Visual models of identities

(9r^2=(3r)^2) and (16=4^2), so it is a difference of squares. In exams first identify square roots.

Dividing the rectangle according to \(x+1\) and \(x+9\) produces four smaller areas: \(x^2\), \(9x\), \(x\), and \(9\). The two middle strips have areas \(9x\) and \(x\), so their sum is \(9x+x=10x\). \(9x\) represents only one middle strip, not their combined area. Exam tip: In a visual model, find each small rectangle’s area separately before combining like terms.

Each side of the square is \(a+b\), so its total area is \((a+b)^2\). The visual model has two squares with areas \(a^2\) and \(b^2\), and two rectangles with area \(ab\) each. Thus, the total area is \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option C is the expansion of \((a-b)^2\), so its middle term is negative. Exam tip: the middle term in \((a+b)^2\) is always \(+2ab\).

After rearranging the cut area into a rectangle, the sides are (a+b) and (a-b). In exams do not confuse it with square of difference.

The large square has area \((a+b)^2\). Its four parts have areas \(a^2, ab, ab, b^2\); the two rectangles together give \(2ab\). In exams, add all partitioned areas carefully.

The total area is \(x^2+xy+xy+y^2=x^2+2xy+y^2\). Thus, when the side of the complete square is \(x+y\), the model represents \((x+y)^2=x^2+2xy+y^2\). Option B has a middle term of \(-2xy\), but both rectangular regions in this model have positive area \(xy\). Exam tip: combine the two equal \(xy\) regions to get \(2xy\).

When each side of a square of side \(a+b\) is divided into lengths \(a\) and \(b\), four regions are formed: one \(a\times a=a^2\) square, one \(b\times b=b^2\) square, and two \(a\times b=ab\) rectangles. Thus the total area is \(a^2+2ab+b^2\), so option A is correct. Option C misses one \(ab\) rectangle. Exam tip: count both cross-rectangles separately in an area model.

The middle part is (2\cdot2u\cdot1=4u) and its sign is negative. In exams take the total of both equal strips.

Since (2+3=5) and (2\cdot3=6), the strips are (2x) and (3x). In exams split the middle term into the correct two parts.

A square of side (a+b) splits into four parts (a^2), (ab), (ab), and (b^2). In exams identify the whole outer square.

A square with side \\(x-7\\) has area \\( (x-7)^2\\). Using \\( (a-b)^2=a^2-2ab+b^2\\) with \\(a=x\\) and \\(b=7\\), we obtain \\(x^2-2(x)(7)+7^2=x^2-14x+49\\). In the visual model, two strips of area \\(7x\\) are removed, producing the middle term \\(-14x\\), and the small corner of area \\(49\\) is added back.

Option A misses one strip and therefore has the wrong middle coefficient. Option C is the difference-of-squares form for \\(x^2-49\\), not the square of \\(x-7\\). Option D has a positive middle term, which would correspond to \\( (x+7)^2\\). Hence option B is the correct expansion.

The governing concept is algebraic comparison of equivalent area expressions. The rectangle has dimensions x + 4 and x + 6, so its area is (x+4)(x+6) = x² + 10x + 24. The square has side x + 5, so its area is (x+5)² = x² + 10x + 25. Subtracting the rectangle’s area from the square’s area gives (x²+10x+25) − (x²+10x+24) = 1. Thus the square is larger by exactly one square unit whenever the stated dimensions are valid lengths. This also illustrates that numbers equally spaced around x+5 produce a slightly smaller product than the square of their average. Therefore option B is correct. The x² and 10x terms cancel, so options C and D cannot be correct.

A square model represents the area of a square whose side is made from algebraic lengths. The expression is a perfect-square trinomial because its first term is the square of x, its last term is the square of 8, and the middle term is twice their product. Therefore, the complete area can be written as \\(x+8\\)^2. The outer side is consequently x+8, so option B is correct.

To verify this, expand the side expression: \\(x+8\\)^2=x^2+2(x)(8)+8^2=x^2+16x+64\\). This gives exactly the stated square model. The value 16 is not the side itself; it is the coefficient produced by the middle-term calculation. Similarly, 64 is only the corner square's area, not the whole side. The negative sign in x−8 would produce −16x, so it cannot match the given expression.

The large square has side \(a+b\). Its four regions have areas \(a^2\), \(ab\), \(ab\), and \(b^2\). Therefore, the total area is \(a^2+2ab+b^2=(a+b)^2\). In \((a-b)^2\), the middle term is \(-2ab\), so it does not match this model. Exam tip: two equal \(ab\) rectangles indicate the middle term \(+2ab\).

The cells are (2x^2), (8x), (3x), and (12), whose sum is (2x^2+11x+12). In exams add all four cells.

The corner is (2^2=4) and the middle area is (-2\cdot3x\cdot2=-12x). In exams pay attention to the subtraction sign.

Since (8+10=18) and (8\cdot10=80), this is the correct rectangle model. In exams match both the middle and constant terms.

((x+2)^2=x^2+4x+4) and ((x+2)(x-2)=x^2-4), so the difference is (4x+8). In exams expand both models separately.

Each equal strip is (6x), so (k=6). In exams distinguish total middle term from each strip.

The side of the square is \(a+2b\), so the diagram can be viewed as two parts with lengths \(a\) and \(2b\) along each side. The rectangular regions involving these two different parts occur twice: one has dimensions \(a\) by \(2b\), and the other has the same dimensions. Each rectangle therefore has area \(a\cdot2b=2ab\).

Adding both rectangular areas gives \(2ab+2ab=4ab\). Equivalently, the cross term in \((a+2b)^2\) is \(2\cdot a\cdot2b=4ab\). The term \(4b^2\) is the square part formed by the \(2b\) sections, while \(a^2\) is the other square part. Hence option B, \(4ab\), is the total rectangular area involving \(b\).