Visual models of identities

Direct answer: Option A is correct. The first three terms form a perfect square: x²+2xy+y²=(x+y)². Therefore the complete expression is (x+y)²−z². This is a difference of two squares, represented visually by starting with a square of side x+y and removing a smaller square of side z. Applying u²−v²=(u−v)(u+v), with u=x+y and v=z, gives (x+y−z)(x+y+z). Option B begins with (x−y)², which has a negative 2xy term. Option C contains too few regions to represent x²+2xy+y²−z². Option D makes a larger square and would introduce additional positive terms instead of subtracting z². The reliable sequence is: identify the perfect square, represent subtraction by removing a square, then use the difference-of-squares identity.

Direct answer: Option A is correct. Divide each side of the large square, whose length is p+q, into lengths p and q. The four resulting regions are a p by p square with area p², a q by q square with area q², and two p by q rectangles, each with area pq. Their total is p²+pq+pq+q²=p²+2pq+q²=(p+q)². Option B has the wrong number of square regions. Option C incorrectly calls pq a square; its two side lengths need not be equal, and the remaining shapes do not have the stated dimensions. Option D assigns q² to rectangles that should have area pq. The central check is to count the two cross-rectangles separately because together they create the 2pq term.

Direct answer: Option A is correct. Start with the a by a square, whose area is a². Removing a strip of width b along one side removes area ab, and removing an equal strip along the adjacent side removes another ab. The corner where the strips meet is b by b, so it was counted as removed twice. Adding it back once gives a²−ab−ab+b²=a²−2ab+b². The remaining figure has side length a−b, so its area is (a−b)². Option B describes adding strips rather than removing them. Option C is the difference of two complete squares and uses a different model. Option D forgets the overlap correction and is therefore missing the +b² term. Memory cue: for a square of a difference, subtract two cross-products and add back the corner square.

Direct answer: Option A is correct. The area of a rectangle equals length times width, so multiply each pair and compare with 2x²+7x+3. For A, (2x+1)(x+3)=2x²+6x+x+3=2x²+7x+3, exactly the required area. Option B gives (2x+3)(x+1)=2x²+2x+3x+3=2x²+5x+3, so its middle coefficient is wrong. Option C gives x²+4x+3, whose leading coefficient is wrong. Option D gives (2x+7)(x+3)=2x²+6x+7x+21=2x²+13x+21, so both the middle and constant terms are wrong. The input key marked B, but direct expansion proves that B is not correct; the corrected answer is A. A useful checking method is to multiply every option rather than comparing only one term.

Answer: B, \((a-b)^2\). Begin with the square of side \(a\), whose area is \(a^2\). Each removed strip has area \(ab\), so removing both seems to subtract \(2ab\). However, the two strips overlap in a square of side \(b\), whose area is \(b^2\). That overlap was subtracted twice, so it must be added once. Therefore the remaining area is \(a^2-2ab+b^2\). By the identity \((a-b)^2=a^2-2ab+b^2\), the model represents \((a-b)^2\). A is the difference of two squares, not this three-part removal. C has a positive middle term and represents addition. D omits the two cross terms. Memory cue: subtracting two equal strips creates a negative \(2ab\) term, while the overlap returns \(+b^2\).

Answer: B, \(16x\). The larger square has area \((x+4)^2\), and the smaller square has area \((x-4)^2\). The required difference is therefore \((x+4)^2-(x-4)^2\). Use the identity \((p+q)^2-(p-q)^2=4pq\), with \(p=x\) and \(q=4\): the difference is \(4\cdot x\cdot4=16x\). Direct expansion confirms it: \(x^2+8x+16-(x^2-8x+16)=16x\). A is only half the result. C is the area of a square with side expression involving 4, not the requested difference. D is a constant and ignores \(x\). Memory cue: the difference between the squares of \(x+r\) and \(x-r\) is \(4xr\).

Answer: B, \(9-x\). First rearrange the terms into descending powers of \(x\): \(81-18x+x^2=x^2-18x+81\). Compare this with \((u-v)^2=u^2-2uv+v^2\). Here \(u=x\) and \(v=9\), because \(2\cdot x\cdot9=18x\) and \(9^2=81\). Hence the expression is \((x-9)^2\). A square has the same value when its side is negated, so \((x-9)^2=(9-x)^2\); the listed suitable side is \(9-x\). A would produce a positive middle term, \(x^2+18x+81\). C and D do not square to the given trinomial. Memory cue: for a negative middle term, use the difference pattern and take half of 18 for the constant side.

Answer: B, 28. Write the first and last terms as squares: \(4x^2=(2x)^2\) and \(49=7^2\). A perfect square with a positive middle term must therefore be \((2x+7)^2\). Apply \((p+q)^2=p^2+2pq+q^2\): the middle term is \(2(2x)(7)=28x\). Thus the coefficient of \(x\) is \(k=28\). A, 14, omits the factor 2 in the cross term. C doubles the correct result and D incorrectly multiplies the constant terms. The positive sign matters: \((2x-7)^2\) would give \(-28x\), not a positive middle term. Memory cue: take the square roots of the end terms, then multiply them and double the result.

Answer: C, \(12ab\). For a square with side \(2a+3b\), split each side into lengths \(2a\) and \(3b\). The two cross rectangles have dimensions \(2a\) by \(3b\). Each rectangle therefore has area \((2a)(3b)=6ab\). Since there are two identical rectangles, their total area is \(6ab+6ab=12ab\). Equivalently, the cross term in \((2a+3b)^2\\) is \(2(2a)(3b)=12ab\). A adds the coefficients without multiplying, and B gives the area of only one rectangle. D ignores both the coefficients and the fact that there are two rectangles. Memory cue: the cross area is always twice the product of the two split lengths.

Direct answer: A. Begin with the large a-by-a square, whose area is a^2. To create the smaller square of side a-b, one strip of width b is removed from the top and another strip of width b from the side. Each strip has area ab. However, the two strips overlap at one corner in a b-by-b square. That corner has area b^2 and is included in both removed strips, so it has been subtracted twice. The desired removal should count it only once. Adding b^2 back once corrects the double subtraction: a^2-ab-ab+b^2=a^2-2ab+b^2=(a-b)^2. Option A states this exact overlap reason. Option B is false because b^2 is generally positive. Option C is false because ab and b^2 are not generally equal. Option D reverses the construction; the original a^2 is the starting area, not an extra amount to subtract. Common warning: “add back” does not mean adding a new region; it corrects an overlap error.

Direct answer: B. A square of side a has area a^2, and a smaller square of side b has area b^2. Removing the smaller square leaves the region a^2-b^2. If the remaining pieces are cut and rearranged without changing their total area, they form a rectangle whose two side lengths are a+b and a-b. Its area is therefore (a+b)(a-b), proving the identity a^2-b^2=(a+b)(a-b). Option A concerns partitioning a different complete square and does not specifically create the difference-of-squares remainder. Option B describes both the subtraction and the rearrangement needed. Option C produces the addition pattern associated with a binomial square. Option D only joins line segments; line lengths alone do not establish an area identity. A reliable memory cue is: remove the small square first, then rearrange the leftover pieces into a rectangle; the rectangle’s sides are the sum and difference of the original side lengths.

Direct answer: C. Multiply the two binomials term by term: x·x=x^2, x·(-2y)=-2xy, 2y·x=2xy, and 2y·(-2y)=-4y^2. In the rectangle model, the two middle rectangles therefore have signed areas -2xy and +2xy. Their magnitudes are equal but their signs are opposite, so their combined area is -2xy+2xy=0. Only x^2-4y^2 remains, which is the difference-of-squares identity. Option A incorrectly adds magnitudes and ignores the opposite signs. Option B has the wrong coefficient and power; the y-by-y corner is -4y^2, not 2y^2. Option C correctly identifies the two cancelling middle rectangles. Option D is impossible because x^2 is a separate corner square and is not cancelled. In signed area models, opposite-oriented or negative tiles cancel equal positive tiles. Memory cue: conjugates (u+v)(u-v) have equal and opposite middle terms.

The correct answer is B. The area of a square is the square of its side, so the required expression is (5x+2y)^2. Use (p+q)^2=p^2+2pq+q^2 with p=5x and q=2y. The large square contributes (5x)^2=25x^2. The two equal cross-rectangles together contribute 2(5x)(2y)=20xy. The small corner square contributes (2y)^2=4y^2. Adding all regions gives 25x^2+20xy+4y^2. Option A includes only one cross-product, 10xy, and forgets the second rectangle. Option C does not square the terms correctly. Option D assigns the wrong coefficients to the mixed and y^2 terms. A useful memory cue is that the middle term in a square always has the factor 2: (p+q)^2=p^2+2pq+q^2.

The correct answer is A. In a rectangle model, the four regions come from multiplying two binomials. The x^2 region comes from x times x. The -4x region comes from x times -4, while the 3x region comes from 3 times x. Finally, 3 times -4 gives -12. Therefore the side labels are (x-4) and (x+3), so the multiplication is (x-4)(x+3). Expanding confirms it: x^2+3x-4x-12=x^2-x-12. Option B would give positive 4x and positive 12, not the displayed signs. Option C would give -3x rather than +3x and positive 12. Option D does not produce the four given products. Always track both the numbers and their signs in signed area tiles.

The correct answer is A. Let the side of the larger square be a and the side of the smaller square be b, with a>b. Removing the b by b square from the a by a square leaves an L-shaped region of area a^2-b^2. By cutting one part of this L-shape and shifting it, the pieces can be rearranged into a rectangle. Its two side lengths are a+b and a-b, so its area is (a+b)(a-b), which equals a^2-b^2. Option B would have area (a+b)^2, which is too large and represents a different identity. Option C does not preserve the required rectangular area relationship. Option D describes a special kind of parallelogram, not the standard rearrangement. Memory cue: difference of squares becomes a rectangle with sum on one side and difference on the other.

The direct answer is B: 11x. In a rectangle model, split the first side into 2x and 1, and the second side into 3x and 4. This creates four smaller regions: (2x)(3x)=6x², (2x)(4)=8x, (1)(3x)=3x, and (1)(4)=4. The x-terms are the two regions containing exactly one factor x, namely 8x and 3x. Adding like terms gives 8x+3x=11x. Option A, 8x, counts only one cross-region and misses 3x. Option B is correct because it combines both linear regions. Option C, 6x, confuses 6x² with an x-term and also loses one x. Option D, 12x, does not come from any correct pair of products. Thus the complete expansion is 6x²+11x+4. Memory cue: the middle term in a binomial product comes from the two cross-multiplications.

The direct answer is B: 20x. A square with side 2x+5 is divided by separating each side into 2x and 5. The corner squares have areas (2x)(2x)=4x² and 5×5=25. The two remaining rectangles each have dimensions 2x and 5, so each has area (2x)(5)=10x. Since there are two such rectangles, their combined area is 10x+10x=20x. Option A gives the area of only one rectangle, not both. Option B correctly counts both middle rectangles. Option C, 25x, incorrectly combines the constant 5 with x rather than multiplying the actual side lengths. Option D doubles the correct total. The full identity is (2x+5)²=4x²+20x+25. Memory cue: in (a+b)², the two middle rectangles together produce 2ab.

The direct answer is A. To model (x−6)², begin with a square of side x and remove a strip of width 6 along one direction and another strip of width 6 along the other direction. Each removed rectangle has area x×6=6x, so together they contribute −6x−6x=−12x. The small corner where the strips overlap has area 6×6=36 and is added back, giving x²−12x+36. Option A correctly identifies both subtracted rectangles. Option B gives only the positive constant term 36, not the negative linear term. Option C represents the original x² square. Option D is incomplete because a side expression alone is not an area and does not show the two cross-regions. The identity is (x−6)²=x²−2(6)x+6². Memory cue: the negative middle term comes from two removed strips, while the corner square is positive because it is restored.

The direct answer is A. Draw a large square whose side is a+b. Divide both its length and width into parts a and b. The four regions are an a² square, a b² square, and two rectangles, each with area ab. Therefore the total area is a²+ab+ab+b²=a²+2ab+b². This shows visually why simply adding a² and b² leaves out the two cross-regions. Option A includes every region of the large square and is correct. Option B concerns a difference of lengths and does not model the square of a sum. Option C changes the geometric figure and gives no explanation of the identity. Option D hides the internal parts, so it cannot reveal the missing middle term. Common confusion: (a+b)² is not a²+b²; squaring a sum creates two cross-products.

The direct answer is A. The large square has side a+b. Divide one side into lengths a and b, and do the same to the adjacent side. The partition produces four regions: the a by a corner is a², the b by b corner is b², and the two remaining a by b regions are rectangles with area ab each. Adding them gives a²+ab+ab+b²=a²+2ab+b². Option A lists exactly these four regions. Option B misses one ab rectangle, so it would give the wrong middle term. Option C invents four square regions; the cross-regions are rectangles, not generally squares. Option D has the wrong number of rectangles and therefore cannot represent the four-part division. Memory cue: a square of a sum has two corner squares and two equal cross-rectangles.

The direct answer is C: a square of area 81. Split the 18x term into two equal rectangles, each with area 9x. Each rectangle can have side lengths x and 9. To complete the large square, the missing corner must then be a 9 by 9 square, whose area is 9²=81. Algebraically, x²+18x+81=(x+9)². Option A uses 18 directly and does not create two 9x rectangles. Option B squares 6, but half of 18 is 9, not 6. Option C correctly uses half the coefficient of x and squares it. Option D squares 18, which is much too large and corresponds to the common mistake of forgetting the half-step. Memory cue: for x²+bx, add (b/2)²; here (18/2)²=9²=81.

The direct answer is B: 9x²−4. This product has the form (A+B)(A−B), where A=3x and B=2. In the visual model, the positive and negative cross-regions have equal area 6x but opposite signs, so they cancel. The remaining areas are (3x)(3x)=9x² and the signed corner 2×(−2)=−4. Therefore the total is 9x²−4. Option A incorrectly adds 4 instead of using the negative corner. Option B correctly applies the difference-of-squares identity A²−B². Option C uses 6x², although (3x)² is 9x². Option D retains a middle term even though the two cross-terms cancel. Memory cue: conjugate factors with the same first part and opposite second parts produce a difference of squares.

The direct answer is B: −10. Split the first dimension into x and 5, and the second into x and −2. The constant corner is formed by multiplying the two constant pieces, 5 and −2. Its signed area is 5×(−2)=−10. Expanding confirms this: (x+5)(x−2)=x²−2x+5x−10=x²+3x−10, so the constant term is exactly −10. Option A, 10, ignores the negative sign in −2. Option B correctly preserves the signs of both constant pieces. Option C is the sum of the two constants, not their product. Option D is their difference and also does not represent the corner product. In signed area models, a negative piece is treated algebraically as a subtraction. Memory cue: the constant term always comes from constant × constant, with the signs included.

The direct answer is A. Compare the expression with the identity (p−q)²=p²−2pq+q². Here p=2a and q=3b. Then p²=(2a)²=4a², q²=(3b)²=9b², and −2pq=−2(2a)(3b)=−12ab. Therefore 4a²−12ab+9b²=(2a−3b)², which is represented by a square whose side is 2a−3b. Option A matches all three terms. Option B would produce 16a²−72ab+81b², so its coefficients are wrong. Option C represents (2a+3b)(2a−3b)=4a²−9b² and has no middle term. Option D produces a²−6ab+9b². Memory cue: check the first and last terms as squares, then verify the middle term is minus twice their roots multiplied.

The direct answer is B: a²+ab+b². The correct area model for a square of side a+b contains four parts: one a² square, two ab rectangles, and one b² square. Its correct total is a²+2ab+b². If the two equal ab rectangles are mistakenly counted only once, one ab contribution is omitted. The written area then becomes a²+ab+b². Option A is the correct complete area, but it is not the result of the stated mistake. Option B accurately represents the undercount. Option C is the difference-of-squares form and does not describe this square. Option D leaves out the b² corner as well, so it is an even larger error. Memory cue: the coefficient 2 in 2ab records the two congruent cross-rectangles; forgetting one changes 2ab to ab.