Algebraic identities

The direct answer is option D: \(9a^2-16b^2\). Start with the identity \((x+y)(x-y)=x^2-y^2\). Here, the common first term is \(x=3a\), and the second term is \(y=4b\). Therefore, \((3a+4b)(3a-4b)=(3a)^2-(4b)^2=9a^2-16b^2\). Option A has a plus sign, but opposite signs in the brackets produce a difference of squares, not a sum. Option B incorrectly changes the coefficients and powers. Option C is the expansion of a form such as \((3a-4b)^2\), so it includes a middle term that cancels here. Option D correctly gives the difference of the two squares. Memory cue: same terms with opposite signs mean square the first term minus square the second term.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Here, \(x^2+12x+36=x^2+2\cdot x\cdot6+6^2\), so it equals \((x+6)^2\). Note that \((x-6)^2=x^2-12x+36\), so it is not correct. Exam tip: use the sign of the middle term to decide whether the binomial has \(+\) or \(-\).

Here, \(96=100-4\) and \(104=100+4\). Therefore, \((100-4)(100+4)=100^2-4^2=10000-16=9984\). The option \(10016\) results from adding \(4^2\) instead of subtracting it. Exam tip: when two numbers are equally spaced around a middle number, use \((a-b)(a+b)=a^2-b^2\).

Using \((a-b)^2=a^2-2ab+b^2\), put \(a=p\) and \(b=q\) to get \(p^2-2pq+q^2\). Option C is the difference of squares. Exam tip: the middle term must be \(-2pq\).

The governing concept is the cube identity followed by extraction of a common factor. Expand the binomial: (a+b)^3=a^3+3a^2b+3ab^2+b^3. Now subtract a^3+b^3. The pure cube terms cancel, leaving 3a^2b+3ab^2. Both remaining terms contain 3ab, so factor it out: 3a^2b+3ab^2=3ab(a+b). Thus option A is correct. Option B has an incorrect minus sign; expanding 3ab(a-b) gives 3a^2b-3ab^2, whereas both remaining terms here are positive. Option C omits the coefficient and variables produced by the expansion. Option D is also wrong because only the pure cubes cancel; the two middle terms remain.

Correct answer: A, 20z. Use the identity (a+b)² − (a−b)² = 4ab. Here a = x+y and b = z. Therefore, (x+y+z)² − (x+y−z)² = 4(x+y)z. Since x+y = 5, substitute 5: 4 × 5 × z = 20z. A is correct. B is wrong because it doubles the coefficient and would require 8(x+y)z, which is not the identity. C is wrong because the two squares are not generally equal; their signs differ in the middle term. D is wrong because the z² terms cancel when the squares are subtracted. A quick expansion confirms this: both expressions contain (x+y)² and z², but one has +2(x+y)z and the other −2(x+y)z; subtracting gives 4(x+y)z. Memory cue: difference of these paired squares gives 4ab.

The direct answer is C. The large square has area p² and the removed small square has area q². Thus the remaining area is p² − q². The difference-of-squares identity says p² − q² = (p−q)(p+q), so option C is the required factor form. Option A adds the two square areas instead of subtracting the removed area. Option B is the square of the sum and would represent (p+q)², not the remaining difference. Option C correctly factors the difference into the product of the difference and the sum. Option D, 2pq, is the middle term associated with (p+q)² and is not the leftover area. Memory cue: for U²−V², write (U−V)(U+V).

The correct answer is A: the rectangle has sides \((x+4)\) and \((x+5)\). Multiply them carefully: \((x+4)(x+5)=x^2+5x+4x+20=x^2+9x+20\). In a tile model, the \(x^2\) tile is the large corner square, the \(5x\) and \(4x\) tiles form the two linear strips, and the unit tiles make the corner area 20. Thus the numbers must have both sum 9 and product 20; 4+5=9 and 4×5=20. B has product 20 but sum 12, so it gives \(x^2+12x+20\). C gives \(x^2+29x+180\), and D gives \(x^2+21x+20\), so neither matches. A complete rectangle requires matching every tile, not merely the constant term. Memory cue: for \(x^2+px+q\), choose numbers adding to p and multiplying to q.

The direct answer is A: $x^2+10x+25=(x+5)^2$. To recognize a perfect-square trinomial, identify the first and last terms as squares: $x^2=x^2$ and $25=5^2$. The middle term must equal $2ab$, where $a=x$ and $b=5$. Indeed, $2×x×5=10x$, so the identity gives $x^2+10x+25=(x+5)^2$. Option B has the same first two terms, but 24 is not 25, so it is not a perfect square of this form. In C, the square roots are x and 5, but $2×x×5=10x$, not 8x. In D, although $x^2$ and 24 do not form the needed square pair, the negative middle term would be associated with $(x-5)^2$ only if the constant were 25; it is not. The safe check is to compare the constant with $b^2$ and the middle term with $2ab$.

Direct answer: A, \(x^2+10x+25=(x+5)^2\). The basic identity is \((x+a)^2=x^2+2ax+a^2\). The constant term 25 has square root 5, so the middle term required for a positive square is \(2\cdot x\cdot5=10x\). Both conditions are satisfied, and therefore the expression is \((x+5)^2\). Option A is correct. Option B has the same first two terms but its constant is 20 rather than 25, so it cannot be \((x+5)^2\). Option C is a difference of squares, not a trinomial, and it factors as \((x-5)(x+5)\). Option D is a sum of two squares and has no middle term, so it does not match the identity. Memory cue: take the square root of the constant and double it to predict the coefficient of \(x\).

Direct answer: A, \(x^2+10x+25=(x+5)^2\). Compare the expression with \(u^2+2uv+v^2\). Here the first square is \(x^2\, so u=x\), and the last term 25 is \(5^2\), so v=5. The required middle term is then \(2uv=2(x)(5)=10x\), exactly as given. Thus A is a perfect-square trinomial. Option B fails because with first and last terms x² and 25, the middle term must be 10x, but its constant is 20. Option C is also a perfect square, \((x-5)^2\), but it has a negative middle term and therefore does not match the specifically stated plus identity. Option D has only two terms, not a trinomial. Memory cue: square the first and last roots, then check whether the middle term is twice their product.