thanks in advance :)

Answer 1

given equation is x^4 + 5x^2 + 9

 let x^2 =a

                         a^2 + 5a + 9=0

 the equation comparing  ax^2 + bx + c=0

    a =1 , b = 5 ,c = 9

the formula is  =( -b (+ or -) squ( b^2 - 4ac))/ 2a

                     = (-5 (+ or -) squ (5^2 -4*1*9))/2*1

                     = ( -5 ( +or -) squ(25 -36))/2

                    =( -5 (+ or -)squ ( -9))/2

                     (-5 + 3i) /2,( -5 -3i)/2     (since  i^2= -1)

                 x^2=   (-5 + 3i)/2  , x^2 =  ( -5 -3i)/2

                 x = +or-squ ( -5 + 3i)/2 ,  x= + or -squ( -5 -3i)/2

the fractors are   x = +or-squ ( -5 + 3i)/2 ,  x= + or -squ( -5 -3i)/2

Comments

The answer is (x² - x + 3)(x² + x + 3).

Answer 2

The expression is x⁴ + 5x² + 9.

Add and subtract x².

= x⁴ + 5x²  + x² + 9 - x²

= x⁴ + 6x² + 9 - x²

= x⁴ + 3x² + 3x² + 9 - x²

= x² (x² + 3) + 3(x² + 3) - x²

= (x² + 3)(x² + 3) - x²

= (x² + 3)²  - x²

This is in the form of a² - b² which is equal to (a + b) (a - b) where a = x² + 3 and b = x.

= (x² + 3 + x )  (x² + 3 - x )