Visual models of identities

(25x^2=(5x)^2) and (1=1^2), so the sides are (5x-1) and (5x+1). In exams take square roots and form sum-difference.

The total middle area is (14xy), so each equal rectangle is (7xy). In exams the middle part of a perfect square splits into two equal rectangles.

The difference in the areas of the larger and smaller squares is \((2a+b)^2-(2a-b)^2\). Using \((x+y)^2-(x-y)^2=4xy\), with \(x=2a\) and \(y=b\), gives \(4\times2a\times b=8ab\). Hence, the remaining visual area is \(8ab\). \(4ab\) is only half of the required result. Exam tip: when symmetric terms \((x+y)\) and \((x-y)\) appear, apply the difference-of-squares identity directly.

First (x^2-4xy+4y^2=(x-2y)^2), then (z^2) is subtracted. In exams identify the perfect-square trinomial first.

Since (6+(-5)=1) and (6\cdot(-5)=-30), the sides are (x+6) and (x-5). In exams use constants with opposite signs for a negative corner.

The difference of squares is ((101-99)(101+99)=2\cdot200=400). In exams turn larger square minus smaller square into a rectangle.

The remaining area is ((x+2)^2-(x-2)^2=8x). In exams subtract the corner square to get the strips.

After removal, the parts involving (a) and (b) remain and make ((a+b)^2). In exams add the remaining small regions to identify the new square.

In ((3x+2)(5x-4)), the cells are (15x^2), (-12x), (10x), and (-8). In exams the sum of cross areas gives (-2x).

This is ((u+c)^2-(u-c)^2=4uc) where (u=a+b), so the area is (4c(a+b)). In exams treat the whole group (a+b) as one side.

Changing the order of parts does not change total area. In exams, treat rearrangement as equal area.

The complete square has side \\(x+7\\), so its area is \\((x+7)^2\\). In the visual model, one part has area \\(x^2\\), the corner has area \\(49=7^2\\), and the two equal rectangles each have area \\(7x\\). Expanding the complete square confirms this: \\((x+7)^2=x^2+2(x)(7)+7^2=x^2+14x+49\\).

The two rectangles together therefore have area \\(7x+7x=14x\\), so option B is correct. The value 7 is the side length of the corner square, and it combines with the side length \\(x\\) to form each rectangle. Option A gives the area of only one rectangle, while option D adds the two square parts instead of finding the combined rectangular area.

When each side is divided into lengths \(p\) and \(q\), the square of side \(p+q\) is split into four regions: one square of area \(p^2\), one square of area \(q^2\), and two rectangles of area \(pq\) each. These two \(pq\) rectangles occupy diagonally opposite positions, so the answer is 2. Option 4 counts all regions of the figure, not just the \(pq\) rectangles. Exam tip: In \((p+q)^2=p^2+2pq+q^2\), \(2pq\) represents the combined area of the two rectangles.

The governing concept is the perfect-square identity (x + y)² = x² + 2xy + y². Compare the given trinomial with this form. Since 81 = 9² and 18m = 2×m×9, we can rewrite m² + 18m + 81 as m² + 2(m)(9) + 9², which equals (m + 9)². If the area of a square is (m + 9)², its side is m + 9, so option D is correct. Option A would expand to m² + 12m + 36, option B would produce m² − 18m + 81 because its middle term is negative, and option C would give m² + 36m + 324. Thus only D reproduces the original square exactly. The visual model represents the two-variable side components and the constant 9, making the identity a geometric as well as algebraic interpretation.

The side parts are (4y) and (5), so the subtracted middle term is (2\cdot4y\cdot5=40y). In exams, use square roots to form strips.

Positive (11r) and negative (11r) cancel because their areas are equal. In exams, look for equal rectangles with opposite signs.

This rectangle model represents the identity \((a+b)(a-b)=a^2-b^2\). Here, \(a=3x\) and \(b=4\), so the area is \((3x)^2-4^2=9x^2-16\). \(9x^2+16\) may look like a sum of squares, but the product of conjugates requires subtracting 16. Exam tip: first identify the common term and the opposite constant terms.

If the rectangle has side lengths \((x+a)\) and \((x+b)\), then \((x+a)(x+b)=x^2+(a+b)x+ab\). Here, we need \(a+b=15\) and \(ab=54\). Since \(6+9=15\) and \(6\times9=54\), the constant strips are \(6\) and \(9\), giving \((x+6)(x+9)\). Although \(3\) and \(18\) have product \(54\), their sum is \(21\). Exam tip: always check both the sum for the middle coefficient and the product for the constant term.

The small corner is (36=6^2) and the large square is (25a^2=(5a)^2). In exams, derive side lengths from areas.

In the area model of \((a+b+c)^2\), both sides of the square are divided into segments \(a\), \(b\), and \(c\). Two rectangles are formed using \(a\) and \(b\): one has area \(ab\) and the other has area \(ba\). Since \(ab=ba\), there are 2 rectangles of type \(ab\), producing the term \(2ab\). Option 1 counts only one of these rectangles. Exam tip: For a mixed term, count regions in both orders, such as \(ab\) and \(ba\).