EXPLANATION
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When you FOIL out two binomials like (x+2)(x+3) you end up with
FIRST x(x) ==> x2
OUTER x(3) ==> 3x
INNER 2(x) ==> 2x
LAST 2(3) ==> 6
The outer and inner combine and you end up with
x2 + 5x + 6
If I wanted to work this process in reverse, and figure out the factored form
of
x2 + 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
x2 + 5x + 6 in factored from must be (x+2)(x+3)
Similarly, to factor
x2 - 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
x2 - 2x - 8 in factored form must be (x+-4)(x+2) [which can also be written as (x-4)(x+2)
When the x2 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 x2
OUTER 5x(3x) ==> 15x
INNER 2(4x) ==> 8x
LAST 2(3) ==> 6
The outer and inner combine and you end up with
20x2 + 23x + 6
If I wanted to work this process in reverse, and figure out the factored form
of
20x2 + 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: 20x2 + 23x + 6
Another way to think about it is this: (mx + n)(qx + r) FOILED OUT is
(mq)x2 + (mr + nq)x + nr