Visual models of identities

In the square model of \((u+v+w)^2\), \(u^2, v^2\), and \(w^2\) are the areas of the three square parts. The mixed parts contain two rectangles each of areas \(uv, vw\), and \(wu\), so their total area is \(2uv+2vw+2wu\). Option A counts each mixed product only once, so it is half the required total. Exam tip: in the square of a three-term expression, every product of two different terms appears twice.

The two linear strips are (8x) and (15x), and the corner is (8\times15=120). In exams, read sides from the linear strips.

The model has two (2x) by (3) rectangles, so the middle term is (12x). In exams, do not stop at one strip.

The two strips are (5t\cdot2) and (5t\cdot2), so total (20t) is subtracted. In exams, include both strips.

The complete area of the square model is \((x+12)^2=x^2+2\times x\times12+12^2\). Besides the \(x^2\) square and the corner square of area \(144=12^2\), two rectangular strips remain. Each strip has area \(12x\), so their combined area is \(12x+12x=24x\). Therefore, option C is correct. Option A gives only one strip and leaves the model incomplete. Exam tip: in the area model for \((a+b)^2\), the middle term is always \(2ab\).

The parts are (2a^2), (-6a), (5a) and (-15), so the linear term is (-a). In exams, add negative rectangles correctly.

Positive and negative tiles leave a difference of (7x). In exams, treat signed tiles as algebraic addition.

In a signed tile model, the constant corner is formed by multiplying the two constant terms: \((-4)\times 9=-36\). Therefore, the constant tiles give \(-36\). The values \(13\) and \(-13\) relate to adding the coefficients \(-4\) and \(9\), not to the constant corner. Exam tip: to find the constant term in a product of binomials, multiply only the constant terms.

The large square has side \(x+y\), so its area is \((x+y)^2\). Adding its parts gives \(x^2+xy+xy+y^2=x^2+2xy+y^2\). Exam tip: always count both \(xy\) rectangles.

The total removed strips are (14y), so each is (7y). In exams, divide the middle term into two equal parts.

One mixed rectangle is (2m\cdot3n=6mn), and there are two such rectangles. In exams, double the mixed rectangles.

Using \((x-y)^2=x^2-2xy+y^2\), with \(x=3a\) and \(y=2b\), we get \((3a-2b)^2=(3a)^2-2(3a)(2b)+(2b)^2=9a^2-12ab+4b^2\). In a visual area model, the two rectangular parts together contribute \(12ab\) to be subtracted, while the corner \(4b^2\) is added back because it was removed twice. Option A has an incorrect coefficient of the middle term. Exam tip: in \((x-y)^2\), the middle term is always negative and includes a factor of 2.

(4x^2=(2x)^2) and (25y^2=(5y)^2). In exams, write difference of squares as two conjugate sides.

The total difference is (4ab), so each of the four equal rectangles is (ab). In exams, divide total area into equal parts.

After removing the squares, only six mixed rectangles remain. In exams, add them as (2xy+2yz+2zx).

The governing concept is the area model for the identity (a + b)² = a² + 2ab + b². The complete square has side a + b, so its area is (a + b)². Within the model, one square contributes a² and another contributes b². The two remaining regions are rectangles, each with side lengths a and b, so each has area ab. Their combined area is therefore ab + ab = 2ab. Equivalently, subtracting the two known square areas from the whole gives (a+b)² − a² − b² = a² + 2ab + b² − a² − b² = 2ab. Hence option B is correct. Option A gives only one rectangle, C belongs to a difference-of-squares expression, and D represents a length rather than an area.

( (a+b)^2 ) has two extra (ab) rectangles along with (a^2+b^2). In exams, view the difference as missing rectangles.

On subtracting the two expansions, (4x-(-4x)=8x) remains. In exams, take the difference of opposite middle terms.

The linear tiles are (4x) and (-2x), so the total is (2x). In exams, add signed linear tiles.

(1\cdot6=6) and (2\cdot3=6), so grouping (x+6x) gives ((2x+1)(x+3)). In exams, verify the middle split by area grouping.