Visual models of identities

Since (-5+3=-2) and (-5\cdot3=-15). Exam tip: interpret factors with signs in the visual model.

Each rectangle is (2a) by (b), so the total is (4ab). Exam tip: handle the changed side (2a) carefully.

The square roots of (25x^2) and (4y^2) are (5x) and (2y), and the middle term (20xy) fits. Exam tip: confirm using the (2ab) term.

The expression represents the difference between two squares: \\(x^2-49=x^2-7^2\\). In a visual area model, the large square has side \\(x\\), while the removed smaller square must have an actual side whose square is \\(49\\). Therefore, a square of side \\(7\\) is the useful piece to remove. This leaves a shape that can be rearranged into a rectangle.

The other choices do not fit the area model. A square of side \\(49\\) would have area \\(49^2\\), not \\(49\\), and a square of side \\(x+7\\) is too large and has the wrong area. Adding a square of side \\(x-7\\) also changes the expression rather than representing the subtraction. Thus option A follows from identifying \\(49\\) as \\(7^2\\).

The difference in the areas of the larger and smaller squares is \((x+4)^2-(x-4)^2\). Using \(a^2-b^2=(a-b)(a+b)\), we get \([(x+4)-(x-4)]\,[(x+4)+(x-4)] = 8\times 2x=16x\). Therefore, \(16x\) is correct. \(x^2-16\) is not the difference of the two square areas. Exam tip: for the difference of two square areas, apply the identity \(a^2-b^2\) first.

The middle term is (6x+8x=14x), so (p=14). Exam tip: the sum of constant side parts gives the middle coefficient.

Dividing each side of a square of side \(a+b\) into lengths \(a\) and \(b\) forms regions \(a^2\), \(ab\), \(ab\), and \(b^2\). The two rectangles give \(2ab\); they are not extra squares. Exam tip: label both dimensions before identifying regions.

In a trinomial square, twice the product of every pair appears. Exam tip: count the pair (x) and (z) too.

The two linear strips give (10x) and (10x), so (m=20). Exam tip: add the two equal strips.

Here (9x^2=(3x)^2) and (1=1^2), so it is a difference of squares. Exam tip: view both terms as perfect squares.

The first side is (x+3) and the second is (x+4), so the difference is (1). Exam tip: find the side before comparing areas.

The total middle term is (12x), and there are two equal strips, so each is (6x). Exam tip: divide the middle term into two equal parts.

Since (25=5^2), and the middle term matches (2\cdot5\cdot x). Exam tip: take the square root of the constant term.

The parts (ax) and (-ax) are opposite linear parts, so they cancel. Exam tip: the middle term is zero in product of sum and difference.

The visual identity is formed by adding areas of parts, not by perimeter. Exam tip: keep area and perimeter separate.

The difference in the areas is \((x+2)^2-(x-2)^2\). Expanding gives \(x^2+4x+4-(x^2-4x+4)=8x\). Hence, the total area of the four strips is \(8x\). The expression \(x^2-4\) equals \((x+2)(x-2)\), not the difference of the two square areas. Exam tip: use \((a+b)^2-(a-b)^2=4ab\); putting \(a=x\) and \(b=2\) gives \(8x\) directly.

Since (7+8=15) and (7\cdot8=56), these strips are correct. Exam tip: match both product and sum.

Since (4a^2=(2a)^2), the difference becomes ((2a-b)(2a+b)). Exam tip: identify the side of the larger square first.

The two squares are (x^2) and (81), while (18x) is the strip area. Exam tip: separate square parts from rectangular parts.

The mistake ignores the two cross rectangles of total area (2xy). Exam tip: never write ((x+y)^2) as (x^2+y^2).