Algebraic identities

The product of two binomials can be expanded by multiplying every term in the first bracket by every term in the second bracket. In \\((x+6)(x-2)\\), the products are \\(x\cdot x=x^2\\), \\(x\cdot(-2)=-2x\\), \\(6\cdot x=6x\\), and \\(6\cdot(-2)=-12\\). Combining the middle terms gives \\(-2x+6x=4x\\).

Therefore, the complete expansion is \\(x^2+4x-12\\), so option B is correct. A useful check is the pattern \\((x+r)(x+s)=x^2+(r+s)x+rs\\). Here, \\(r+s=6-2=4\\) and \\(rs=6(-2)=-12\\). Option A incorrectly adds 6 and 2, ignoring the negative sign.

The expression is a difference of two squares, so use \\(a^2-b^2=(a+b)(a-b)\\). Taking \\(a=104\\) and \\(b=96\\), we get \\(104^2-96^2=(104+96)(104-96)=200\\times8=1600\\). This is faster and safer than calculating both large squares separately.

Therefore, option C is correct. The key is to preserve the subtraction and use the sum and difference of the two numbers. The other values do not result from this product: \\(800\\) is only the difference multiplied by 100, \\(1200\\) is too small, and \\(2000\\) incorrectly uses the sum without the correct difference. The identity directly confirms 1600.

The expansion of \((a+b)^2\) contains the middle term \(+2ab\), twice the product of \(a\) and \(b\). In \((a-b)^2\), the middle term is \(-2ab\). Exam tip: the sign between the terms determines the sign of the middle term.

The direct answer is B: the term 6ab comes from the pair product 2 × a × 3b. For three terms, use the identity \\(p+q+r\\)^2=p^2+q^2+r^2+2pq+2qr+2rp\\). Here the three terms are a, 3b, and c. The pair a and 3b produces 2(a)(3b)=6ab. Thus the required mixed term is formed by doubling the product of those two terms. Option A, a², is only the square of the first term and has no b. Option B is correct because 2 × a × 3b simplifies to 6ab. Option C gives 2(3b)c=6bc, not 6ab, because it uses the second and third terms. Option D gives (3b)²=9b², which is a square term, not the mixed term involving a and b. The same expansion also contains a², 9b², c², 6ab, 6bc, and 2ac. Memory cue: in a square of several terms, every pair contributes twice its product. Decision: pass; the algebra and supplied option B are correct.

(2\times2x\times3=12x) gives the linear (x) term. Exam tip: identify the mixed term formed with the constant separately.

\(x^2-25=x^2-5^2\), so it is a difference of two squares and factorises as \((x-5)(x+5)\). Option D is a perfect-square trinomial. Exam tip: check for two squared terms separated by a minus sign.

The difference-of-squares identity is \(a^2-b^2=(a-b)(a+b)\), so option A is correct. Option B gives \((x-y)^2=x^2-2xy+y^2\), not \(x^2-y^2\). Exam tip: use conjugate factors for a difference of squares.

(87\times93=(90-3)(90+3)=90^2-3^2=8091). Exam tip: use (a^2-b^2) for numbers around the same centre.

The direct answer is option B: \(16x^2\). This is an opposite-sign pair, so use \((a-b)(a+b)=a^2-b^2\). Taking \(a=4x\) and \(b=3\), \((4x-3)(4x+3)=(4x)^2-3^2=16x^2-9\). The extra \(+9\) cancels the \(-9\): \(16x^2-9+9=16x^2\). Option A stops before adding the final 9, so it is incomplete. Option B includes the cancellation and is correct. Option C adds 9 instead of cancelling it. Option D changes a quadratic expression into an unrelated linear expression, so it cannot be right. Exam cue: whenever two brackets have the same first terms and opposite signs, check difference of squares first.

To find both requested values, first expand the product or use the pattern for two binomials. The constant term comes from multiplying the two constants, 2 and 7, giving 14. The coefficient of x comes from the two middle products: x multiplied by 7 and 2 multiplied by x. These give 7x and 2x, whose sum is 9x. Therefore, the constant term and coefficient of x are 14 and 9 respectively, so option A is correct.

In full form, \\( (x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14 \\). The number at the end is the constant term, while the number attached to x is its coefficient. Option B reverses these two values, and options C and D use individual numbers rather than the required product and sum. Keeping the sum for the middle coefficient and the product for the constant avoids this common error.

(121=11^2) and the middle term is (-22x=-2\times x\times11). Exam tip: identify ((a-b)^2) from the negative middle term.

Use the identity \((x+y)^2-(x-y)^2=4xy\). Here, \(x=7a\) and \(y=2b\), so the value is \(4\times 7a\times 2b=56ab\). \(49a^2-4b^2\) is the value of \((7a+2b)(7a-2b)\), not the difference of the two squares given here. Exam tip: first identify the pattern \((x+y)^2-(x-y)^2\).

Using the identity \((a+b)^2+(a-b)^2=2a^2+2b^2\), take \(a=2p\) and \(b=3q\). This gives \(2(2p)^2+2(3q)^2=8p^2+18q^2\). The mixed terms involving \(12pq\) cancel when the two squares are added, so option C is not correct. Exam tip: recognise this identity first to avoid lengthy expansion.

The expansion of \((a-b)^2\) is \(a^2-2ab+b^2\), so option A is correct. Option B represents the square of a sum. Exam tip: always check the sign of the middle term.

The direct answer is option A: 38809. Choose a nearby round number: 197 = 200 - 3. Use (a - b)^2 = a^2 - 2ab + b^2. Thus, 197^2 = (200 - 3)^2 = 200^2 - 2(200)(3) + 3^2 = 40000 - 1200 + 9 = 38809. Option A is correct because the calculation gives 38809. Option B, 38800, is close but leaves out the final +9 or uses an incomplete middle-term calculation. Option C, 39409, is too large and does not follow the subtraction-square identity; using 200 + 3 would give 203^2, not 197^2. Option D, 39809, is also incorrect and does not result from the correct three-term expansion. Another check is ordinary multiplication: 197 × 197 = 197 × (200 - 3) = 39400 - 591 = 38809. The important point is that a number just below a convenient round number can be squared quickly by subtracting twice the product and adding the small square. Memory cue: for (near number - small number)^2, square the near number, subtract the double product, then add the small square.

( (x+y)^2=x^2+2xy+y^2 ), so subtracting (x^2+y^2) leaves (2xy). Exam tip: identify the middle term.

( (x-y)^2=x^2-2xy+y^2 ), subtracting (x^2+y^2) gives (-2xy). Exam tip: watch the negative middle term.

In the square of three terms, the extra mixed terms are (2ab+2bc+2ca). Exam tip: remember all three pairs.

Using the distributive property, \((x+4)(x-9)=x^2-9x+4x-36=x^2-5x-36\). Therefore, option A is correct. In option B, the sign of the middle term is incorrect because \(-9x+4x=-5x\), not \(+5x\). Exam tip: multiply the first, outer, inner and last terms, then combine like terms.

((-8)+(-3)=-11), so the coefficient of (x) is (-11). Exam tip: the sum of two negative numbers remains negative.

Since \(144=12^2\), we have \(x^2-144=x^2-12^2\). Using \(a^2-b^2=(a+b)(a-b)\), its factor form is \((x+12)(x-12)\). The expansion of \((x-12)^2\) is \(x^2-24x+144\), so it is not correct. Exam tip: first check whether the constant term is a perfect square.

In \((a+b)(a-b)\), the middle terms \(-ab\) and \(+ab\) cancel, leaving \(a^2-b^2\). Options B and C are identities for squaring a binomial. Exam tip: recognise “sum × difference” as a difference of squares.

The direct answer is option A: the mixed terms are \(4ab\) and \(4ac\). Use \((p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr\), taking \(p=2a\), \(q=b\), and \(r=c\). The three mixed products are \(2(2a)(b)=4ab\), \(2(2a)(c)=4ac\), and \(2(b)(c)=2bc\). Since \(2bc\) is already mentioned, the other two are \(4ab\) and \(4ac\). Option A is therefore correct. Option B gives \(2ab\) and \(2ac\), but it forgets that the first term is \(2a\), not a. Option C lists square terms, not mixed terms, and even \((2a)^2\) equals \(4a^2\). Option D contains an unsuitable three-variable product and a non-mixed linear term. Memory cue: every pair of different terms contributes twice their product.

The expression can be simplified by using the difference of squares. Let the two squares be formed around the same middle term. Direct expansion gives \\(x+5\\)^2=x^2+10x+25 and \\(x-5\\)^2=x^2-10x+25. Subtracting the second expression from the first cancels both \\(x^2\\) and 25, leaving \\(20x\\). The condition therefore becomes \\(20x=80\\).

Dividing both sides by 20 gives \\(x=4\\). Hence option B is correct. The same result follows from \\(a^2-b^2=(a-b)(a+b)\\): here the difference is 10 and the sum is \\(2x\\), so the product is \\(20x\\). Options A, C and D do not satisfy the equation; substituting 4 gives exactly 80. The supplied answer B is therefore verified.

The identity for the square of a sum is \((a+b)^2=a^2+2ab+b^2\). In this question, \(a=x\) and \(b=3\). The middle term is produced by twice the product of the two parts, so it is \(2\times x\times3=6x\). The constant term is the square of 3, which is 9.

Therefore, expanding the left side gives \((x+3)^2=x^2+6x+9\). Comparing this with the given expression \(x^2+kx+9\), the coefficients of the corresponding \(x\)-terms must be equal. Hence \(k=6\), so option B is correct. The value 3 is only the added number, not the coefficient of the middle term. The value 9 is the constant term, and 12 would result from an incorrect multiplication rather than the identity.