how to factor a trinomial with FOIL in Reverse. Demonstrate with an example

Genrally speaking:

a trinomial X^2 + bx + c

Have to find 2 numbers whose sum is - b, and their product is c

X^2 - 10X +21

here -b = 10 c = 21

So the 2 numbers are 7, and 3 whose sum are 10 and product is 10

then, X^2 -10x + 21 = (X-7)(X-3)

if we work back ward

(x -3)(X-7) = ( X-3)X -7( x-3)

X^2 -3X -7x +21 = X^2-10X +21

That was for coefficient of X^2 equals 1

for 2X^2 -23X +38 to factor we have to find 2 numbers such that their Sum is 23 and their product is 2(38)=76

To find that we write 76= 2(2)(19)

we see that factors of 4,19 can be added to 23

2X^2-4x -19x +38= 2X (x -2) -19 (x -2)

then factoring by grouping 2X^2 -23 x + 38 = ( 2X -19 )(X- 2 )