Visual models of identities

The direct answer is option A, 7. In the area model for (x+7)^2, divide each side into lengths x and 7. The two equal cross-rectangles then have dimensions x by 7, so each has area 7x. Together their area is 7x+7x = 14x, exactly as stated. The remaining small corner region has both side lengths equal to 7, so it is a square with side 7 and area 7^2 = 49. Option A asks for the side length, so 7 is correct. Option B, 14, is the coefficient in the total area of the two rectangles, not the small square’s side. Option C, 49, is the area of the small square, not its side. Option D, x, describes the side of the large x-by-x square in the model. Memory cue: in (x+b)^2, the small corner square has side b and area b^2; do not confuse the side with its area.

The direct answer is B: a square-removal and rearrangement model. Start with a large square of side \(a\), whose area is \(a^2\). Remove a smaller square of side \(b\), whose area is \(b^2\). The remaining region therefore has area \(a^2-b^2\). By cutting or rearranging this remaining L-shaped region, it can be formed into a rectangle whose two side lengths are \(a-b\) and \(a+b\). Its area is consequently \((a-b)(a+b)\), proving the identity \(a^2-b^2=(a-b)(a+b)\). Option A represents adding square regions, not subtracting one. Option C is associated with a square such as \((a+b)^2\) and includes a middle term. Option D considers only boundary length, whereas this question concerns area. Memory cue: difference of two squares becomes difference times sum.

The direct answer is A, 624. The two factors are equally spaced around 25: \(24=25-1\) and \(26=25+1\). Apply the difference-of-squares identity \((u-v)(u+v)=u^2-v^2\). Thus, \(24\cdot26=(25-1)(25+1)=25^2-1^2=625-1=624\). We can also check directly: \(24\times26=24(20+6)=480+144=624\). Option B, 625, is only \(25^2\); it forgets to subtract \(1^2\). Option C is one more than the correct result and has no identity-based justification. Option D is an incorrect multiplication. The useful mental-math rule is: when two numbers are equally far from a middle number, multiply by squaring the middle number and subtracting the square of the distance. Here the distance is 1.

The direct answer is B, \(12a\). Split each side of the square into lengths \(2a\) and 3. The area model then contains one \(2a\) by \(2a\) square, one 3 by 3 square, and two identical mixed rectangles. Each mixed rectangle has area \((2a)(3)=6a\). Since there are two of them, their combined area is \(6a+6a=12a\). This is also the middle term in \((2a+3)^2=(2a)^2+2(2a)(3)+3^2=4a^2+12a+9\). Option A counts only one mixed rectangle. Option C is the area of the square with side \(2a\), not the mixed parts. Option D is the area of the small square with side 3. A helpful check is to remember that the two cross-regions always produce the factor \(2uv\) in \((u+v)^2\).

The direct answer is A: \(4x^2+12x+9\). A square with side \(2x+3\) can be divided into four regions. The large square has side \(2x\), so its area is \((2x)^2=4x^2\). Each of the two mixed rectangles has area \((2x)(3)=6x\), giving a combined area of \(12x\). The small square has side 3 and area \(3^2=9\). Adding all regions gives \(4x^2+6x+6x+9=4x^2+12x+9\). Option B incorrectly changes \((2x)^2\) to \(2x^2\). Option C counts only one mixed rectangle, giving 6x instead of 12x. Option D uses 3 rather than \(3^2=9\) for the constant square. The safe method is to use \((u+v)^2=u^2+2uv+v^2\).

The direct answer is A. The whole figure is a square whose side is made by joining lengths \(a\) and \(b\), so its side is \(a+b\) and its area is \((a+b)^2\). The model is divided into one region of area \(a^2\), one of area \(b^2\), and two congruent rectangles, each of area \(ab\). Adding these areas gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Therefore \((a+b)^2=a^2+2ab+b^2\). Option B would require negative mixed regions, which are not shown. Option C describes a difference of squares and would involve subtracting one square. Option D is false because the product of a sum and a difference gives \(a^2-b^2\), not \(a^2+b^2\). Memory cue: two positive \(ab\) rectangles mean \(+2ab\).

The direct answer is C, \(7x\). Use the distributive property and identify every region in the area model: \((3x+1)(x+2)=3x\cdot x+3x\cdot2+1\cdot x+1\cdot2\). These products are \(3x^2\), \(6x\), \(x\), and 2. The terms containing exactly one x are \(6x\) and \(x"); combining them gives \(6x+x=7x\). Option A is not obtained from the expansion. Option B counts only \(3x\cdot2\) and misses the rectangle \(1\cdot x\). Option D incorrectly adds an extra x-term. The full expansion is \(3x^2+7x+2\). A reliable habit is to draw the four products first, then combine only like terms; \(x^2\), \(x\), and constants must not be mixed.

The direct answer is B, 6. In the area model, multiply every part of the first side by every part of the second side: \((a+1)(a+6)=a\cdot a+a\cdot6+1\cdot a+1\cdot6=a^2+6a+a+6=a^2+7a+6\). A constant part contains no variable, so it comes from multiplying the two constant pieces, 1 and 6. Thus the constant term is \(1\cdot6=6\). Option A is only one of the constants, not their product. Option C, 7, is the sum of 1 and 6 and becomes the coefficient of the middle term after combining \(6a+a\). Option D still contains a, so it is not constant. Memory cue: in \((a+p)(a+q)\), the constant term is always \(pq\).

The direct answer is B: \((x+2)(x+3)\). To factor \(x^2+5x+6\), find two numbers whose sum is 5 and whose product is 6. Those numbers are 2 and 3. Therefore, \((x+2)(x+3)=x^2+3x+2x+6=x^2+5x+6\). In the rectangle model, the x-by-x region gives \(x^2\), the two side strips give \(3x ext{ and }2x\), and the corner gives 6. Option A gives \(x^2+7x+6\), because 1+6=7. Option C gives \(x^2+11x+30\), so both the middle and constant terms are wrong. Option D gives \(x^2+6x+5\). A useful factorisation check is: for \(x^2+bx+c\), the two numbers must add to b and multiply to c.

The direct answer is C, 1024. First combine the two parts of the side: \(30+2=32\). The area of a square is side multiplied by itself, so the total area is \(32^2=32\times32=1024\). The area model gives the same result: \((30+2)^2=30^2+2(30)(2)+2^2=900+120+4=1024\). Option A is only \(30^2\), so it ignores the extra strips and small corner. Option B, 960, does not include all the area-model parts. Option D is too large and results from an incorrect addition of components. The important distinction is between the side length, 32, and its square, 1024. Memory cue: for a square, never stop after adding the side parts; square the complete side.

The direct answer is A. The complete side of the square is formed by joining lengths \(a\) and \(b\), so it is \(a+b\). Its total area is therefore \((a+b)^2\). The model partitions that area into four parts: \(a^2\), \(ab\), another \(ab\), and \(b^2\). Adding them gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B has \(-2ab\), which would correspond to subtracting mixed regions, not the two positive rectangles shown. Option C is the difference-of-squares identity and has a subtraction between the square areas. Option D is a cube identity and would require volume-like terms, not four regions of a square. Memory cue: a square split into a and b with two positive cross-rectangles gives plus, plus two-ab.

The correct answer is A: \((x+y)^2=x^2+2xy+y^2\). The whole square has side length \(x+y\), so its area is \((x+y)^2\). Divide the square using the x-part and the y-part on both directions. This creates four regions: one square of area \(x^2\), two rectangles each of area \(xy\), and one square of area \(y^2\). Adding them gives \(x^2+xy+xy+y^2=x^2+2xy+y^2\). Option A matches all four regions. Option B represents a square with side difference, not a side sum. Options C and D are equivalent forms of the difference-of-squares identity and involve subtraction, so they do not describe this four-part sum model. Memory cue: the two identical middle rectangles produce the term \(2xy\).

The correct answer is A: \(a^2,ab,ab,b^2\). A square with side \(a+b\) is divided into an a-by-a square, an a-by-b rectangle, a b-by-a rectangle, and a b-by-b square. Their areas are respectively \(a^2\), \(ab\), \(ab\), and \(b^2\). Adding these four regions gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\), which is the expansion of \((a+b)^2\). Option B leaves out one of the two middle rectangles. Option C combines terms while also giving the wrong last term, and it does not list the actual regions. Option D lists three b² regions, but the model contains only one b² square. Remember: an area model shows the separate pieces before like terms are combined.

The correct answer is A. Begin with a large square of side a, whose area is \(a^2\). To create the smaller square of side \(a-b\), remove a horizontal strip of area \(ab\) and a vertical strip of area \(ab\). These two strips overlap in a corner square of side b, having area \(b^2\). The overlap is included in both subtractions, so it has been removed twice. It should be removed only once; therefore, add \(b^2\) back once: \(a^2-ab-ab+b^2=a^2-2ab+b^2=(a-b)^2\). B is false because area can decrease. C is false because \(a^2\) is generally nonzero. D is false because the two \(ab\) strips do form. Memory cue: subtracting overlapping regions means adding their overlap once.

The correct answer is B, \(4x\). In the model for \((x+2)^2\\), the two middle regions are rectangles with dimensions x and 2. Each rectangle therefore has area \(x\times2=2x\). Since there are two such rectangles, their combined area is \(2x+2x=4x\), or equivalently \(2\times x\times2=4x\). Option A gives the area of only one middle rectangle. Option C adds lengths rather than calculating area, so it is not appropriate. Option D squares x and does not represent either middle rectangle. The complete expansion is \(x^2+4x+4\). Memory cue: in \((a+b)^2\), the two middle rectangles together always give \(2ab\).

The correct answer is A. The labels across the top combine to give the total length \(x+3\), and the labels down the side combine to give the total width \(x+2\). The grid’s area is therefore \((x+3)(x+2)\). Multiply each part: \(x\cdot x=x^2\), \(x
ot
eed o2=2x\), \(3
ot
eed ox=3x\), and \(3
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eed o2=6\). Adding them gives \(x^2+2x+3x+6=x^2+5x+6\). A is correct. B incorrectly swaps the middle coefficient and constant. C omits the contributions involving 3 and 6. D includes the right terms but assigns the wrong coefficients. Memory cue: in a grid, multiply every top label by every side label, then combine like terms.

The correct answer is A. Since \(9=3^2\), we can write \(x^2-9=x^2-3^2\). The term \(x^2\) is the area of a large square with side x, and \(3^2=9\) is the area of a smaller square with side 3. Removing the smaller square from the larger one leaves exactly the difference \(x^2-9\). B represents \((x+3)^2\), which is a larger square area, not a subtraction. C represents \((x-3)^2=x^2-6x+9\), which contains extra terms and is not equal to \(x^2-9\). D represents an area of \(9x\), not \(x^2-9\). Memory cue: a square’s area is the square of its side, so 9 must be viewed as \(3^2\).

The correct answer is C: \(a^2-16\). The rectangle has side lengths \(a+4\) and \(a-4\). Using the area model, multiply each part: \(a\cdot a=a^2\), \(a\cdot(-4)=-4a\), \(4
ot
eed oa=4a\), and \(4
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eed o(-4)=-16\). The middle terms cancel because \(-4a+4a=0\), leaving \(a^2-16\). This is also the identity \((p+q)(p-q)=p^2-q^2\). A is the expansion of \((a+4)^2\), while B is the expansion of \((a-4)^2\). D has the wrong sign for the constant term; the product of 4 and -4 is negative. Memory cue: conjugate factors cancel the middle terms and leave a difference of squares.

The correct answer is B, 12. In the identity \((a+b)^2=a^2+2ab+b^2\), the two middle rectangles together have area \(2ab\). Here a=3 and b=2, so their combined area is \(2\times3\times2=12\). Equivalently, one rectangle has dimensions 3 and 2, so its area is 6; there are two such rectangles, giving \(6+6=12\). A is only the area of one middle rectangle. C is \(2^2\), the area of the small square, not the two rectangles. D is the whole square’s area, \((3+2)^2=25\). Memory cue: the two cross-rectangles give the middle term \(2ab\), not just ab.

The correct answer is C, 4. Write \(7^2\) as \((5+2)^2\). A square whose side is split into 5 and 2 is divided in both directions. This creates four regions: a square of area \(5^2\), one rectangle of area \(5\times2\), a second rectangle of area \(5\times2\), and a square of area \(2^2\). Thus the total area is \(5^2+5\cdot2+5\cdot2+2^2=25+10+10+4=49=7^2\). A and B do not count all four pieces. D invents an extra region; the two cross-regions are already exactly two rectangles. Memory cue: every square-of-a-sum area model has one first square, two cross-rectangles, and one second square—four regions.

The correct answer is B: an individual rectangular strip of area \(ab\) is subtracted twice. Start with the large square of area \(a^2\). To obtain a square of side \(a-b\), remove one horizontal strip of area \(ab\) and one vertical strip of area \(ab\). Thus the subtraction contains \(-ab-ab=-2ab\). The two strips overlap in a corner square of area \(b^2\), so that overlap is counted in both removals and must be added back once. Option A is the starting area, not a removed strip. C is the overlap that is restored, not the type of strip subtracted twice. D is not an area at all and does not describe the construction. Memory cue: the coefficient 2 in \(-2ab\) counts the two removed strips.

The direct answer is B: x²+8x+16. A square with side x+4 has area (x+4)². Use the identity (a+b)²=a²+2ab+b², taking a=x and b=4. Then (x+4)²=x²+2(x)(4)+4²=x²+8x+16. In the visual model, the large x by x square contributes x², two equal rectangles of dimensions x by 4 contribute 4x+4x=8x, and the small 4 by 4 square contributes 16. Option A has only 4x, so it counts one rectangle instead of two. Option B includes all three regions correctly. Option C belongs to a subtraction identity such as (x−4)², so its middle sign is wrong. Option D has the middle and constant terms confused. Memory cue: in a square of side x+b, the two mixed rectangles always give 2bx.

The direct answer is A: x+6. The x² region is a square whose side is x, because x×x=x². The small square has area 36, so its side is 6, since 6×6=36. Each middle rectangle has area 6x, which is exactly x×6. Therefore the two side parts forming the outside edge have lengths x and 6, and the complete outer side is x+6. The full area confirms this: x²+6x+6x+36=x²+12x+36=(x+6)². Option B incorrectly adds 6 twice to the outside length; the two rectangles are placed on different sides, not end-to-end along one outer side. Option C uses subtraction and would correspond to a model involving a missing region or x−6. Option D is not a side length obtained from the pieces and mixes an area-like product with a length. Memory cue: identify the side of each square first, then add the adjacent side segments.

The direct answer is C: 10a. Split the side a+5 into lengths a and 5. The visual square then contains an a by a square, two equal rectangles, and a 5 by 5 square. Each rectangle has dimensions a and 5, so its area is a×5=5a. Since there are two such rectangles, their combined area is 5a+5a=10a. This is also the middle term in (a+5)²=a²+10a+25. Option A gives the area of only one rectangle, not both. Option B incorrectly multiplies 25 by a; 25 is the area of the small 5 by 5 square and is not part of the rectangle total. Option C correctly doubles 5a. Option D is the area of the a by a square, so it identifies a different region. Memory cue: two identical mixed rectangles contribute twice the product of their side labels.

The direct answer is A: x²+7x+10. Draw or imagine a rectangle with one side split into x and 2 and the other side split into x and 5. The four smaller regions have areas x×x=x², x×5=5x, 2×x=2x, and 2×5=10. Adding all regions gives x²+5x+2x+10=x²+7x+10. Option A combines the like terms correctly. Option B uses 10 as the coefficient of x and 7 as the constant, confusing the sum of 2 and 5 with their product. Option C uses 3x, which is only (2+1)x and does not represent the two cross-products. Option D has the correct x coefficient but changes the constant 10 to 7. A useful check is (x+m)(x+n)=x²+(m+n)x+mn; here m+n=7 and mn=10.