1. ## Factoring trinomials

This method of factoring trinomials was shown to me
. . by one of my students many years ago.
You may find it as fascinating as I did.

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Example: $6x^2 - 35x + 36$

Write the two pairs of parentheses and "split the x's": . $(\;x\qquad)(\:x\qquad)$

Use the first coefficient twice: . $(\overbrace{6x}^\downarrow\qquad)(\overbrace{6x}^\ downarrow\qquad)$

Multiply the first coefficient by the last coefficient: . $6 \times 36 \,=\,216$

We will factor $216$ into two parts.
Note the sign of the last term.
. . If "+", think sum.
. . If "-", think difference.

Since the last sign is "+", factor $216$ into two parts
. . whose sum is the middle coefficient $(35)$.

To factor $216$ into all possible pairs, divide by 1, 2, 3, ...
. . keeping those that "come out even".

$216\:=\:\begin{Bmatrix}1\cdot216 \\2\cdot108\\3\cdot72 \\ 4\cdot54\\6\cdot36\\8\cdot27\\9\cdot24\\12\cdot18\ end{Bmatrix}$

The pair with a sum of 35 is: $8$ and $27.$

Since the middle coefficient is $-35$, we will use: $-8$ and $-27.$

Insert them into the parentheses: . $(6x \overbrace{-\, 8}^\downarrow)(6x \overbrace{-\, 27}^\downarrow)$

Factor out all common factors and discard them: . $\not{2}(3x - 4)\not{3}(2x - 9)$

Answer: . $(3x - 4)(2x - 9)$

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Example: . $18x^2 - 11x - 24$

We have: . $(\underbrace{18x}\qquad)(\underbrace{18x}\qquad)$

Then: . $18 \times 24 \,= \,432$

The last term is negative, think difference.
Factor $432$ into two factors whose difference is $11.$

$432 \:=\:\begin{Bmatrix}1\cdot432 \\ 2\cdot216\\3\cdot144\\ \vdots \\ 16\cdot27\\18\cdot24\end{Bmatrix}$

The pair with a difference of 11 is $16$ and $27.$

Since we want $-11$, we will use: $+16$ and $-27.$

Insert them into the parentheses: . $(18x \underbrace{+\, 16})(18x \underbrace{-\, 27})$

Factor out common factors and discard them: . $\not{2}(9x+ 8)\not{9}(2x-3)$

Answer: . $(9x + 8)(2x - 3)$

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At this point, someone will say, "Why not use the Quadratic Formula?"
. . Why not, indeed?
In fact, why bother teaching Factoring at all?
. . And I don't have an answer to that question.

But it would be a shame to run through the Quadratic Formula
. . to factor, say, $x^2 - 9$ or $2x^2 + 6x$

2. It reminds me of the "box method"

If you have a trinomial, $ax^2+bx+c$ then write out a box like this:
Code:
:-----:-----:
:  a  :     :
:-----:-----:
:     :  c  :
:-----:-----:
Now you need to fill in the two empty squares. The trick is that they add to "b" and multiply to "ac", I'll use $2x^2+5x+3$ as an example:

Code:
:-----:-----:
:  2  :  2  :
:-----:-----:
:  3  :  3  :
:-----:-----:
Then find the greatest common factor of each row and collum:

Code:
     1      1
:-----:-----:
2 :  2  :  2  :
:-----:-----:
3 :  3  :  3  :
:-----:-----:
So: $2x^2+5x+3=(2x+3)(x+1)$