Algebraic identities

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.

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.

\(a^2+2ab+b^2=(a+b)^2\), 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.

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

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.

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

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.

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.

\((x+6)(x-1)=x^2-x+6x-6=x^2+5x-6\). 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.

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\\).

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

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.

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.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), 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.

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.

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

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.

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.

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

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, \(-2ab\).

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.

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.

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.”

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.

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.