# The Sum and Product Game

Lesson Plans > Mathematics > Mental Math## The Sum and Product Game

I have a game I like to play with my Algebra students, and I wish sometimes that this game had been played with students in the years leading up to Algebra. I call the the "Sum and Product Game." It works best with small groups of students, I think, although there are probably ways it can be modified to work with larger groups. First I'll explain how the game is played, and then in the comments below, I'll explain why it's helpful, and some of the strategies I teach my students for getting better at the game.

## How to Play

I begin by writing the following on the board:

____ + ____ = _____

____ x ____ = _____

Then I say, "I'm looking for two numbers that add to 12 and multiply to 32." While I'm saying that, I write the two numbers on the board:

____ + ____ = 12

____ x ____ = 32

At this point, the students begin trying to find a pair of numbers that matches the criteria. As soon as they find a pair of numbers, they call them out. They don't raise their hand and wait to be called on, because I don't want to have to try to judge who had their hand up first. It's easier to simply listen for the voice I hear first. In the example given, of course, the pair of numbers is 8 and 4.

The scoring goes as follows:

- If the student is correct, they gain one point.
- If they are incorrect (or if they only give a partial answer) they lose a point*
- Students can't have a score less than zero, so there is no penalty for a wrong answer if you have zero points.

* This rule is because students will sometimes start to answer, and then realize their answer doesn't work. Once they've started an answer, they're committed to it, and are penalized if they can't finish within a reasonable time frame.

Additional rules which you might use:

- Students who give an incorrect answer cannot give a second answer for the same round
- Whoever gets one answer must sit out the next round (or next two rounds). This keeps one or two students from completely monopolizing the game.
- Try it with students allowed to use calculators, and try it without calculators
- Do a team competition by splitting the team into groups. Each group works in a corner of the room, and is allowed to confer to get an answer

Do remember that you can't just pick any two numbers; not all pairs of numbers are solvable. Usually I work backwards by picking the solution I want, and then figuring out the "adds to" and "multiplies to" numbers.

Don't forget to do cominations of positives and negatives. Most students find this game much more challenging with negatives.

## Strategies to Teach

- Don't focus on the "numbers that add" portion; focus instead on the product. After all, the number of pairs of factors is fairly limited. Consider the example of two numbers that add to 10 and multiply to 16. If you focus on pairs that add to 10, you've got (without thinking about negatives) 6 pairs of numbers (0 and 10, 1 and 9, 2 and 8, 3 and 7, 4 and 6, 5 and 5). However there are only three pairs that multiply to 16 (1 and 16, 2 and 8, 4 and 4).
- If the numbers given are both positive, both numbers in the answer will be positive.
- If the "multiplies to" number is negative, then one of the answers will be negative and the other positive.
- If the "multiplies to" number is positive, but the "adds to" number is negative, the numbers in the answer will both be negative.
- If the "multiplies to" number is negative, and the "adds to" number is close to zero, the two numbers will be close to the square root of the opposite of the "multiplies to" number. For example, find two numbers that add to 1 and multiply to -72. The square root of 72 is between 8 and 9, so my first guess would be 9 and -8.

## Why Play This Game?

For someone who teaches algebra, the reason for playing this game will probably be fairly evident: this method is used in factoring quadratics. If you are factoring x^{2} + 12x + 32 = 0, you find two numbers that add to 12, and multiply to 32. Then you use those numbers in the factorization:

(x + 8)(x + 4) = 0

If you are factoring 3x^{2} + 11x + 6 = 0, you look for two numbers that add to 11 and multiply to 18 (the coefficient of x^{2} times the constant). Then you use the two numbers (9 and 2) like this:

3x^{2} + 11x + 6 = 0

3x^{2} + **9x** + **2x** + 6 = 0

3x(x + 3) + 2(x + 3) = 0

(3x + 2)(x + 3) = 0

Students who are very good at playing this game will be very quick at factoring quadratics.

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