EXPLANATION

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When you FOIL out two binomials like (x+2)(x+3) you end up with

FIRST x(x) ==> x^{2}

OUTER x(3) ==> 3x

INNER 2(x) ==> 2x

LAST 2(3) ==> 6

The outer and inner combine and you end up with

x^{2} + 5x + 6

If I wanted to work this process in reverse, and figure out the factored form
of

x^{2} + 5x + 6

I would look for a pair of numbers that multiply to 6 that add to 5.

My only choices for pairs that multiply to 6 are 6 and 1 and then 2 and 3.

Of those choices only 2 and 3, add up to 5 so

x^{2} + 5x + 6 in factored from must be (x+2)(x+3)

Similarly, to factor

x^{2} - 2x - 8

I need to think of all the factors of -8 that add to -2. The possible
pairs are

-8 and 1

8 and -1

-4 and 2

4 and -2

Only one these pairs add to -2 so

x^{2} - 2x - 8 in factored form must be (x+-4)(x+2)
[which can also be written as (x-4)(x+2)

When the x^{2 }has a coefficient it is a little more
complicated.

When you FOIL out two binomials like (5x+2)(4x+3) you end up with

FIRST 5x(4x) ==>20 x^{2}

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

INNER 2(4x) ==> 8x

LAST 2(3) ==> 6

The outer and inner combine and you end up with

20x^{2} + 23x + 6

If I wanted to work this process in reverse, and figure out the factored form
of

20x^{2} + 23x + 6

I would look for a pair of numbers that multiply to 6 another pair of numbers that multiply to 20 AND the way that I arrange these numbers have to add to 23.

factors of 6 are 6 and 1 and 2 and 3

factors of 20 are 20 and 1, 2 and 10, 4 and 5

These are all the possible combinations:

(20x + 6) (1x + 1)

(20x + 1) (1x + 6)

(20x + 2) (1x + 3)

(20x + 3) (1x + 2)

(2x + 6) (10x + 1)

(2x + 1) (10x + 6)

(2x + 2) (10x + 3)

(2x + 3) (10x + 2)

(4x + 6) (5x + 1)

(4x + 1) (5x + 6)

(4x + 2) (5x + 3)

(4x + 3) (5x + 2)

ONLY THE LAST ONE GIVES YOU: 20x^{2} + 23x + 6

Another way to think about it is this: (mx + n)(qx + r) FOILED OUT is

(mq)x2 + (mr + nq)x + nr