1. Factoring Problem

I was tutoring a student in Algebra today, and came across an interesting factoring problem. The student had just finished a unit on factoring techniques (sum/difference of cubes, difference of squares, trinomials, etc.) Where I was tutoring the common rules of thumb taught are, after factoring the gcf, for binomials check difference of squares, etc.; for trinomials try factoring by finding two appropriate binomials in the usual way, and for 4-termed polynomials try grouping. This example goes against these rules of thumb taught, so I was wondering, although I know how to do this problem, if anyone here knows a better way to explain how to do it. The problem was to factor

$x^2-6x-y^2+9$

I told the student that, to better see what to do, rewrite as

$x^2-6x+9-y^2$

and factor the trinomial on the left

$(x-3)^2-y^2$

resulting in a difference of two squares.

$[(x-3)-y][(x-3)+y]$

Any advice on how to get an average algebra student to see the appropriate first steps?

2. An idea

Originally Posted by Diamondlance
I was tutoring a student in Algebra today, and came across an interesting factoring problem. The student had just finished a unit on factoring techniques (sum/difference of cubes, difference of squares, trinomials, etc.) Where I was tutoring the common rules of thumb taught are, after factoring the gcf, for binomials check difference of squares, etc.; for trinomials try factoring by finding two appropriate binomials in the usual way, and for 4-termed polynomials try grouping. This example goes against these rules of thumb taught, so I was wondering, although I know how to do this problem, if anyone here knows a better way to explain how to do it. The problem was to factor

$x^2-6x-y^2+9$

I told the student that, to better see what to do, rewrite as

$x^2-6x+9-y^2$

and factor the trinomial on the left

$(x-3)^2-y^2$

resulting in a difference of two squares.

$[(x-3)-y][(x-3)+y]$

Any advice on how to get an average algebra student to see the appropriate first steps?
To mention, among my experience, I used to tutor in a math workshop.

You can explain that this is really two problems in one where you have to isolate the $y^2$ term from the other terms before factoring, in this case.

Other problems of this nature would require examination to see how the terms should be grouped before factoring. The rest is practice.

3. You could also relate this to the workings of "conjugates".

Conjugates normally involve the changing of the sign of the imaginary part
of a complex number before calculating the product (though the sign may be changed on the real part also),
or the changing of the sign on the square root part of a compound surd.

The "difference of squares" follows the same route.

$(x+y)(x-y)=x(x-y)+y(x-y)=x^2-xy+xy-y^2$

which cancels the centre terms as in the above mentioned cases, giving

$x^2-y^2$

Hence

$x^2-y^2-6x+9=(x+y)(x-y)-6x+9=\left[(x+y)-3\right]\left[(x-y)-3\right]$

since the "y" terms will vanish in the centre due to the opposite signs.

Then that can be compared to the "completing the square" procedure as you've done

$x^2-6x+9-y^2=(x-3)^2-y^2$

which is the differences of squares again.

By drawing comparisons, familiarity is built up.