**So far** we have completed three parts of the place value series:

Part 1: The Chocolate Factory which covered the regrouping or trading aspect of place value and explored regrouping in base ten and other bases.

Part 2: Base Ten for Young Students which introduced several games and trading activities to help young children acquire a solid foundation in place value.

Part 3: The Bake Sale demonstrates the role of place value in long division.

Today, Part 4: Geometry of Place Value will explore place value within a quadratic equation. We will further show that each monomial can be modeled geometrically. **Review**

The expression, 5x^{2} + 6x + 3, appeared in The Chocolate Factory, as a summary of the chocolate packing activity. Five cases and six boxes were packed with three leftover chocolates. Where x stood for the number of chocolates per box, five cases and six boxes could represent different absolute numbers of chocolates. If x =10, or 10 chocolates per box, then ((5 times 100) + (6 times 10) + 3) chocolates, or 563 chocolates came down the conveyor belt, I Love Lucy style. In fact, this episode of I Love Lucy was the inspiration for the math activity.

If the chocolates are packed in boxes of five then the 563 means ((5 times 25) + (6 times 5) + 3) or 158 chocolates came down the conveyor belt. So a quadratic equation can be thought of as an expression of place value in any base. In fact, a polynomial of any degree can be seen as an expression of place value. Missing terms are represented by zeros. So 2x^{6} + 5x^{5}+ 3x^{2} + 7x + 2 would be 2,500,372_{base x}.

Now we can see where the analogy to place value breaks down. If x = 6, then a term like 7x would be “illegal.” Once six “boxes” had been packed, those six boxes would immediately be packed into one “case,” so in base 6 the last three terms would properly be 4x^{2} + x + 2, either way, the last three terms represent 152 “chocolates.” Obviously I have just been speaking to adults initiated into the joys of algebra, not children.**What? No Fourth Dimension**

Obviously you can use standard base ten blocks to model quadratic equations. If we assume that x represents base 10, for 4x^{3} + 3x^{2} + 7x + 2, we would use 4 large cubes, 3 flats, 2 rods, and 2 small cubes to model the expression. However, if all I meant by geometry was geometric solids, I would not have meant much. The geometry is more interesting, and becomes clearer when you look at a set of blocks in a different base, say, base 5, the small cube looks the same as a base ten small cube, but the rod is five cubes long, the flat is a square of 25 cubes, the large cube is 5 flats stacked or 125 cubes.

So the rod of any set of base blocks determines the base of the set. If we look at the large cube of any base, we see that any one of the 12 edges shows x^{1}, the first dimension, any one of the six faces shows x^{2}, the second dimension, and the whole cube shows x^{3}, all three dimensions. Now we run smack into a physical limitation of manipulatives; not one can show more than three dimensions. The power of math is that math is the language of imagination. We can imagine a fourth dimension, x^{4} and beyond, even if we cannot model it. How fun is that?

Interestingly, we can also show x^{0} on the cube. Remember for any x, x^{0} = 1. Cubes of varying bases are all different sizes, or volume. The x^{1}, x^{2}, and x^{3} is different on each cube. But since x^{0} = 1 for any base, it stands to reason that x^{0} or 1 would have an identical appearance no matter the size of the cube. In fact, it does. You can find x^{0} at any vertex, that is to say, the corner shows 1, the zero-th dimension, if you will. In fact, the vertex is a geometric point, described as having no length, width or height.**Multiplication with Base Ten Blocks**

So far we have spent a great deal of time establishing that x^{2} means x times x, and that we can show x times x geometrically, by using a flat from a base block set. The flat has a square shape which we would expect from an expression like x-squared. But let's consider a rectangle shape. Now we are not multiplying the same number by itself, x times x, the very definition of squaring, “the product obtained when a number or quantity is multiplied by itself”.

With a rectangle, we are multiplying two different numbers, x times y (or length times width, the formula for the area of a rectangle). Using the cubes from a base blocks set, we can model 5 x 3.

Math educators call this type of diagram a multiplication array. Now lets try 13 x 11.

To show the factor 13 along the top, I used a rod and three small cubes. The factor 11 is along the side with a rod and one small cube. One rod times one rod equals one flat (square, and you expected a square, right?), one rod times three small cubes equals three rods or three lengths. Then, one small cube times one rod equals one length, and one small cube times three small cubes equals three small cubes.

Combining like terms, that is, similar objects, together, we have one flat (10^{2} or 100), four rods ((3 times 10) + (1 times 10)) or 40, and three small cubes (1 times 3, or 3) for a total of 143 which I could express as(1 x 10^{2}) + (4 x 10) + (3 x 1) .

What if we wanted to multiply (x+3)(x+1). I am using the magenta to stand for x, a number we don't know, also called a variable.

The product is 1x^{2} + 4x +3, and geometrically, the product is the picture of a quadratic equation showing both its factors above and to the left of the crossbars. I recommend manipulatives that elucidate the geometry of quadratic equations, available, for example, the Montessori Binomial Cube and Creative Publications Algebra Lab Gear. Remember we have shown that the magenta rod could stand for any value, that is, for any base.

We can show three factors and therefore three dimensions with the same model by standing a rod and/or stacking small cubes vertically in the corner where the crossbars intersect. If I were to stack four small cubes in that intersection, I would be modeling (4)(x+3)(x+1) or by multiplying the x-factors first, (4)(x^{2} + 4x +3). You could think of it as stacking four layers of the x-factor product. In fact, the formula for volume is height times base, or height layers of the base.

In terms of base blocks, the product would be modeled with 4 flats, 16 rods, and 12 small cubes. If we are working in base ten, we would have 400 + 160 + 12. We can exchange 10 of the rods for a flat, and 10 of the small cubes for a rod, ending up with 5 flats, 7 rods, and 2 small cubes or 572.

If I were to stand a rod in the intersection, I am modeling 10 layers of 143 or 1430. In the upper left hand corner of the product there would be ten flats stacked which I can exchange for a large cube worth 10x10x10. Completing any other exchanges, the product would consist of 1 large cube, 4 flats, 3 rods and 0 small cubes. If I were to stand a magenta rod in that intersection. I would be modeling (x)(x+3)(x+1). The product would have 1 magenta cube, 4 magenta flats, 3 magenta rods and 0 small cubes or x^{3} + 4x^{2} + 3x.

## Sunday, August 30, 2009

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