solve 5x^2+13x=6 by factoring

Question

I have tried to figure the problem out and I am lost

Answer 1

I do it like the followings:

 

    5 X^2  + 13X - 6 = 0

 

     Have to verify and see if, there exist  2 integers whose product is 5(-6) = -30 , and Sum is +13

 

       On scratch paper, not here, I write  30 as product of all its factor:   30 = 2 * 15 = 6* 5 

       

           How about  +15 , -2    15 -2 = 13 ,  15( -2) = -30

     

            I break   13X = 15X -2

 

       5 X^ 2 + 15X - 2X  - 6        / Now we factor by grouping

 

        5 X( X + 3) - 2 ( X +3 ) =0

 

         ( X+3) ( 5X -2 ) =0

 

           X +3 = 0   X = -3          5X -2 = 0      X = 2/5

 

           In the case of   : a X² + bX + c

 

              The equation of  m+ n = b

                                       m.n = ac

                                                     does not have integer as solution, then have to use quadratic formula

                                                      to obtain the irrational numbers as the solution. 

 

 

                     

Answer 2

5x^2+13x=6

5x^2+13x-6=0

 

5=1*5 and 6=1*6 or 6=2*3

use 1 and 5 together with 2 and 3

3*5=15, 1*2=2, and 15-2=13

(5x-2)(x+3)=0

5x-2=0

5x=2

x=2/5

x+3=0

x=-3

Answer 3

Hi Chandler;

5x²+13x=6

First, let's move the 6 to the other side.  It goes from being positive to negative.

5x²+13x-6=0

For the FOIL...

FIRST...(5x)(x).

OUTER and INNER must add-up to 13x.

LAST must be (3)(2) or (2)(3) or (6)(1) or (1)(6) and one number must be negative.

(5x-2)(x+3)

FOIL...

FIRST...(5x)(x)=5x²

OUTER...(5x)(3)=15x

INNER...(-2)(x)=-2x

LAST...(-2)(3)=-6

5x²+15x-2x-6=0

5x²+13x-6=0