Factorise 6x^2+8x-3

Question

Must be easy....... Stumped

Answer 1

 To find out if a quadratic is factorable to its rational roots is:

 

   like in the example:

 

   f(X ) = 6X^2 + 8X - 18   

 

           = 2 ( X^2 + 4X - 9)     in order to have a factors of integers  - 9 has to be factored to numbers whose

                                            difference is 4 ,  

                                                     9 = 9 *1 = 3 * 3, doesn't have such factors.

                                                    So the quadratic doesn't have rational roots.

 

     Have to factor it with completing squares.

 

       f( X ) = 2 ( X^2 + 4X - 9)

                = 2 ( X^2 +4X +4 - 4 -9)

                = 2 [ ( X + 2 ) ^2 - 13 ]

                 = 2 [ ( X + 2 + √13) ( X +2 - √13 ) ]

 

    Quadratic formula developed by factoring by completing square of a quadratic in a generalized form of:

 

  a X^2 + bX + c.

      

Answer 2

f(x) = 6x^2 + 8x - 3

 

f(x) = 0 if:

 

x = (-8 ± √(8^2-4(6)(-3)))/(2*6)   <== USING QUADRATIC FORMULA!

 

x = (-4 ± √(2*8-(6)(-3)))/6

 

x = (-4 ± √(2(8+9))/6

 

x = (-4 ± √(34))/6

 

f(x) = 6(x - (-4 - √(34))/6)(x - (-4 + √(34))/6)

 

check:

 

f(x) = (1/6) (6x-(-4 - √(34)) ) (6x-(-4 + √(34)) )

 

= (1/6) (36x^2 -6x(-4 + √(34) -4 - √(34)) + (-4 - √(34))(-4 + √(34)) )

= (1/6) (36x^2 -6x(-8) + 16 - 34)

= (1/6) (36x^2 + 48x - 18)

= 6x^2 + 8x - 3   √