Algebraic identities

Using the identity \((a-b)^2=a^2-2ab+b^2\), we get \((x-9)^2=x^2-18x+81\). A constant term has no variable, so the constant term is 81. The term \(-18x\) contains x and is therefore not constant. Exam tip: in the square of a binomial, the final constant term is the square of the constant part.

\(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\), so it is a perfect square of a binomial. In option B, the constant should be 9, not 8. Exam tip: match the middle term with \(2ab\).

Using \((a+b)^2=a^2+2ab+b^2\), with \(a=50\) and \(b=1\), we get \((50+1)^2=50^2+2\times50\times1+1^2=2500+100+1=2601\). Option 2600 is close, but it misses the \(1^2\) term. Exam tip: while expanding a square, include both the middle term \(2ab\) and the final term \(b^2\).

Since \(48=50-2\), it is convenient to write \(48^2\) as \((50-2)^2\). Using \((a-b)^2=a^2-2ab+b^2\), we get \((50-2)^2=50^2-2\times50\times2+2^2\). In contrast, \((48+2)(48-2)\) equals \(48^2-2^2\), not \(48^2\) directly. Exam tip: for quick squaring, express a number near 10, 50, or 100.

The difference-of-squares identity is \(a^2-b^2=(a-b)(a+b)\). On multiplying, the middle terms cancel, leaving \(a^2-b^2\). In contrast, \((a-b)^2\) contains a \(-2ab\) term. Exam tip: first check whether both terms are perfect squares.

Use the difference-of-squares identity \\( (u+v)^2-(u-v)^2=4uv\\). In this expression, \\(u=x\\) and \\(v=5\\). Hence the value is \\(4(x)(5)=20x\\). Direct expansion gives \\(x^2+10x+25-(x^2-10x+25)\\); the equal square and constant terms cancel, leaving \\(20x\\). Both methods confirm the same result.

Therefore option C is correct. Option A, \\(10x\\), keeps only one of the two linear contributions, while option B incorrectly treats the expression as a product involving 25x. Option D ignores the variable terms and is not obtained by the identity. Recognising the plus-minus square pattern makes the calculation quick and reduces sign errors.

Adding both expansions cancels middle terms and gives (2a^2+2b^2). Exam tip: add like terms.

In the identity \((a+b)^2=a^2+2ab+b^2\), the last term is \(b^2\). Here, \(a=2x\) and \(b=3y\), so the last term is \((3y)^2=9y^2\). \(12xy\) is the middle term because \(2(2x)(3y)=12xy\). Exam tip: in a square expansion, the first and last terms are the squares of the two individual terms.

\(x^2-9=x^2-3^2\), so it matches \(a^2-b^2\) and factorises as \((x+3)(x-3)\). \(x^2+9\) is a sum of squares, not a difference. Exam tip: identify perfect squares first.

Using the distributive property, \((m+4)(m+6)=m^2+6m+4m+24=m^2+10m+24\). Hence, option A is correct. In option B, the middle-term coefficient and constant term are interchanged, while option C incorrectly adds \(6m+4m\). Exam tip: in \((m+a)(m+b)\), the middle term is \((a+b)m\) and the constant term is \(ab\).

\((x-2)(x-5)=x^2-7x+10\). Hence, the term without \(x\), i.e. the constant term, is \(10\). Here, \(-7\) is the coefficient of \(x\), not the constant term. Exam tip: in the product of two binomials, multiply the constant terms to find the constant term.

Using the distributive property, \((x-2)(x-5)=x^2-5x-2x+10=x^2-7x+10\). Therefore, the coefficient of \(x\) is \(-7\). The number \(10\) is the constant term, not the coefficient of \(x\). Exam tip: in \((x-a)(x-b)\), the coefficient of \(x\) is \(-(a+b)\).

This expression uses the identity \((a+b)(a-b)=a^2-b^2\). Taking \(a=z\) and \(b=9\), \((z+9)(z-9)=z^2-9^2=z^2-81\). \(z^2+81\) is incorrect because this is a difference of squares, not a sum. Exam tip: for binomials with opposite signs, the middle terms cancel.

In ( (a+b+c)^2 ), (2ab), (2bc), and (2ca) appear. Exam tip: remember the three pairwise product terms.

A square term is produced when a quantity is multiplied by itself. In the expression \\( (x+y+z)^2 \\), the individual quantities are squared, giving \\(x^2\\), \\(y^2\\), and \\(z^2\\). These are called the square terms. When the expression is expanded, it also contains the mixed products \\(2xy\\), \\(2yz\\), and \\(2zx\\), but those are not square terms because they involve two different variables.

The identity is \\( (x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx\\). Thus option B lists exactly the three terms in which each variable is squared. Option A lists unsquared products, option C lists doubled cross-products, and option D is the original sum rather than a list of terms. The distinction is based on the exponent and the variables present.

The square of three terms has three square terms and three double product terms. Exam tip: write all terms with (2).

The square of three terms contains the square of each term and twice the product of every distinct pair. For the terms in \\(x+2+y\\), the pair consisting of \\(x\\) and \\(y\\) contributes \\(2xy\\). Other pairwise products also occur, such as \\(4x\\) and \\(4y\\), along with \\(x^2\\), \\(4\\), and \\(y^2\\).

Thus, option B is correct. The term \\(2xy\\) does not appear because x and y are square terms; it appears because their product is counted twice when the whole sum is multiplied by itself. The number 2 does not disappear, and the expansion definitely contains product terms. Pairing the three terms prevents omissions.

Using the identity \((a+b)^2=a^2+2ab+b^2\), the given expression is \((a+b)^2\); hence option C is correct. In \((a-b)^2\), the middle term is \(-2ab\), so it cannot match. Exam tip: use the sign of the middle term to distinguish between the two square identities.

On expanding \((a+b)^2\), the terms obtained are \(a^2\), \(2ab\), and \(b^2\), so option A is correct. Option B is the identity for the square of a difference. Exam tip: the middle term is positive in the square of a sum.

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