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 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.
We can even make this mechanical.
- Example: $y=x^2+6x+5$
- Set a to be half the middle term.
- Take (x+a)^2
- Find the error:
- $E=(x+a)^2-y=(x^2+6x+9)-(x^2+6x+5)=4$
- Take the square root of the error:
- The values to use are a-2 and a+2, which are 1 and 5.
- Answer
This seems a long drag, but it starts with some really useful
number familiarisation. By playing with squares and seeing how
products are more-or-less close to the square of the average,
students can get a feel for
numbers. We can even demonstrate
by drawing a square, and seeing how the area varies if we
increase one side and decrease the other. The area is nearly
constant, and the error is itself a square.
Diagrams to follow if someone asks.
This method can even lead to the standard formula. Writing
that slightly differently than usual, the solutions are -a+e
and -a-e, where a is half the coefficient of x and e
is the square root of the error. If the error is negative you
have no real solutions.
Other ideas?
CategoryMaths