Visual models of identities

Dividing each side into lengths \(a\) and \(b\) gives regions \(a^2\), \(b^2\), and two rectangles of area \(ab\). Hence the full area is \(a^2+2ab+b^2\). Exam tip: do not miss the second \(ab\) rectangle.

Both strips are (6y) and (6y), so the total is (12y). In exams, count equal rectangles separately.

A square with side \\(a+5\\) has area \\( (a+5)^2\\). A visual model divides it into one large square of area \\(a^2\\), two rectangular strips, each of area \\(5a\\), and one small corner square of area \\(5^2=25\\). Adding these regions gives \\(a^2+5a+5a+25=a^2+10a+25\\). The two equal strips are important because both contribute to the middle term.

The constant corner is correctly shown as 25, but the middle term must be \\(10a\\), not \\(5a\\). Equivalently, the identity \\( (a+b)^2=a^2+2ab+b^2\\) with \\(b=5\\) gives \\(a^2+2(a)(5)+25\\). Thus option C is the complete expansion; omitting one strip would produce an incomplete expression.

Two (3m) strips are removed and the (9) corner is added. In exams, add back the corner in subtraction models.

Split the rectangle into four parts: \(p\times p=p^2\), \(p\times 7=7p\), \(4\times p=4p\), and \(4\times 7=28\). Therefore, the total area is \(p^2+7p+4p+28=p^2+11p+28\). Option C incorrectly combines \(4p\) and \(7p\). Exam tip: when multiplying two binomials, add the coefficients of the middle like terms carefully.

This is a difference of squares model, so (8^2=64) is subtracted. In exams, identify ((r+8)(r-8)) directly.

The strips are (9x) and (2x), and the corner is (9\cdot2=18). In exams, form sides from strip widths.

The combined middle area is (2\cdot2x\cdot5=20x). In exams, write the square of (2x) as (4x^2).

In the area model of \((a+b)^2\), there is one \(a^2\) square, one \(b^2\) square, and two \(a\times b\) rectangles. Their areas add as \(ab+ab=2ab\). Exam tip: identify each region by multiplying its side lengths.

The two \(xy\) rectangles together have area \(2xy\). Thus, the whole square has area \(x^2+2xy+y^2\) and side \(x+y\). Exam tip: use the sign of the middle term to distinguish square identities.

The expression can be read as a perfect-square identity: \\(z^2+16z+64=(z+8)^2\\). In a visual area model, the large square has side length \\(z+8\\). It can be divided into a \\(z\\)-by-\\(z\\) square, two rectangular parts whose combined area is \\(16z\\), and one small corner square with side length \\(8\\). The corner area is therefore \\(8\times8=64\\).

The constant term represents the area that does not contain \\(z\\). Thus it corresponds to the small fixed square at the corner, giving option A. The term \\(16z\\) represents the combined area of the two rectangles, while \\(z^2\\) represents the large variable square. Therefore, the correct choice is 64, not 16z or \\(z^2\\).

The two strips are (5n) and (5n), so total (10n) is removed. In exams, add both removed parts.

The area of a square is the square of its side, so the area is \((q-4)^2\). Using \((a-b)^2=a^2-2ab+b^2\), with \(a=q\) and \(b=4\), gives \(q^2-2\times q\times4+16=q^2-8q+16\). In option A, the factor 2 is missing from the middle term. Exam tip: the middle term in \((a-b)^2\) is always negative, \(-2ab\).

The area of the rectangle is the product of its sides: \((s+3)(s-3)\). Using the identity \((a+b)(a-b)=a^2-b^2\), the area is \(s^2-3^2=s^2-9\). Although \(s^2+9\) also has no middle term, its constant term has the wrong sign. Exam tip: whenever you see \((a+b)(a-b)\), apply the difference-of-squares identity directly.

To factor \(t^2+7t+12\), we need two numbers whose sum is \(7\) and product is \(12\). Since \(3+4=7\) and \(3\times4=12\), \(t^2+7t+12=(t+3)(t+4)\). Option C also gives constant term \(12\), but its middle term is \(8t\), not \(7t\). Exam tip: multiply the binomials or check both the sum and product before choosing an answer.

The large square has side \(a+b\), so its area is \((a+b)^2\). Adding its parts gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B would require subtraction, not two positive \(ab\) rectangles. Exam tip: count both rectangles.

Dividing a square of side a+b at lengths a and b gives squares of areas a² and b². The remaining two regions are ab rectangles, so the area is a²+2ab+b². Exam tip: always count both ab rectangles.

Dividing a square of side a+b into lengths a and b produces areas a², b², and two a×b rectangles. Their total is a²+2ab+b². One a×b rectangle would miss the 2ab term. In diagrams, always count both congruent rectangles.

(36=6^2) and (-12c=-2\cdot c\cdot6). In exams, identify the subtractive side from the negative middle term.

For a square whose side is \\(4x+3\\), the visual model separates the variable part \\(4x\\) from the constant part \\(3\\). The small corner is formed where the constant part meets itself, so both of its side lengths are \\(3\\). Its area is therefore \\(3\times3=3^2=9\\). It is a constant square, not a strip involving \\(x\\).

The quantities \\(16x^2\\), \\(12x\\), and \\(24x\\) describe other possible regions or combined terms in an expansion, not the small constant corner. Indeed, the complete square would expand as \\( (4x+3)^2=16x^2+24x+9\\), and the final term is exactly the corner area. Hence option D is correct.