Visual models of identities

The correct answer is A: 2x+7. The large part has area 4x², so its side is 2x because (2x)²=4x². The corner has area 49, so its side is 7 because 7²=49. The two equal strips together have area 2×(2x×7)=28x, which agrees with the information in the question. Therefore the full side is 2x+7. Option B uses 4x as though it were the side of the first square, but its square would be 16x². Option C combines unlike quantities incorrectly, and option D would give only x² for the large part. The reliable check is (2x+7)²=4x²+28x+49.

The direct answer is B: 10x. The large square has side x+5, so its area is (x+5)²=x²+10x+25. The displayed square regions already account for x² and 25. The part left over is therefore 10x. Visually, the two equal rectangles have dimensions x by 5, so each has area 5x; together their area is 5x+5x=10x. Option A gives the area of only one rectangle, not both. Option C multiplies x by 25 and does not represent either rectangle’s dimensions. Option D is an addition of unlike quantities rather than an area calculation. The middle term in a square expansion comes from the two equal cross-rectangles. Memory cue: in (x+a)², the two strips together always contribute 2ax; here 2·x·5=10x.

The direct answer is C: 100. The area a² is 49, so the positive side length a is √49=7. Similarly, b²=9 gives b=√9=3. The whole square has side a+b=7+3=10, so its area is 10²=100. The identity also confirms this: (a+b)²=a²+2ab+b²=49+2(7)(3)+9=49+42+9=100. Option A, 64, is 49+9 and forgets the two rectangular parts 2ab. Option B, 80, does not follow from the given square sides. Option D, 112, uses an incorrect cross-term calculation. In a geometric model, the whole square contains two known smaller squares plus two rectangles, so simply adding 49 and 9 is incomplete. Memory cue: first take square roots of the square areas, add the side lengths, then square the sum.

The direct answer is A: 7. Compare the expression with the perfect-square identity (x+k)²=x²+2kx+k². Here the constant term 49 represents the area of the small square, so its side is √49=7. The middle term provides a check: 2·x·7=14x, exactly as required. Therefore the expression is (x+7)², and the small square has side 7. Option B, 14, is the coefficient of x, not the side of the small square. Option C, 49, is the small square’s area, not its side; a square with side 49 would have area 2401. Option D, 2, does not produce either the constant term 49 or the middle term 14x. A reliable method is to take the square root of the constant term and then verify it using half the middle-term coefficient. Memory cue: in x²+2kx+k², k is the small-square side.

The direct answer is A: the sides are p-q and p+q. The original square has area p², and the removed square has area q², so the remaining area is p²-q². Using the difference-of-squares identity, p²-q²=(p-q)(p+q). Thus the rearranged rectangle must have side lengths p-q and p+q. Option B gives the sides of the two original squares, but not the dimensions of the rearranged remaining region. Option C would give area (p+q)², which is too large because it includes more area than remains. Option D gives area (p-q)² and does not equal p²-q² in general. The visual rearrangement changes the shape but preserves area. Memory cue: difference of two square areas becomes a rectangle whose sides are the difference and the sum of the original side lengths.

The direct answer is B: x-4. Use the identity (x-a)²=x²-2ax+a². Since the constant term is 16=4², the small square has side 4. The negative middle term is -8x=-2·4·x, so the cross-rectangles are subtracted and the complete square is (x-4)²=x²-8x+16. Option A, (x+4)², would produce a positive middle term +8x. Option C, (x-8)², would produce -16x and a constant term 64. Option D, (x+8)², would produce +16x and 64. The sign of the middle term determines whether the model uses addition or subtraction. Memory cue: a negative middle term means the smaller side is subtracted; half of 8 gives 4, so the side is x-4.

The direct answer is B: 160. In the identity (a-b)²=a²-2ab+b², the two removed rectangular regions together have area 2ab. Here a=40 and b=2, so their total area is 2·40·2=160. The full calculation is 38²=(40-2)²=40²-2·40·2+2²=1600-160+4=1444. Option A, 80, is only one rectangle’s area, 40·2, not the total of two equal rectangles. Option C, 4, is the area of the small 2-by-2 square. Option D, 1520, incorrectly combines terms and does not represent the removed regions. In the visual model, the rectangles are subtracted from the large 40-by-40 square, so their areas are counted together as 2ab. Memory cue: in (a-b)², the two side rectangles always total 2ab and are subtracted.

The direct answer is A: x²+4x-12. The area of a rectangle is length multiplied by breadth, so A=(x+6)(x-2). Apply the distributive property term by term: x·x=x², x·(-2)=-2x, 6·x=6x, and 6·(-2)=-12. Combining like terms gives x²-2x+6x-12=x²+4x-12. Option B incorrectly makes the two middle products add to 8x; one product is -2x, so the correct combination is 4x. Option C reverses the signs of the middle and constant terms. Option D correctly has the middle coefficient but incorrectly treats 6·(-2) as +12. A signed area model may show pieces with different signs, but algebraic addition still requires preserving every sign. Memory cue: multiply every part of the first factor by every part of the second, then combine like terms.

Direct answer: Option A, (x + 7)(x − 3), is correct. In an area model, one side is divided into lengths x and 7, while the other side is divided into lengths x and −3. Multiplying every row length by every column length gives four signed rectangular parts: x·x = x², 7·x = 7x, x·(−3) = −3x, and 7·(−3) = −21. Their sum is x² + 7x − 3x − 21. Option B gives −7x and +3x instead, so its signs do not match. Option C makes both cross-products positive, giving +7x and +3x. Option D makes both cross-products negative, giving −7x and −21? More precisely, it gives −7x, −3x, and +21, so it also fails. Memory cue: the signs of the four parts come directly from multiplying the signs on the two sides.

Direct answer: Option A, 2x + 3, is correct. The area of a square is the square of its side. We look for a binomial whose square equals 4x² + 12x + 9. Since 4x² = (2x)² and 9 = 3², test (2x + 3)². Using (p + q)² = p² + 2pq + q², we get (2x + 3)² = (2x)² + 2(2x)(3) + 3² = 4x² + 12x + 9. Thus the whole square has side 2x + 3. Option B has square 16x² + 24x + 9, so its first term is wrong. Option C gives 4x² + 36x + 81, and option D gives 16x² + 72x + 81. A useful check is to compare both the first and last square terms before checking the middle term.

Direct answer: Option B, (3a − 4b)², is correct. Use the identity (p − q)² = p² − 2pq + q². Here, 9a² is (3a)² and 16b² is (4b)². The middle term must therefore be −2(3a)(4b) = −24ab, which agrees exactly with the expression. So 9a² − 24ab + 16b² = (3a − 4b)². Option A has a plus sign and would produce +24ab. Option C begins with (9a)² = 81a², not 9a². Option D gives a middle term −48ab and a final term 64b², so it also fails. Memory cue: when the two square terms are identified, the sign of the middle term tells whether the binomial uses plus or minus.

Direct answer: Option B, 2ab, is correct. Divide each side of the large square into lengths a, b, and c. The rectangular pieces formed by the pair a and b occur twice: one has dimensions a by b, and the other has dimensions b by a. Each has area ab because rectangle area is length multiplied by breadth. Their combined area is ab + ab = 2ab. Option A counts only one rectangle and misses its matching rectangle in the opposite position. Option C incorrectly suggests that three such rectangles occur; the third subdivision involving c produces ac or bc terms, not another ab term. Option D describes two square areas, not the two ab rectangles. Memory cue: in (a + b + c)², every pair of different terms appears twice: 2ab, 2ac, and 2bc.

Direct answer: Option A, 2·2x·5 = 20x, is correct. A square with side 2x + 5 can be divided into a square of side 2x, a square of side 5, and two equal rectangles. Each rectangle has dimensions 2x and 5, so each has area (2x)(5) = 10x. Since there are two such rectangles, their total, or middle, area is 10x + 10x = 20x. This is also the middle term in (p + q)² = p² + 2pq + q². Option B gives the area of only one rectangle. Option C incorrectly adds unlike quantities and does not represent an area term. Option D is the sum of the two square areas, 4x² and 25, not the middle rectangles. Memory cue: the middle term of a square is twice the product of the two side parts.

Direct answer: Option A is correct. Imagine a large square whose side is a + b. Divide both its length and width into parts a and b. The four regions are an a by a square with area a², a b by b square with area b², and two rectangles, each measuring a by b. Each rectangle has area ab, so together they contribute ab + ab = 2ab. Option B represents only the a² region. Option C represents only the b² region. Option D represents the entire large square, whose area is (a + b)², not just the middle part. A common confusion is to count one rectangle and write ab; the second rectangle is essential, which is why the coefficient is 2.

Direct answer: Option B, 9, is correct. The term 6x can be represented by two equal rectangles, each with area 3x. One side of each rectangle is x, so the other side is 3. When these pieces are arranged around the x² square, a small corner remains empty. That corner is a square with side 3, so its area is 3² = 9. Algebraically, x² + 6x + 9 = x² + 2(x)(3) + 3² = (x + 3)². Option A is the full coefficient of x, not the missing corner area. Option C doubles the required corner area. Option D squares 6 instead of half of 6. Memory cue: to complete x² + bx, take half of b and square it: (b/2)².

Direct answer: Option C, 25, is correct. To complete x² − 10x, compare it with (x − k)² = x² − 2kx + k². The middle coefficient satisfies 2k = 10, so k = 5. The missing corner in the area model is therefore a square of side 5, with area 5² = 25. Thus x² − 10x + 25 = (x − 5)². Option A is only half of the numerical coefficient and is a side length, not an area. Option B is the original coefficient and does not complete the square. Option D is 10², but the required value comes from squaring half the coefficient. The minus sign belongs to the side expression x − 5; the missing area itself is positive. Memory cue: halve the coefficient of x, ignore its sign while squaring, and square the result.

Direct answer: Option B, 2ab, is correct. A square of side a + b is partitioned into four regions: an a by a square with area a², a b by b square with area b², and two rectangles whose dimensions are a and b. The question shades only those two rectangles. Each shaded rectangle has area ab, so the total shaded area is ab + ab = 2ab. Option A counts only one rectangle. Option C is the total area of the two unshaded squares, not the shaded portion. Option D is the difference-of-squares expression and does not describe the sum of the two rectangles. As a check, subtract the unshaded areas from the whole square: (a + b)² − a² − b² = 2ab. Memory cue: the two cross-shaped rectangles in a square expansion always combine to 2ab.

Direct answer: Option A is correct. The difference-of-squares identity is x² − y² = (x + y)(x − y). Therefore, a rectangle with length x + y and breadth x − y has area (x + y)(x − y), exactly equal to x² − y². The condition x > y ensures that x − y is positive, so it can represent a physical breadth. Option B has area (x − y)², which is a square of the difference, not the difference of the squares. Option C has area x(x − y) = x² − xy. Option D has area (x + y)². None of these equals x² − y² in general. Memory cue: difference of squares factors into conjugate factors with the same middle term and opposite signs.

The direct answer is A. Begin with a square of side \(a+b\), whose area is \((a+b)^2\), and compare it with a square of side \(a-b\), whose area is \((a-b)^2\). When the smaller square is subtracted from the larger square, the two \(a^2\) parts cancel and the two \(b^2\) parts also cancel. What remains consists of four rectangular regions, each having side lengths \(a\) and \(b\), so each has area \(ab\). Their total area is therefore \(ab+ab+ab+ab=4ab\). Option B repeats an \(a^2\) region but does not represent the subtraction. Option C removes only one \(b^2\) square, so it cannot produce four \(ab\) rectangles. Option D concerns halving a length, not comparing areas. Memory cue: in a difference-of-squares picture, cancel matching square pieces and count the remaining cross-rectangles.

The direct answer is C, \(5x-3\). To identify a square side, take the square roots of the first and last terms: \(25x^2=(5x)^2\) and \(9=3^2\). The expression must therefore have the form \((5x\pm3)^2\). Check the middle term. Using \((u-v)^2=u^2-2uv+v^2\), \((5x-3)^2=25x^2-2(5x)(3)+9=25x^2-30x+9\), exactly as required. A gives a positive middle term, \(+30x\), so its sign is wrong. B does not square to the required leading term and is not the appropriate binomial side. D would produce a constant term 81 and middle term \(-90x\), both incorrect. Memory cue: take the roots of the end terms, then use the middle-term sign to choose plus or minus.

The direct answer is A: the rectangle has sides \(x+y\) and \(x-y\). The original large square has area \(x^2\), and removing the smaller square of area \(y^2\) leaves \(x^2-y^2\). The difference-of-squares identity says \(x^2-y^2=(x+y)(x-y)\), so the rearranged pieces must form a rectangle with those two side lengths. Option B has area \(y(x+y)\), which is not generally \(x^2-y^2\). Option C has area \(x(x-y)=x^2-xy\), missing the required \(-y^2\) term. Option D has area \((x-y)^2=x^2-2xy+y^2\), a different expression. The rearrangement changes the shape but not the total area. Memory cue: removing one square from another turns the difference of their side lengths into the two factors \(x+y\) and \(x-y\).

Direct answer: Option A, 7 × 17, is correct. The large square has area 12² and the removed square has area 5². Therefore, the remaining area is 12² − 5². The difference-of-squares identity says a² − b² = (a − b)(a + b). Substituting a = 12 and b = 5 gives (12 − 5)(12 + 5) = 7 × 17. Option A follows exactly. Option B multiplies the two given side lengths, but that does not represent the difference between their square areas. Option C uses 12 instead of the difference 12 − 5. Option D uses 5 instead of the required sum 12 + 5. A check confirms that 144 − 25 = 119 and 7 × 17 = 119. Memory cue: difference of squares means difference times sum.

Direct answer: Option A is correct. Divide both sides of the large square, each of length a+b, into parts of lengths a and b. This creates four regions: an a by a square with area a², a b by b square with area b², and two rectangles, each measuring a by b and having area ab. Adding the four regions gives a² + ab + ab + b² = a² + 2ab + b², which is the expansion of (a+b)². Option B has two copies of each square and does not match the four-part division. Option C contains only one ab rectangle, so it misses one ab term. Option D has no a² or b² square. The key warning is that the middle term 2ab comes from two separate equal rectangles.

Direct answer: Option A, 4a+5b, is correct. Recognise the expression as a perfect square. The first term is 16a²=(4a)², and the last term is 25b²=(5b)². For a square of side 4a+5b, the middle term must be 2×4a×5b=40ab, which is exactly the given middle term. Therefore 16a²+40ab+25b²=(4a+5b)², so the side of the modelled square is 4a+5b. Option B would produce a negative middle term, −40ab. Option C is not formed by taking the square roots of the first and last terms and would produce different powers and coefficients. Option D is twice the correct side, so its square would be four times the given area. A useful method is to take square roots of the two square terms and verify the middle term.

Direct answer: Option C, 36, is correct. Begin with the x by x square, whose area is x². Removing two strips of width 6 removes areas 6x and 6x. Their corner overlaps in a 6 by 6 square, so that corner has been removed twice and must be added back once. The added-back area is 6×6=36. This produces x²−6x−6x+36=x²−12x+36=(x−6)². Option A gives only the side of the small square, not its area. Option B is twice the side and corresponds to the coefficient in the middle term, not the corner area. Option D is the area of one variable rectangle, not the constant square. The common confusion is to use 6 or 12 instead of squaring the side 6.