How do people teach factoring quadratics?

What are the different methods used?

Here's a sketch of one idea:

• First, compare the results of 23*23, 22*24, 21*25, 20*26, and so on.
• See how the successive answers diverge from the originals by the squares.
• 529, 529-1, 529-4, 529-9, etc.
• Try again, starting with a different number:
• 38*38, 37*39, 36*40, 35*41, 34*42, ...
• 1444, 1444-1, 1444-4, 1444-9, 1444-16, ...
• Draw out the conclusion that the product of two numbers is a bit less than the square of the average ...
• and the error is the square of the distance to the average.
• 60*66 is roughly 63*63, and the error is $(63-60)^2$
• So $60*66=63^2-3^2=3960$
• So a product, any product, every product, is a square minus an error.
Now let's look at $x^2+12x+27.$ This should be a square minus the error. Let's suppose it's $(x+a)^2.$ That's $x^2+2ax+a^2.$ We use whatever value of a make the middle term work, so we want $2ax=12x.$ So let's set a=6.

That gives us $(x+6)^2=x^2+12x+36,$ which is too big by 9. That's the square of the amount that a is the wrong thing to use. The square root of 9 is 3. That means we should use a+3 and a-3, which is 6+3 and 6-3, 9 and 3.

• $x^2+12x+27=(x+9)(x+3).$
We can even make this mechanical.

• Example: $y=x^2+6x+5$
• Set a to be half the middle term.
• $a=\frac{6}{2}=3$
• Take (x+a)^2
• $(x+3)(x+3)=x^2+6x+9$
• Find the error:
• $E=(x+a)^2-y=(x^2+6x+9)-(x^2+6x+5)=4$
• Take the square root of the error:
• $\sqrt{E}=\sqrt{4}=2$
• The values to use are a-2 and a+2, which are 1 and 5.
• $(x+1)(x+5)$