Visual models of identities

The direct answer is B. The rectangle’s area is length times breadth: (a+b)(a−b). Multiply each part: a·a=a², a·(−b)=−ab, b·a=ab, and b·(−b)=−b². The two mixed terms, −ab and +ab, cancel, leaving a²−b². In an area model this cancellation can be shown by equal positive and negative mixed regions. A is the expansion of (a+b)², not conjugate factors. C fails to cancel the mixed terms completely. D is the expansion of (a−b)², where both mixed terms are negative. Memory cue: opposite signs in the two brackets produce the difference of the squares.

The direct answer is A. Each side of the large square is divided into adjacent lengths a and b, so the full side is a+b. The total area is therefore (a+b)². Adding the four regions gives a²+ab+ab+b², and the two like terms ab combine to 2ab. Thus (a+b)²=a²+2ab+b². B would require the mixed regions to be subtracted, not shown as two positive areas. C uses side lengths a+b and a−b and gives a²−b², so its middle terms cancel. D is a cubic identity involving a three-dimensional-style product and four cubic terms, not this square area model. Memory cue: two positive ab rectangles produce the middle term +2ab.

The direct answer is B: (p+3)²=p²+6p+9. Split the square into a p-by-p square of area p², two strips each measuring p by 3 and therefore having area 3p, and a 3-by-3 corner square of area 9. Add all parts: p²+3p+3p+9=p²+6p+9. A counts only one strip, so its middle term is too small. C invents 9p, which is neither the area of one strip nor the corner. D uses 3 instead of the corner area 9. The identity (u+v)²=u²+2uv+v² explains the same result: u=p and v=3. Memory cue: the middle term comes from two identical strips.

The direct answer is A: (y+5)^2 = y^2 + 10y + 25. The area of a square is side multiplied by side, so the large square has area (y+5)(y+5). Split each side into lengths y and 5. The y by y central square has area y^2. There are two equal rectangles: each is y by 5, so each has area 5y and together they contribute 10y. The corner is a 5 by 5 square, whose area is 25. Adding all four regions gives y^2+5y+5y+25=y^2+10y+25. A therefore matches the complete area model. B counts only one strip and misses 5y. C puts the numbers in the wrong roles: 25 is not the coefficient of y and 10 is not the corner area. D has the correct middle term but incorrectly uses 5 instead of 25 for the corner. A useful memory cue is (u+v)^2=u^2+2uv+v^2: the two equal strips create 2uv, not just uv.

The direct answer is B: subtract 4m and add 4. Begin with a square of side m, whose area is m^2. To change each side from m to m-2, remove a strip of width 2 along one direction and another strip of width 2 along the other direction. Each strip has area 2m, so the total first subtraction is 2m+2m=4m. However, the 2 by 2 corner lies in both removed strips, so it has been subtracted twice. It must be restored once; its area is 2×2=4. Thus the remaining area is m^2-4m+4, which is exactly (m-2)^2. A subtracts only one strip and is incomplete. B correctly handles both strips and the overlap. C reverses the roles of the removed strips and the corner. D uses 2 rather than the corner area 4. Memory cue: (m-c)^2=m^2-2cm+c^2.

The direct answer is B: (z-7)^2=z^2-14z+49. A square with side z-7 has area (z-7)(z-7). Multiply the two binomials in separate parts: z×z gives z^2; z×(-7) gives -7z; (-7)×z gives another -7z; and (-7)×(-7) gives +49. Combining the two equal linear terms gives -7z-7z=-14z, so the full expansion is z^2-14z+49. A includes only one cross-product, so it misses half of the middle term. B includes both negative strips and is correct. C has the wrong sign for the middle term; subtracting 7 makes both cross-products negative. D uses incorrect coefficients and does not follow multiplication. The identity (u-v)^2=u^2-2uv+v^2 is a reliable check: the middle term is negative, while the final square is positive.

The direct answer is A: r^2+8r+12. The area of the rectangle is length multiplied by breadth, so calculate (r+6)(r+2). In an area model, split each side into its variable and constant parts. The four regions have areas r×r=r^2, r×2=2r, 6×r=6r, and 6×2=12. Adding them gives r^2+2r+6r+12=r^2+8r+12. Therefore A is correct. B incorrectly uses 12r as the middle term; 12 is the product of the constants and must be the constant term, not a coefficient of r. C adds the constant terms instead of combining the two variable rectangles. D has the right middle coefficient but the wrong constant product: 6×2 is 12, not 8. A useful check is to multiply symbolically: (r+6)(r+2)=r^2+2r+6r+12. In a rectangle model, every one of the four smaller regions must be counted exactly once.

The direct answer is B: the regions 4n and -4n cancel. Expand the rectangle’s area: (n+4)(n-4)=n×n+n×(-4)+4×n+4×(-4)=n^2-4n+4n-16. The two middle regions have equal magnitude, 4n, but opposite signs: one is -4n and the other is +4n. Their sum is zero, so they cancel. The remaining expression is n^2-16, the difference-of-squares identity. A is wrong because n^2 is the main square and 16 is the constant corner; they do not cancel and have different algebraic forms. C pairs a variable term with a constant term, so cancellation is not possible. D also pairs unlike terms and ignores that -4n cancels with +4n, not with 16. A useful warning is that cancellation requires like terms with opposite signs. Equal-looking sizes alone are not enough; the algebraic type and sign must both match.

The direct answer is C: t^2-25. The rectangle’s area is the product of its sides, (t+5)(t-5). Use the difference-of-squares identity (u+v)(u-v)=u^2-v^2, with u=t and v=5. This gives t^2-5^2=t^2-25. The same result appears in the area model: the four regions are t^2, -5t, +5t, and -25. The two middle regions cancel because -5t+5t=0, leaving t^2-25. A is wrong because the corner product 5×(-5) is negative, not positive. B and D incorrectly retain a middle term; opposite cross-products cancel completely. B also has the wrong constant sign, while D has both the wrong middle term and wrong constant sign. Memory cue: conjugate binomials, with opposite signs in the constants, produce a difference of squares and no middle term.

The correct answer is B. The area is \((2x+3)^2\). Expanding gives \((2x)^2+2(2x)(3)+3^2=4x^2+12x+9\). The middle term includes both equal rectangular parts, so it is \(12x\).

The direct answer is B: 12a. Divide each side of the square into parts 3a and 2. The area model then contains two middle rectangles. Each rectangle has area (3a)(2)=6a. Since both rectangles belong to the middle strips, their combined area is 6a+6a=12a. This is also the middle term in (3a+2)², because 2(3a)(2)=12a. Option A is only the area of one rectangle, not both. Option C is (3a)²=9a², the large corner square. Option D is 2²=4, the small constant corner. Memory cue: in (u+v)², the total middle area is 2uv, so it always includes two equal cross-rectangles.

A is correct. The expression 4r² + 20r + 25 is a perfect square: (2r + 5)² = 4r² + 2(2r)(5) + 25 = 4r² + 20r + 25. Therefore, the square's side can be 2r + 5.

The correct answer is A, 3x + 4. Since 9x² + 24x + 16 = (3x)² + 2(3x)(4) + 4² = (3x + 4)², the expression is the square of 3x + 4. Thus the side length is 3x + 4, assuming this expression is non-negative as a length must be.

The correct answer is A. The area is (5y − 1)². Using (p − q)² = p² − 2pq + q² with p = 5y and q = 1 gives 25y² − 10y + 1. The middle term is negative because the binomial contains subtraction.

The direct answer is A: (5k−3)². Recognize the expression as a square identity. Since 25k²=(5k)² and 9=3², the large square and the small corner correspond to squares of 5k and 3. The two removed strips represent the negative middle term −2(5k)(3)=−30k. Therefore the complete expression is (5k)²−2(5k)(3)+3²=(5k−3)². Option B would produce a positive middle term +30k, which conflicts with subtraction. Option C incorrectly uses 25k as the side instead of its square root 5k. Option D would give a constant term 81, not 9, and its middle term is also different. Memory cue: subtracting the two cross-strips signals the minus sign in (u−v)².

The direct answer is B: 4a²−9. The rectangle’s area is length times breadth, so calculate (2a+3)(2a−3). This has the form (u+v)(u−v)=u²−v², with u=2a and v=3. Hence the area is (2a)²−3²=4a²−9. In an area model, the two cross-regions have equal size but opposite signs, so they cancel. Option A incorrectly adds 3² instead of subtracting it. Option C changes the leading square incorrectly: (2a)² is 4a², not 2a². Option D is the expansion of (2a−3)², not the product of opposite-sign binomials. Memory cue: conjugate binomials produce a difference of squares and no middle term.

The direct answer is C: the linear term in m will not remain. Multiply the side expressions: (3m+7)(3m−7). Using (u+v)(u−v)=u²−v², with u=3m and v=7, the result is (3m)²−7²=9m²−49. If the rectangle is split into parts, the cross-products are 21m and −21m. They have equal magnitude and opposite signs, so their sum is zero. Thus the m-term cancels. Option A, 9m², remains as the product of the variable parts. Option B, −49, remains as the constant product 7×(−7). Option D is a category: the constant term does remain, specifically −49. Memory cue: conjugate factors remove the middle term.

The correct answer is C: 100. The area of the rectangle is found by multiplying its length and breadth: (x+10)(x-10). This matches the difference-of-squares identity (p+q)(p-q)=p²-q². Here p=x and q=10, so the area is x²−10²=x²−100. In a visual model, the large x by x square has area x², while a 10 by 10 corner square is removed; that corner has area 100. Option A is only the side length 10, not its area. Option B is 10+10 and does not represent the removed square. Option D contains x, so it is not a constant. Memory cue: opposite binomials produce a difference of squares.

The correct answer is A: (x+4)(x+5). To factor the trinomial, find two numbers whose product is the constant term 20 and whose sum is the coefficient of x, namely 9. The numbers 4 and 5 work because 4×5=20 and 4+5=9. Therefore (x+4)(x+5)=x²+5x+4x+20=x²+9x+20. In a tile model, the x² tile, 9 x-tiles, and 20 unit tiles can be arranged into a rectangle with side lengths x+4 and x+5. Option B has product 20 but sum 12, giving x²+12x+20. Option C has sum 21, and D does not give the required middle term. Check both sum and product, not just one.

The correct answer is B: \(v+4\) and \(v+9\). Their product is \((v+4)(v+9)=v^2+9v+4v+36=v^2+13v+36\). The numbers 4 and 9 multiply to 36 and add to 13, giving the required constant and middle terms.

The correct answer is A: a square of side x+6. Compare the expression with the perfect-square identity (x+k)²=x²+2kx+k². Since 36=6², take k=6. Then 2x×6=12x, so x²+12x+36=(x+6)². A visual model therefore has one x² square, two equal 6x rectangles, and one 36-unit square, forming a larger square. Option B would produce x²+24x+144. Option C gives x²+15x+36 because 3+12=15. Option D gives x²+13x+36 because 4+9=13. Memory cue: in a perfect-square trinomial, the constant is a square and the middle coefficient is twice its square root.

Direct answer: Option B, a−4. The expression is a perfect-square trinomial. Use the identity (p−q)²=p²−2pq+q². Here p=a and q=4, so (a−4)²=a²−2(a)(4)+4²=a²−8a+16. Therefore, the complete visual square has side length a−4. In an area model, the a² tile is the large square, the two cross-parts together contribute −8a, and the small corner square contributes 16. Option A, a+4, would give (a+4)²=a²+8a+16, so its middle-term sign is wrong. Option C would give a²−16a+64, not the given expression. Option D would give a²+16a+64, with both the middle term and constant term wrong. A useful memory cue is that a negative middle term in a square identity indicates subtraction inside the bracket: (a−4)², not (a+4)². The side length is an algebraic expression; the model assumes the dimensions are meaningful in the stated context.

The correct answer is B: 12b. A square of side b+6 has area (b+6)². Split each side into lengths b and 6. The model contains the large b by b square with area b², two rectangular strips, each with dimensions b and 6, and a corner square with area 6²=36. Each rectangle therefore has area 6b, so together they have area 6b+6b=12b. This is also the middle term in b²+12b+36. Option A counts only one rectangle. Option C uses the wrong multiplication, and option D adds the two square areas rather than finding the strips. The factor 2 in 2×b×6 represents the two equal rectangles.

The correct answer is A: c+9. The c² part is a square with side c because c×c=c². Each 9c strip has dimensions c and 9, so the added width is 9. The corner area is 81=9², confirming that its side is 9. Therefore the complete side of the large square is c+9. Algebraically, the model represents c²+9c+9c+81=c²+18c+81=(c+9)². Option B incorrectly adds the strip width twice to one side. Option C mixes area-like and length-like quantities and is not obtained from the model. Option D would describe a difference model and would produce negative middle terms. Add lengths along one side, not the two strip areas.

The correct answer is A: (x+y)(x−y). The original square has area x² and the removed square has area y², so the remaining area is x²−y². The difference-of-squares identity factors this as x²−y²=(x+y)(x−y). A visual model shows the same result by rearranging the leftover parts into a rectangle whose sides are x+y and x−y. Option B is the square of a sum and is larger than the required expression. Option C is the area of a square with side x−y, not generally the leftover region. Option D adds the two areas instead of subtracting the removed one. The phrase “removed square” is a strong clue for a difference of squares.