Visual models of identities

In the sum square model, two rectangular strips and a small square are added. In exams add area parts for quick calculation.

The complete side lengths are \(x+4\) and \(x-6\). The four smaller areas are \(x^2\), \(-6x\), \(4x\), and \(-24\). Their sum is \(x^2-6x+4x-24=x^2-2x-24\). In \(x^2+10x-24\), the negative sign of \(-6x\) has been incorrectly changed. Exam tip: determine the sign of every small rectangle before combining like terms.

The total area of the four parts equals the square with side (a+b). In exams identify the whole outer shape first.

The difference between (a^2-b^2) and ((a-b)^2=a^2-2ab+b^2) is (2ab-b^2). In exams keep difference of squares and square of difference separate.

The total middle term is (6x), so each strip is (3x). In exams divide the middle term into two equal parts in a perfect square.

The four parts of ((2x+1)(x+3)) are (2x^2), (6x), (x), and (3). In exams multiply each cell of the grid.

The area of the whole square is \((a+b)^2\). Adding its four parts gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\), so A is correct. Option B would require negative middle terms. Exam tip: count both \(ab\) rectangles.

((x+4)^2-(x+2)^2=4x+12). In exams subtract the smaller square from the larger and cancel equal (x^2) terms.

Two (xy) rectangles and two squares make a square of side (x+y). In exams see (2xy) as two equal rectangles.

Since (16p^2=(4p)^2) and (25=5^2), the outer side is (4p+5). In exams take square roots when finding side length from area parts.

The large square has area \(a^2\), while the removed small square has area \(b^2\). Rearranging the leftover pieces gives a rectangle with sides \((a+b)\) and \((a-b)\). \((a-b)^2\) misses the conjugate factor \((a+b)\). Exam tip: recognise difference of squares as conjugate factors.

When the square is split using lengths \(a\) and \(b\), two mixed regions are formed: \(a\times b\) and \(b\times a\). Each has area \(ab\), giving \(2ab\). Exam tip: always count both rectangles.

The area of the rectangle is the product of its length and breadth: \((a+b)(a-b)\). In the area model, the \(a^2\) part remains, while the \(+ab\) and \(-ab\) parts cancel each other. Hence, the area is \(a^2-b^2\). Option A is the expansion of \((a+b)^2\), so it does not apply here. Exam tip: recognise \((x+y)(x-y)=x^2-y^2\) as the difference-of-squares identity.

The large square has side \(a+b\), so its area is \((a+b)^2\). Adding the four regions gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Exam tip: count both equal \(ab\) rectangles before choosing the identity.

The two strips have total area (6p) and the small square is (9). In exams, write ((3)^2) as (9).

The two strips are (5y+5y), so (10y) appears. In exams, keep the small square area as (25).

Two (2m) strips are subtracted, so total (4m) is subtracted. In exams, add back the corner square.

In ((z-7)^2), the middle term is (-2\cdot z\cdot7). In exams, count the two removed strips.

In the area model, the rectangle is split into four parts with areas \(r\times r=r^2\), \(r\times 2=2r\), \(6\times r=6r\), and \(6\times 2=12\). Therefore, the total area is \(r^2+2r+6r+12=r^2+8r+12\). Option B has an incorrect coefficient of the middle term. Exam tip: when multiplying two binomials, add the areas of all four smaller rectangles.

In the area model, the constant region is formed where the two constant lengths meet. Hence its area is \(3\times8=24\). The regions \(s\times3\) and \(s\times8\) contain the variable \(s\), so they are not constant parts. Exam tip: to find the constant term, multiply only the constant terms of the two binomials.