Algebraic identities

Correct answer: A. \((x-7)(x+7)\)

Use the identity \(a^2-b^2=(a-b)(a+b)\). Since \(49=7^2\), \(x^2-49=x^2-7^2=(x-7)(x+7)\). The close distractor \((x-7)^2\) expands to \(x^2-14x+49\), so it is not equal to the given expression. Exam tip: find the square root of the constant term before applying the difference-of-squares identity.

Correct answer: A. \((11+2y)(11-2y)\)

The expression is a difference of two squares: \(121=11^2\) and \(4y^2=(2y)^2\). Using \(a^2-b^2=(a+b)(a-b)\), we get \(121-4y^2=(11+2y)(11-2y)\). The option \((11-2y)^2\) would also produce a middle term \(-44y\), so it is not correct. Exam tip: identify the square roots of both terms before applying the difference-of-squares identity.

Correct answer: A. \(a^2+2ab+b^2\)

\(a^2+2ab+b^2=(a+b)^2\), so it is a perfect-square trinomial. In option B, the middle term is not \(2ab\). Exam tip: take square roots of the first and last terms, then verify the middle term.

Correct answer: A. First is (4p^2-12p+9) and second is ((2p+3)(2p-3))

One is a perfect square and the other is difference of squares. Exam tip: distinguish bracket square from difference of squares.

Correct answer: B. (2ab+2bc+2ca)

The direct answer is option B: \\(2ab+2bc+2ca\\). Expand the square by multiplying \\(a+b+c\\) by itself. Each term first multiplies every term in the second bracket: the square terms are \\(a^2,b^2,c^2\\). The unlike pairs occur twice: \\(ab+ba=2ab\\), \\(bc+cb=2bc\\), and \\(ca+ac=2ca\\). Therefore, \\(a+b+c\\)^2=a^2+b^2+c^2+2ab+2bc+2ca\\). Option B is correct because it contains the double product for every pair. Option A gives only one copy of each pair and therefore misses a factor of 2. Option C is wrong because \\(abc\\) is a product of three terms and belongs to cubic-type expressions, not this square expansion. Option D is wrong because the original sum itself is not the extra part produced by squaring. Remember: square of three terms means three squares plus twice each of the three pairwise products.

Correct answer: B. \((a+b)(a-b)=a^2-b^2\)

\((a+b)(a-b)=a^2-b^2\) is the difference-of-squares identity because the \(+ab\) and \(-ab\) terms cancel on multiplication. Option A is the identity for the square of a sum. Exam tip: look for binomials with opposite signs.

Correct answer: A. \(x^2+10x+25\)

Using \((a+b)^2=a^2+2ab+b^2\), put \(a=x\) and \(b=5\). The middle term is \(2\times x\times5=10x\), and the last term is \(25\). Hence A is a perfect-square trinomial. Exam tip: check whether the middle coefficient is twice the product of the terms.

Correct answer: A. \(a^2-2ab+b^2\)

Expanding \((a-b)(a-b)\) gives \(a^2-ab-ab+b^2=a^2-2ab+b^2\). Option B is the identity for the square of a sum. Exam tip: always check the sign of the middle term.

Correct answer: B. -6

\((x+6)(x-1)=x^2-x+6x-6=x^2+5x-6\). Hence, the term without \(x\), namely \(-6\), is the constant term. The number \(5\) is the coefficient of the middle term \(5x\), not the constant term. Exam tip: in a product of two binomials, multiply the constant terms to find the constant term.

Correct answer: A. (5)

To find the coefficient of \\(x\\), expand the product and collect the terms containing exactly one \\(x\\). In \\((x+6)(x-1)\\), multiplying each term gives \\(x\cdot x+x\cdot(-1)+6\cdot x+6\cdot(-1)\\). The two terms containing \\(x\\) are \\(-x\\) and \\(6x\\). Their sum is \\(5x\\), so the coefficient of \\(x\\) is 5.

Equivalently, for \\((x+r)(x+s)\\), the coefficient of \\(x\\) is \\(r+s\\). Here, \\(r=6\\) and \\(s=-1\\), so \\(r+s=6-1=5\\). Therefore option A is correct. The number 6 is only one constant from the first bracket, and \\(-6\\) is the constant term after multiplication, not the coefficient of \\(x\\).

Correct answer: A. ( (1000+2)^2 )

Since (1002=1000+2), the square of sum identity is useful. Exam tip: use a nearby base for large numbers.

Correct answer: C. \((a+b)^2=a^2+2ab+b^2\)

The correct identity is \((a+b)^2=a^2+2ab+b^2\). Its middle term is twice the product of the two terms, \(2ab\). Option A is for the square of a difference. In exams, always check the sign of the middle term.

Correct answer: A. \((1000-1)^2\)

Since \(999=1000-1\), the most convenient form is \((1000-1)^2\). It can be evaluated directly using \((a-b)^2=a^2-2ab+b^2\). The form \((1000+1)^2\) represents the square of 1001, not 999. Exam tip: use a nearby base such as 10, 100, or 1000 with an algebraic identity.

Correct answer: A. \(x^2+10x+25\)

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square trinomial. Option D is also a square but represents \((x-5)^2\); use the sign of the middle term as an exam check.

Correct answer: A. p² − q²

The difference-of-squares identity is p² − q² = (p + q)(p − q), so option A is correct. p² + q² does not factorise in this form. Exam tip: check for a minus sign between two squares.

Correct answer: B. (80x)

In ( (a+b)^2-(a-b)^2=4ab ), (a=5x) and (b=4), so (80x) is obtained. Exam tip: take the full (a).

Correct answer: A. \(a^2-b^2=(a-b)(a+b)\)

A difference of squares factorises as \(a^2-b^2=(a-b)(a+b)\); on expansion, the middle terms cancel. Option C misses the \(-2ab\) term. Exam tip: look for two squared terms joined by a minus sign.

Correct answer: A. 1

The governing concept is the constant term of a polynomial expression. A constant term contains no variable, so in (3u + 2v + 1)² it comes from squaring the constant part 1: 1² = 1. Other terms, such as 4v² and 6uv, contain variables and therefore are not constant terms. Hence option A is correct; option B is merely the unsquared coefficient and is not the required result.

Correct answer: B. (2xy)

( (x+y)^2=x^2+2xy+y^2 ), so (x^2) and (y^2) cancel leaving (2xy). Exam tip: expand and subtract.

Correct answer: B. \(-2xy\)

Using the identity \((x-y)^2=x^2-2xy+y^2\), the expression becomes \(x^2-2xy+y^2-x^2-y^2\). The \(x^2\) and \(y^2\) terms cancel, leaving \(-2xy\). The result \(2xy\) would arise from \((x+y)^2\), not \((x-y)^2\). Exam tip: the middle term in \((a-b)^2\) is always negative, \(-2ab\).

Correct answer: A. The first is a difference of squares, whereas the second gives a perfect-square trinomial

The first expression contains conjugate binomials:

(a+b)(a-b)

. Hence, using

(a+b)(a-b)=a^2-b^2

,

(2x+3)(2x-3)=4x^2-9

; there is no middle x-term. The second expands as

(2x-3)^2=4x^2-12x+9

, which is a perfect-square trinomial and has the middle term

-12x

. Therefore, option A is correct; option C reverses the two identities. Exam tip: recognise

(a+b)(a-b)

as a difference of squares and

(a-b)^2

as a perfect square.

Correct answer: A. \((m+n)(m-n)\)

In \((m+n)(m-n)\), the middle terms cancel: \(m^2-mn+mn-n^2=m^2-n^2\). Hence it is a difference of squares. Exam tip: look for binomials with opposite signs.

Correct answer: A. When an expression has to be expanded, simplified or factorised quickly

The direct answer is A: algebraic identities are most useful when an expression must be expanded, simplified, or factorised quickly. An identity is a rule that remains true for all allowed values of its variables. For example, \((a+b)^2=a^2+2ab+b^2\), \((a-b)^2=a^2-2ab+b^2\), and \(a^2-b^2=(a-b)(a+b)\). The method is simple: first compare the expression with a known identity, match the terms, and then apply the rule carefully. This saves time and reduces repeated multiplication. Option A is correct because these are exactly the main uses of identities. Option B, colouring a figure, is an artistic activity and does not require algebraic identities. Option C, saying there is no term, is not a meaningful use; identities work with algebraic terms and expressions. Option D is wrong because identities do not remove mathematical symbols; they rearrange or transform expressions while preserving their value. Memory cue: “recognise the pattern, apply the identity, check by expansion.”

Correct answer: A. 4x² + 12x + 9

Use the identity
(a+b)² = a² + 2ab + b²
. Here, a = 2x and b = 3. Therefore, (2x+3)² = (2x)² + 2(2x)(3) + 3² = 4x² + 12x + 9. In option B, the middle term 2ab has been calculated incorrectly; it should be 12x. Exam tip: In a squared binomial, the sign of the middle term follows the sign inside the binomial.

Correct answer: C. (x^2+5x-14)

The direct answer is option C: \(x^2+5x-14\). Expand by multiplying every term in the first bracket by every term in the second: \(x\cdot x=x^2\), \(x\cdot(-2)=-2x\), \(7\cdot x=7x\), and \(7\cdot(-2)=-14\). Combining like terms gives \(x^2-2x+7x-14=x^2+5x-14\). Option A has \(9x\), which would require the middle terms to add to 9, but they add to 5. Option B has \(-5x\), so the sign of the combined middle terms is wrong. Option C has the correct middle coefficient and negative constant. Option D has the wrong sign for the constant, because a positive 7 times a negative 2 is negative. Memory cue: use FOIL and remember the last product keeps the signs of both numbers.