Visual models of identities

(x^2-y^2=(x-y)(x+y)), so if the height is (x-y), the length is (x+y). In exams identify the other side of the rearranged rectangle.

The required area is the difference between the areas of the larger and smaller squares: \((x+2y)^2-(x-y)^2\). Using the difference of squares, \([(x+2y)-(x-y)]\,[(x+2y)+(x-y)] = (3y)(2x+y)=6xy+3y^2\). Hence, option C is correct. \(3xy+3y^2\) has an insufficient coefficient of the \(xy\) term. Exam tip: use \(a^2-b^2=(a-b)(a+b)\) to avoid errors in expansion.

From the total area ((a+b)^2), removing the shaded (2ab+b^2) leaves (a^2). In exams shaded and unshaded parts add to the whole square.

The removed small square is (64), and the remaining part rearranges into ((x-8)(x+8)). In exams keep difference and square-difference models distinct.

((3x+5)(4x-1)=12x^2-3x+20x-5=12x^2+17x-5). In exams the sum of cross strips gives the middle term.

The border strip is the part left after the smaller square is removed from the larger square. Its area is therefore found by subtracting the area of the inner square from the area of the outer square. This is a visual way to understand the difference-of-squares identity, rather than treating the strip as a single unfamiliar shape.

The outer square has area \\(p+q)^2\\), while the inner square has area \\(p-q)^2\\). Thus the strip has area \\( (p+q)^2-(p-q)^2 \\). Expanding gives \\(p^2+2pq+q^2-p^2+2pq-q^2=4pq\\). Therefore the complete border area is \\(4pq\\), so option B is correct. Option A represents only half of this result, while option C is a different identity.

Since (6\cdot7=42) and (6+7=13), the strips are (6x) and (7x). In exams match both the corner and the middle conditions.

It is ((2x)^2-3^2=4x^2-9). In exams treat a rectangle with same center and opposite offset as a difference of squares.

Dividing each side into lengths \(p\) and \(q\) creates areas \(p^2, pq, pq, q^2\). Their sum is \(p^2+2pq+q^2\). Exam tip: do not miss the two equal \(pq\) rectangles.

Both rectangles have sides (a) and (b), so they have equal area (ab). In exams a rotated rectangle keeps the same area.

Both have (x^2+8x) common, and the only difference is (16-15=1). In exams cancel common parts while comparing.

The middle term is (-2\cdot7a\cdot5b=-70ab), so the side is (7a-5b). In exams the sign of the middle term fixes the identity.

The middle rectangles in the grid are (7x) and (8x), whose sum is (15x). In exams write separate strips first and then add them.

The area difference is (12x), equal to four (3x) strips. In exams think of four equal strips in the difference of symmetric squares.

((2x-5)(x+3)=2x^2+6x-5x-15=2x^2+x-15). In exams strips with opposite signs subtract to form the middle term.

First (a^2+2ab+b^2=(a+b)^2), then (c^2) is subtracted. In exams group first and then apply difference of squares.

The border area is \((x+4)^2-(x-4)^2=16x\). In exams use outer square minus inner square.

The cells are (3x^2), (6x), (-5x), and (-10), whose sum is (3x^2+x-10). In exams include the sign of every cell.

The full model has two (xy) rectangles, so missing one gives (x^2+xy+y^2). In exams always check the two parts of the middle term.

Adding two (ab) strips and a (b^2) corner to the (a) square forms ((a+b)^2). In exams distinguish addition and removal models.