Investigate this question by building rectangles with algebra tiles for the following expressions. For each one, write the area as a sum and as a product. If you cannot build a rectangle, be prepared to convince the class that no rectangle exists (and thus the expression cannot be factored). \( \begin{array}{ll}\text { a. } 2 x^{2}+7 x+6 & \text { b. } 6 x^{2}+7 x+2 \\ \text { c. } x^{2}+4 x+1 & \text { d. } 2 x y+6 x+y^{2}+3 y\end{array} \)

We are going to “build” rectangles whose total area is given by the polynomial. With algebra‐tiles (or a drawn diagram) you want to “decompose” the area into parts (smaller areas corresponding to the “tiles”) that add up to the full expression. But if a rectangle cannot be built from rectangular pieces with sides of the form (ax + b) (or in the two‐variable case, with terms in x and y) then the expression does not factor nicely (over the integers) into two binomials. In what follows we show one acceptable answer for each part. ───────────────────────────── Part (a) 2x² + 7x + 6 Step 1. Write the area as a sum of four parts so that two of the pieces combine to make a “side” of the rectangle and the other two pieces combine to make the other side. (In a rectangle whose sides are expressed as binomials, the area is given by the product.) One way:   2x² + 7x + 6 = 2x² + 3x + 2x + 6. Step 2. Group the terms:   (2x² + 3x) + (2x + 6). Factor each group:   First group: 2x² + 3x = x(2x + 3)   Second group: 2x + 6 = 2(x + 3) — but notice: In our goal we wish to have a common binomial factor. With a little rearrangement it is more convincing to group as:   2x² + 7x + 6 = 2x² + 3x + 2x + 6     = (2x² + 3x) + (2x + 6)     = x(2x + 3) + 2(x + 3). That grouping is not “compatible” because one factor is (2x + 3) and the other is (x + 3), which are different. (This is a hint to re-split the middle term the “correct” way.) A better method is to use the idea: We need two numbers that multiply to (2·6 = 12) and add to 7. (The numbers 3 and 4 work since 3 · 4 = 12 and 3 + 4 = 7.) So, rewrite:   2x² + 7x + 6 = 2x² + 3x + 4x + 6. Now group:   = (2x² + 3x) + (4x + 6)   = x(2x + 3) + 2(2x + 3)   = (2x + 3)(x + 2). Thus we have both a “sum” decomposition:   Area = 2x² + 3x + 4x + 6, which corresponds to subrectangles of areas 2x², 3x, 4x, and 6, and a “product” form:   Area = (2x + 3)(x + 2). On an algebra tile model one might see a large rectangle whose length is (2x + 3) and width is (x + 2). The four pieces inside then have areas:  • (2x)(x) = 2x²,  • (2x)(2) = 4x,  • (3)(x) = 3x,  • (3)(2) = 6. ───────────────────────────── Part (b) 6x² + 7x + 2 Here we use the same strategy. Multiply the coefficient of x² and the constant:  6 · 2 = 12. We need two numbers that multiply to 12 and add to 7. Again 3 and 4 work. Rewrite:  6x² + 7x + 2 = 6x² + 3x + 4x + 2. Group:  = (6x² + 3x) + (4x + 2)  = 3x(2x + 1) + 2(2x + 1)  = (2x + 1)(3x + 2). Thus the area can also be viewed as made up of four parts:  • (2x)(3x) = 6x²,  • (2x)(2) = 4x,  • (1)(3x) = 3x,  • (1)(2) = 2. So the “sum” form is:   Area = 6x² + 3x + 4x + 2, and the “product” form is:   Area = (2x + 1)(3x + 2). ───────────────────────────── Part (c) x² + 4x + 1 For an area to be represented as the product of two binomials with integer coefficients, we would try to write it as (x + a)(x + b) so that:  x² + (a + b)x + ab = x² + 4x + 1. That would require:  a + b = 4 and ab = 1. The only integer factors of 1 are 1 and 1 but 1 + 1 = 2, not 4. (We might also consider –1 and –1, but –1 + (–1) = –2.) Since there is no pair of integers a and b that fit, we conclude that x² + 4x + 1 does not factor into two binomials with integer coefficients. (Its roots are irrational.) Thus, using algebra tiles (which only come in whole, fixed sizes), you will not be able to build a neat rectangle with sides of the form (x + constant) to match this expression. In other words, no “nice” rectangle exists and so the expression is not factorable in this context. ───────────────────────────── Part (d) 2xy + 6x + y² + 3y This expression involves two variables. We can try grouping by looking for a common binomial factor. Group the terms as follows:  Group 1: 2xy + 6x = 2x(y + 3)  Group 2: y² + 3y = y(y + 3) Then,  2xy + 6x + y² + 3y = 2x(y + 3) + y(y + 3)   = (y + 3)(2x + y). Thus the product form is:   Area = (y + 3)(2x + y). To see the sum form in terms of areas (if you imagine a rectangle with side lengths (y + 3) and (2x + y)):  • one subrectangle has area 2x · y = 2xy,  • a second has area 2x · 3 = 6x,  • a third has area y · y = y²,  • a fourth has area y · 3 = 3y. So the sum form is just the original:   Area = 2xy + 6x + y² + 3y. ───────────────────────────── Summary of Answers a. 2x² + 7x + 6 = 2x² + 3x + 4x + 6 = (2x + 3)(x + 2) b. 6x² + 7x + 2 = 6x² + 3x + 4x + 2 = (2x + 1)(3x + 2) c. x² + 4x + 1 cannot be factored as a product of two binomials with integer coefficients (so no rectangle can be built using standard algebra tiles). d. 2xy + 6x + y² + 3y = 2xy + 6x + y² + 3y = (y + 3)(2x + y) Each factorization shows two equivalent pictures:  • As a sum (a breakdown into smaller areas corresponding to tiles) and  • As a product (the area of a rectangle with the given side lengths). These are acceptable answers for the investigation.