Algebraic identities

\((a-b)^2=a^2-2ab+b^2\) because the middle term in this square identity is \(-2ab\). In \((a+b)^2\), the middle term is \(+2ab\). Exam tip: check the sign of the middle term first.

\(a^2-b^2\) is the difference of two squares, so it factorises as \((a+b)(a-b)\). Option C misses the middle term \(-2ab\). Exam tip: distinguish a difference of squares from the square of a difference.

The identity is \((a+b)^2=a^2+2ab+b^2\), so A is correct. The middle term \(2ab\) is positive for a sum. Option B is the expansion of \((a-b)^2\). In exams, always check the sign of the middle term.

When a sum is squared, the middle term is twice the product of the terms: \((a+b)^2=a^2+2ab+b^2\). Option B is the expansion of a difference. Exam tip: check the sign of the middle term first.

The identity for a square of a difference is \((a-b)^2=a^2-2ab+b^2\), so B is correct. Option A is the square of a sum. Exam tip: a minus sign between terms gives the middle term \(-2ab\).

The difference of squares identity is \(a^2-b^2=(a-b)(a+b)\), so B is correct. Options A and C are squares of binomials. Exam tip: two squared terms with opposite signs factor as \((a-b)(a+b)\).

Since \(63=60+3\) and \(57=60-3\), we can write \(63\times57=(60+3)(60-3)\). Using \((a+b)(a-b)=a^2-b^2\), the value is \(60^2-3^2=3600-9=3591\). Option A represents \(63\times63\), while option B represents \(57\times57\). Exam tip: use the difference-of-squares identity when two factors are equally spaced from the same middle number.

The difference-of-squares form is \(a^2-b^2\), which factorises as \((a-b)(a+b)\). Option B is a perfect-square trinomial. Exam tip: look for two square terms separated by a minus sign.

Using the identity \((a+b)^2=a^2+2ab+b^2\), put \(a=x\) and \(b=6\). Then \((x+6)^2=x^2+2\times x\times6+36=x^2+12x+36\). Therefore, the blank is \(12x\). \(36x\) is incorrect because 36 is the constant term obtained from \(6^2\). Exam tip: the middle term in the square of a binomial is always \(2ab\).

Use the identity \((a-b)^2=a^2-2ab+b^2\). Here, \(a=z\) and \(b=8\), so \((z-8)^2=z^2-2(z)(8)+8^2=z^2-16z+64\). Therefore, the blank is 64. The number 16 is related to the coefficient of the middle term, not the last term. Exam tip: in \((a-b)^2\), the last term is always \(b^2\), so it is positive.

In the plus square, the middle term is (2rs). Exam tip: do not write ( (r+s)^2 ) as only (r^2+s^2).

The square of a difference is \((a-b)^2=a^2-2ab+b^2\), so its middle term is negative. Option B is the identity for a square of a sum. Exam tip: check the sign of the middle term first.

Using a^2-2ab+b^2 with a=4x and b=1,

(4x-1)^2=16x^2-8x+1. The term without x is called the constant term, so it is 1. The term -8x is a linear term, not a constant term. Exam tip: in the square of a binomial, the constant term is the square of its constant part.

\(x^2-25=x^2-5^2\), so using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-5)(x+5)\). Option C is a perfect square instead. Exam tip: check for two squares separated by a minus sign.

Since \((a+b)^2=(a+b)(a+b)\), multiplication gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B represents \((a-b)^2\). Exam tip: always check the sign of the middle term.

The difference-of-squares identity is \(a^2-b^2=(a+b)(a-b)\). Therefore, \((a+b)(a-b)\) is the correct factor form. In contrast, \((a-b)^2=a^2-2ab+b^2\), so it is not equal to \(a^2-b^2\). Exam tip: whenever you see \(x^2-y^2\), factor it as \((x+y)(x-y)\).

( (x+9)(x-9)=x^2-81 ), so the middle term cancels. Exam tip: opposite sign terms add to zero.

\((a+b)^2=(a+b)(a+b)=a^2+ab+ab+b^2=a^2+2ab+b^2\), so A is correct. In \((a-b)^2\), the middle term is \(-2ab\). Exam tip: always check the sign of the middle term.

Expanding, \((x-2)(x-7)=x^2-7x-2x+14=x^2-9x+14\). Therefore, the term without \(x\), the constant term, is \(14\). Note that \(-9\) is the coefficient of the middle term, not the constant term. Exam tip: the product of two negative numbers is positive.

\((x-4)(x+6)=x^2+6x-4x-24=x^2+2x-24\). Therefore, the coefficient of \(x\) is \(2\). The number \(-24\) is the constant term, not the coefficient of \(x\). Exam tip: simplify the expression by combining like terms before identifying a coefficient.