Originally Posted by

**Soroban** Hello, sparky!

I use a primitive approach.

It takes a bit of time, but it is *direct*.

Factor 180 into pairs of factors, and note the difference.

. . $\displaystyle \begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$

How do we find these pairs?

Divide 180 by 1, 2, 3, . . .

Some of the divisions do not "come out even", of course.

When can we stop?

We can stop at the integer part of $\displaystyle \sqrt{180} \,\rightarrow\, 13$

We see that the pair $\displaystyle (9,20)$ has a difference of 11.

(Of course, we can stop listing the moment we find it.)

We have: .$\displaystyle 12x^2 \quad 9x\quad 20x - 15$

We want the middle term to be +11x, so we will use -9x and +20x.

So we have: .$\displaystyle 12x^2 - 9x + 20x - 15$

Factor "by grouping": .$\displaystyle 3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$