Visual models of identities

Adding the areas of the four parts gives \((a+2)(a+9)=a^2+9a+2a+18\). The two linear terms, \(9a\) and \(2a\), add up to \(11a\), so the middle term is \(11a\). \(18\) is the constant term, while \(7a\) comes from subtracting the coefficients. Exam tip: add the two rectangular linear-area terms to find the middle term.

One (4n) positive part and one (-4n) negative part cancel. In exams, identify equal terms with opposite signs.

The sides of the rectangle are \((t+5)\) and \((t-5)\), so its area is \((t+5)(t-5)\). Using the difference-of-squares identity, \((a+b)(a-b)=a^2-b^2\), with \(a=t\) and \(b=5\), we get \(t^2-25\). The expression \(t^2+25\) ignores the cancellation of the middle terms. Exam tip: when two binomials have opposite signs, their middle terms cancel.

The side of the whole square is \(a+b\), so its area is \((a+b)^2\). The two rectangular regions of area \(ab\) together give \(2ab\), making the total \(a^2+2ab+b^2\). Exam tip: always count both \(ab\) regions.

The total area is \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Since the outer figure is a square of side \(a+b\), its area is \((a+b)^2\). Option B would require negative middle terms. Exam tip: combine the two equal \(ab\) rectangles to get \(2ab\).

The rectangle has side lengths \((q+1)\) and \((q+6)\), so its area is \((q+1)(q+6)\). The smaller regions have areas \(q^2\), \(6q\), \(q\), and \(6\). Thus, \(q^2+6q+q+6=q^2+7q+6\). In option B, the \(q\)-terms have been combined incorrectly. Exam tip: in an area model, add the areas of all smaller regions and then combine like terms.

The area of the square is \((2x+3)^2\). In the area model, \(2x\times2x=4x^2\), the two rectangles give \(2x\times3+3\times2x=6x+6x=12x\), and the small square gives \(3\times3=9\). Hence, the expansion is \(4x^2+12x+9\). Option A includes only one \(6x\) term. Exam tip: in \((a+b)^2=a^2+2ab+b^2\), do not miss the middle term \(2ab\).

In the area model, split the side \(3a+2\) into \(3a\) and \(2\). This creates two middle rectangles, each with area \((3a)(2)=6a\). Hence, the total middle strip area is \(6a+6a=12a\). \(6a\) is the area of only one middle rectangle, while \(9a^2\) is the area of the larger square part. Exam tip: in \((x+y)^2\), the middle term is always \(2xy\).

When the side of the square is split as \(2p+5\), the small constant corner is a square with side \(5\). Therefore, its area is \(5\times5=25\). \(10p\) is the middle term obtained from the two rectangles, not the area of the constant corner. Exam tip: in a visual identity model, square the constant part to find the constant corner.

The expression contains three area parts:
\(4r^2\), \(20r\), and \(25\). These match the identity \((u+v)^2=u^2+2uv+v^2\). Here, \(4r^2=(2r)^2\) and \(25=5^2\). Thus the two side lengths used in the square are \(2r\) and \(5\), and the complete side is \(2r+5\).

Checking the middle part confirms the result: \(2(2r)(5)=20r\). Therefore \((2r+5)^2=4r^2+20r+25\). Option A gives this side. The other choices would produce a different leading term, constant term, or both. A useful method is to identify the square roots of the first and last terms, then verify the middle term using twice their product.

(9x^2=(3x)^2) and (16=4^2), with middle term (2\cdot3x\cdot4=24x). In exams, identify perfect squares.

The middle term is \(-2\cdot5y\cdot1=-10y\). In exams, count the negative sign in both strips.

(25k^2=(5k)^2) and (9=3^2), so the side is (5k-3). In exams, identify the minus sign from removed strips.

Area of a rectangle = length × breadth. Hence, the area is \((2a+3)(2a-3)\). This matches \((x+y)(x-y)=x^2-y^2\), where \(x=2a\) and \(y=3\). Therefore, the area is \((2a)^2-3^2=4a^2-9\). Option A incorrectly adds the squares instead of subtracting them. Exam tip: The product of two identical binomials with opposite signs is a difference of squares.

\((3m+7)(3m-7)=(3m)^2-(7)^2=9m^2-49\). In the rectangle model, the two parts \(21m\) and \(-21m\) are equal in magnitude and opposite in sign, so they cancel. Hence, no linear term in \(m\) remains in the final expression. The constant term \(-49\) does remain. Exam tip: In the identity \((a+b)(a-b)=a^2-b^2\), the middle or linear term always cancels.

The rectangle has sides (x+10) and (x-10). Its area is found by multiplying these expressions. Using the difference-of-squares identity, (x+a)(x-a)=x^2-a^2, the area becomes (x+10)(x-10)=x^2-10^2=x^2-100. Thus the constant part removed in the visual model is the square of 10.

The subtracted constant is therefore 100, so option C is correct. The number 10 is the side length of the smaller square, but its area is 10^2=100, not 10. The value 20 comes from adding the two strips and is not the constant area being subtracted; 10x is also not constant.

(4+5=9) and (4\times5=20), so the sides are (x+4) and (x+5). In exams, check both sum and product.

The product of the rectangle’s sides must equal the given area. \((v+4)(v+9)=v^2+9v+4v+36=v^2+13v+36\), so the sides are \(v+4\) and \(v+9\). In option A, the constants multiply to \(36\), but \(3+12=15\), so it produces a middle term of \(15v\), not \(13v\). Exam tip: find two numbers whose product is the constant term and whose sum is the coefficient of \(v\).

(36=6^2) and (12x=2\cdot x\cdot6). In exams, identify a perfect square trinomial by square model.

(16=4^2) and (-8a=-2\cdot a\cdot4). In exams, identify the subtractive side from the negative middle term.