Algebraic identities

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.

Identities make long calculations shorter and systematic. Exam tip: identify the identity first, then solve.

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 coefficient of (x) is (7-2=5) and the constant term is (7\cdot(-2)=-14). Exam tip: watch the signs.

Use the identity \((x+y)(x-y)=x^2-y^2\), where \(x=5a\) and \(y=2b\). Therefore, \((5a+2b)(5a-2b)=(5a)^2-(2b)^2=25a^2-4b^2\). Option A incorrectly uses the sum of squares; the product of two binomials with opposite signs gives a difference of squares. Exam tip: whenever you see \((p+q)(p-q)\), apply \(p^2-q^2\) directly.

Since \(98=100-2\), writing \(98^2=(100-2)^2\) is the most convenient form. Using \((a-b)^2=a^2-2ab+b^2\) with \(a=100\) and \(b=2\) makes the calculation quick. Although \((90+8)^2\) and \((50+48)^2\) also equal \(98^2\), they do not use the advantage of the nearby round number 100. Exam tip: express a number in terms of a nearby multiple of 10 or 100 when finding its square.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Here, \(x^2+16x+64=x^2+2\times x\times 8+8^2\), so it equals \((x+8)^2\). If it were \((x-8)^2\), the middle term would be \(-16x\), not \(+16x\). Exam tip: match the middle term with \(2ab\) and the last term with \(b^2\).

Using the identity \((a-b)^2=a^2-2ab+b^2\), put \(a=p\) and \(b=9\). Then \((p-9)^2=p^2-18p+81\). Hence, the correct factor form is \((p-9)^2\). The expression \((p+9)^2\) has middle term \(+18p\), so it is not correct. Exam tip: for a perfect-square trinomial, halve the coefficient of the middle term and check whether its square equals the constant term.

Here, \(103=100+3\) and \(97=100-3\). Thus the product is of the form \((100+3)(100-3)\), so we use \((a+b)(a-b)=a^2-b^2\). Hence, \(103\times97=100^2-3^2=10000-9=9991\). The identity \((a-b)^2\) is for the square of one binomial, so it does not fit here. Exam tip: when two numbers are equally spaced from the same number, look for the difference-of-squares identity.