Visual models of identities

Answer: B, \(2x^2+11x+5\). The area of a rectangle is length multiplied by breadth, so calculate \((2x+1)(x+5)\). Multiply every term in the first bracket by every term in the second: \(2x\cdot x=2x^2\), \(2x\cdot5=10x\), \(1\cdot x=x\), and \(1\cdot5=5\). Add the four regions: \(2x^2+10x+x+5=2x^2+11x+5\). Therefore B is correct. A is missing the contribution \(1\cdot x=x\), so it has \(10x\) instead of \(11x\). C incorrectly combines the middle products as \(10x-3x\). D places the constant and linear terms incorrectly and does not come from multiplication. In an area model, every row part must meet every column part. Memory cue: use FOIL or a four-box model, then combine only like terms.

Answer: A, \(14y\). Multiply the two side expressions: \((3y+2)(y+4)\). The four area pieces are \(3y\cdot y=3y^2\), \(3y\cdot4=12y\), \(2\cdot y=2y\), and \(2\cdot4=8\). The terms containing exactly one factor of \(y\) are the two mixed products, \(12y\) and \(2y\). They are like terms, so add their coefficients: \(12y+2y=14y\). Thus the complete expansion is \(3y^2+14y+8\). Option A is correct. B is the constant corner, obtained from \(2\cdot4\). C is only one mixed contribution, from \(3y\cdot4\). D is the other mixed contribution, from \(2\cdot y\), not their total. A common mistake is to choose one rectangle instead of combining both. Memory cue: for \((ay+b)(y+d)\), the linear term comes from \(ady+by\), so both mixed rectangles must be counted.

Answer: A, \(14y\). An area model for \((3y+2)(y+4)\) has four products. The product \(3y\cdot y\) gives \(3y^2\), and the product \(2\cdot4\) gives the constant \(8\). The two remaining mixed rectangles are \(3y\cdot4=12y\) and \(2\cdot y=2y\). Both contain one \(y\), so both contribute to the total linear term. Adding them gives \(12y+2y=14y\). Therefore the full product is \(3y^2+14y+8\). Option A is correct. Option B, \(6y\), is not produced by the required multiplication. Option C is only the first mixed rectangle, while D is only the second mixed rectangle. The word “total” is important: it asks for both mixed parts together, not one part alone. Memory cue: the linear term is always the sum of the two cross-products.

Answer: A, \(10x\) and \(-10x\). Expand the product: \((x+10)(x-10)=x\cdot x+x\cdot(-10)+10\cdot x+10\cdot(-10)\). These four regions are \(x^2\), \(-10x\), \(10x\), and \(-100\). The two mixed regions have equal magnitude but opposite signs: \(10x-10x=0\), so they cancel. The remaining area expression is \(x^2-100\), which is the difference-of-squares identity. Option A is correct. B is wrong because \(x^2\) and \(-100\) do not cancel; they are different kinds of terms and remain in the result. C is wrong because the two constants are not both positive; one is \(-100\). D lists pieces rather than the two opposite linear products. Memory cue: conjugate factors \((a+b)(a-b)\) remove the middle terms and leave \(a^2-b^2\).

Answer: B, \(25a^2+20a+4\). A square’s area is side multiplied by itself, so the required expression is \((5a+2)^2\). Use the identity \((u+v)^2=u^2+2uv+v^2\), with \(u=5a\) and \(v=2\). Then \(u^2=25a^2\), the two mixed rectangles together give \(2uv=2(5a)(2)=20a\), and the small square gives \(v^2=4\). Adding these parts gives \(25a^2+20a+4\). Thus B is correct. A has only \(10a\), as though one mixed rectangle had been counted instead of two. C incorrectly squares \(5a\) as \(5a^2\) instead of \(25a^2\). D uses 2 instead of the constant square \(2^2=4\). Memory cue: the middle term of a binomial square is twice the product of the two parts.

Answer: C, \(4p^2-28p+49\). The square’s area is \((2p-7)^2\). Apply \((u-v)^2=u^2-2uv+v^2\), where \(u=2p\) and \(v=7\). The first square is \((2p)^2=4p^2\). Two strips are removed, each with area \((2p)(7)=14p\), so together they contribute \(-28p\). The small corner square has area \(7^2=49\). Therefore the result is \(4p^2-28p+49\), so C is correct. A has only one strip’s contribution and therefore uses \(-14p\). B has the wrong positive sign; subtraction creates a negative middle term. D incorrectly treats \((2p)^2\) as \(2p^2\) instead of \(4p^2\). Memory cue: in \((u-v)^2\), the first and last terms are positive, but the middle term is negative and doubled.

Answer: B, \(k=4\). In the square model for \((k+3)^2\), the two mixed strips have areas \(k\cdot3=3k\) each. Therefore their combined area is \(3k+3k=6k\), which is also the middle term in \(k^2+6k+9\). The question says this combined area is 24, so set up \(6k=24\). Divide both sides by 6: \(k=24/6=4\). Hence option B is correct. Option A would give a combined strip area of 12, not 24. Option C would give 36, and option D would give 48. The key is not to use just one strip; the phrase “both strips” requires \(2\times k\times3\). Memory cue: for \((k+c)^2\), the combined mixed area is \(2kc\).

The direct answer is A: 10x+25. A border is the region inside the larger square but outside the smaller square, so subtract the inner area from the outer area. The outer square has area (x+5)² = x²+10x+25. The inner square has area x². Therefore, border area = (x²+10x+25)−x² = 10x+25. Option A is correct. Option B, 5x+25, includes only one 5x contribution instead of the two side strips. Option C, x²+25, incorrectly leaves the inner-square term and misses the 10x contribution. Option D, 10x+5, uses 5 instead of 5²=25 for the corner part. The same result comes from (x+5)²−x² = [(x+5)−x][(x+5)+x] = 5(2x+5). Memory cue: border area always means outer area minus inner area.

The direct answer is B: 4ab. Divide the square of side a+2b into an a-by-a square, a 2b-by-2b square, and two mixed rectangles. Each mixed rectangle has side lengths a and 2b, so its area is a×2b=2ab. Since there are two such rectangles, their total area is 2ab+2ab=4ab. Option A gives the area of only one mixed rectangle, not both. Option C is the area of neither required mixed rectangle; it also has the wrong type of term because (2b)²=4b² is the small square’s area. Option D combines the two square areas, a² and 4b², and leaves out the mixed rectangles. This is also visible in (a+2b)²=a²+4ab+4b²: the middle term 4ab is precisely the combined area of the two mixed rectangles. Memory cue: two identical cross-parts double one rectangle’s area.

The direct answer is A: 4a²−4ab+b². The area of a square with side 2a−b is (2a−b)². Use the identity (p−q)²=p²−2pq+q² with p=2a and q=b. Then p²=(2a)²=4a², the middle term is −2(2a)(b)=−4ab, and q²=b². Thus the expansion is 4a²−4ab+b². Option B has the wrong middle coefficient; it forgets that two cross-rectangles contribute 2×2ab=4ab. Option C incorrectly changes (2a)² into 2a². Option D uses a plus sign in the middle, which belongs to (2a+b)², not a difference. The visual model explains the negative middle contribution when subtraction is used. Memory cue: square the first term, double the product with the sign, then square the second term.

The direct answer is B: 9x²−4. Rectangle area equals length×width, so calculate (3x+2)(3x−2). These are conjugate binomials of the form (p+q)(p−q)=p²−q², with p=3x and q=2. Therefore, the area is (3x)²−2²=9x²−4. Option A incorrectly adds the two squares; the conjugate identity requires subtraction. Option C changes (3x)² into 3x², but the coefficient must also be squared: (3x)²=9x². Option D is the expansion of (3x−2)², not the product of one plus and one minus expression. The middle terms +6x and −6x cancel, which is why no x-term remains. For x=1, the area is 5×1=5, and 9−4=5, confirming the result. Memory cue: matching binomials with opposite signs give difference of squares.

The direct answer is B: 4x. The two areas are (x+1)² and (x−1)², so their difference is (x+1)²−(x−1)². Apply A²−B²=(A−B)(A+B), taking A=x+1 and B=x−1. Then A−B=(x+1)−(x−1)=2, while A+B=(x+1)+(x−1)=2x. Their product is 2×2x=4x. Option A gives only one factor of 2 and is therefore half the correct difference. Option C is (x+1)(x−1), the product of the two side expressions, not the difference of their areas. Option D ignores the variable-dependent part. For x=3, the areas are 16 and 4, whose difference is 12; 4x also gives 12. Memory cue: area difference means square first, then subtract—or use difference of squares.

The direct answer is A: 8x+16. The remaining region is the outer square minus the removed inner square. Its area is (x+4)²−x². Expand the larger square: (x+4)²=x²+8x+16. Subtracting x² leaves 8x+16. Geometrically, the remainder consists of two strips, each contributing 4x, plus a corner square of area 4²=16. Option B includes only one 4x strip and misses the second. Option C leaves an x² term that should have been cancelled, so it is not the border area. Option D has the correct 8x part but uses 4 instead of the corner area 16. The difference-of-squares form also gives [(x+4)−x][(x+4)+x]=4(2x+4)=8x+16. Memory cue: the constant corner is the square of the added width.

The direct answer is A: a²+b²+c². Split the side of the large square into three lengths a, b, and c. The visual model then contains three small square regions: one with side a, one with side b, and one with side c. Their areas are a², b², and c² because square area is side×side. Adding only these requested regions gives a²+b²+c². Option B lists the areas of three types of mixed rectangles, not the small squares. Option C gives the doubled mixed terms that appear in the full expansion (a+b+c)², but the question excludes those rectangles. Option D is only the side length, not an area, and it also has the wrong degree. The complete expansion is a²+b²+c²+2ab+2bc+2ca; this helps separate square terms from cross terms. Memory cue: single-part regions produce squares; two-part regions produce products.

The correct answer is B: 2ab+2bc+2ca. Imagine a large square whose side is divided into three lengths a, b, and c. The three diagonal square regions have areas a², b², and c². The remaining regions are mixed rectangles. The pair formed by lengths a and b gives two rectangles, each with area ab, so together they give 2ab. Similarly, the a-c pair gives 2ac, and the b-c pair gives 2bc. Therefore, the total mixed-rectangle area is 2ab+2ac+2bc, written here as 2ab+2bc+2ca. Option A counts only one rectangle from each pair. Option C counts the three pure square regions, not the mixed rectangles. Option D has the wrong type and degree of terms. Memory cue: in a square of three parts, every mixed pair appears twice.

The correct answer is A: x²−2x−15. Use the distributive property and multiply every term in the first binomial by every term in the second. The four signed regions are x·x=x², x·(−5)=−5x, 3·x=3x, and 3·(−5)=−15. Now combine like terms: x²−5x+3x−15=x²−2x−15. Option B incorrectly treats the middle terms as +5x and +3x, producing +8x. Option C places the coefficient and constant incorrectly; −15x is not the product of the two constants. Option D changes the signs of the middle and constant terms and therefore does not follow the given factors. Memory cue: use FOIL carefully, and remember that a positive times a negative is negative while two unlike x-terms combine algebraically.

The correct answer is A: 8x+24. The two square areas are (x+5)^2 and (x+1)^2, so their difference is (x+5)^2-(x+1)^2. This is a difference of squares. Let P=x+5 and Q=x+1. Using P^2-Q^2=(P+Q)(P-Q), we get [(x+5)+(x+1)] [(x+5)-(x+1)]=(2x+6)(4)=8x+24. The same result can be checked by expansion: (x+5)^2=x^2+10x+25 and (x+1)^2=x^2+2x+1. Subtracting gives x^2+10x+25-x^2-2x-1=8x+24. Option B incorrectly handles the multiplication by 4 and gives the wrong coefficient of x. Option C does not come from either correct factoring or expansion. Option D combines terms as though the two areas should be added, rather than finding their difference. Thus A is the only consistent answer. Memory cue: the difference of two square areas equals the sum of their side expressions multiplied by their difference.

The correct answer is B: x²+6x. The model already contains one large x-by-x square with area x². It also contains two strips, each with dimensions x and 3, so each strip has area 3x. Since the question says that the corner is not added, do not include the small 3-by-3 region, whose area would be 9. Add only the included parts: x²+3x+3x=x²+6x. Option A counts only one strip and leaves out the second. Option C includes the corner but omits both strips. Option D is the complete square (x+3)² and therefore adds the excluded 9. Memory cue: first identify exactly which regions are present; an incomplete model does not automatically equal the full square.

The correct answer is D: 25. In a visual completion-of-square model, the middle term 10x is formed by two equal strips. Each strip must therefore have coefficient 5x, because 5x+5x=10x. The missing corner has side lengths 5 and 5, so its area is 5×5=25. Algebraically, x²+10x+25=(x+5)². Option A is only half of the middle coefficient and gives a length, not the corner area. Option B uses the full coefficient without halving. Option C doubles the coefficient and has no matching square construction. Option D correctly squares half of 10. Memory cue: for x²+bx, add (b/2)²; always halve the coefficient of x before squaring.

The correct answer is B: 9. The middle term −6x is made from two equal negative strips, each of size −3x, because −3x−3x=−6x. The missing corner has side lengths −3 in the signed model, and its area is (−3)(−3)=9. Thus x²−6x+9=(x−3)², a perfect square. Option A is the absolute value of the full coefficient and does not represent a square corner. Option C doubles the magnitude instead of halving it. Option D squares the full coefficient 6, but the corner is based on half the coefficient. Memory cue: ignore the sign while finding the area, halve the coefficient first, then square: (−6/2)²=9.

The correct answer is A: the rectangle has side lengths b+c and b−c. The large square has area b², and the removed square has area c². Therefore the remaining area is b²−c². The difference-of-squares identity says b²−c²=(b+c)(b−c). Thus the remaining pieces can be rearranged to form a rectangle with one side b+c and the other side b−c. Multiplying confirms it: (b+c)(b−c)=b²−bc+bc−c²=b²−c². Option B gives area bc, not the required difference. Option C gives c(b+c), and option D gives c(b−c); neither equals the stated remaining area in general. Memory cue: a²−b² factors as (a+b)(a−b), not as a single sum or difference.

The correct answer is A: x^2, 3x, 3x, and 9. Split each side of the large square, whose side is x+3, into lengths x and 3. This creates four regions: an x-by-x square with area x^2; one x-by-3 rectangle with area 3x; another 3-by-x rectangle with area 3x; and a 3-by-3 square with area 9. Adding them gives x^2 + 3x + 3x + 9 = x^2 + 6x + 9, which is (x+3)^2. Option A lists exactly these four geometric parts. Option B combines the two rectangles into 6x and then incorrectly lists an extra 9, so it does not show four correct regions. Option C includes 12, which is not the area of any required part. Option D gives x as though it were an area of a strip, but the rectangle must have area 3x. Memory cue: for (x+3)^2, the two mixed rectangles are each 3x, not 6x.

The direct answer is option B: y+4. A square model with outer side y+4 can be divided into a y by y square, two rectangles measuring y by 4, and a 4 by 4 corner square. Their areas are y², 4y, 4y, and 16. Adding them gives y²+4y+4y+16=y²+8y+16, exactly the three given parts. Therefore, the side is y+4. Option A, y+8, incorrectly treats the complete middle coefficient 8 as a side length; its square would contain 16y as the middle term. Option C, y−4, gives y²−8y+16, so the sign of the middle term is wrong. Option D, y+16, treats the corner area as though it were a length. A useful memory cue is: in a²+2ab+b², the outer side is a+b, and the last term is the square of the smaller side.

The direct answer is option C: 25. Begin with a square of side a, whose area is a². To make the remaining figure have side a−5, remove a strip of width 5 along one side and another strip of width 5 along the adjacent side. Each strip has area 5a, so subtracting both gives a²−5a−5a=a²−10a. However, the two strips overlap at one corner. That 5 by 5 corner has area 25 and has been subtracted twice, although it should be removed only once. We therefore add 25 back, giving a²−10a+25=(a−5)². Option A is only the area of one strip, not the final correction. Option B is the combined area of the two strips before correcting the overlap. Option D is the area of the original square, not the small corner. Memory cue: in a subtraction model, subtract the two strips and add back the overlap once.

The direct answer is option D: x²−36. The area of a rectangle is length multiplied by breadth, so here it is (x+6)(x−6). This is a product of a sum and a difference. Using (a+b)(a−b)=a²−b² with a=x and b=6 gives x²−6²=x²−36. In a visual model, the x by x square has area x², while the two side portions removed or rearranged together have total area 36; the x terms cancel because one contribution is positive and the other is negative. Option A, x²+36, has the wrong sign and would correspond to adding the square parts rather than taking their difference. Option B is (x−6)², the square of a difference, not this rectangle. Option C is (x+6)², the square of a sum. A common confusion is to see two brackets and automatically use a square identity; first check whether the signs are opposite.