Algebraic identities

\(x^2-25=x^2-5^2\), so it has the form \(a^2-b^2\). In contrast, \(x^2+25\) is a sum of squares. Exam tip: check whether each term is a perfect square.

\(x^2-9=x^2-3^2\), so it is a difference of two perfect squares and factors as \((x-3)(x+3)\). \(x^2+9\) is a sum of squares. Exam tip: first check whether both terms are perfect squares.

\(x^2-y^2\) is the difference of two perfect squares and factors as \((x-y)(x+y)\). Options B and D are perfect-square trinomials. In exams, check for two square terms separated by a minus sign.

The first is a perfect square and the second is difference of squares. Exam tip: keep bracket square and difference of squares separate.

This expression has the form \((p+q)(p-q)=p^2-q^2\). Here, \(p=2a\) and \(q=3\), so \((2a+3)(2a-3)=(2a)^2-3^2=4a^2-9\). Option A incorrectly adds the two squares. Exam tip: for conjugate binomials, the middle terms cancel, leaving the difference of the squares.

In \((a+b)(a-b)\), the middle terms \(+ab\) and \(-ab\) cancel, leaving \(a^2-b^2\). Option C also contains the term \(-2ab\). Exam tip: conjugate binomials give the difference of two squares.

The identity is \((a+b)^2=a^2+2ab+b^2\). Its middle term is twice the product of the two terms. Option B represents \((a-b)^2\). Exam tip: always check the sign of the middle term.

\((a+b)^2=(a+b)(a+b)\). Multiplying gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B is the expansion of \((a-b)^2\). Exam tip: check the sign of the middle term in square identities.

In both, (x^2) and (64) remain same; only the sign of the middle term changes. Exam tip: compare signs.

Algebraic identities make calculation shorter and systematic. Exam tip: identify the identity and apply it directly.

On multiplying \((a+b)(a+b)\), we get \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B is the expansion of \((a-b)^2\), so its middle term is \(-2ab\). Exam tip: in the square of a binomial, the middle term is twice the product of the two terms.

In the subtraction square identity the middle term is negative. Exam tip: focus on ( -2ab ).

Using ( (a+b)^2=a^2+2ab+b^2 ), (2\cdot x\cdot5=10x). Exam tip: multiply the middle term carefully.

The identity \((a+b)^2=a^2+2ab+b^2\), so option A is correct. Option B represents \((a-b)^2\). Exam tip: in a squared binomial identity, check that the middle term has coefficient \(2\).

On multiplying \((p+q)(p-q)\), we get \(p^2-pq+pq-q^2\). The middle terms \(-pq\) and \(+pq\) cancel, leaving \(p^2-q^2\). \(p^2+q^2\) is incorrect because the product of a sum and a difference gives a difference of squares. Exam tip: remember \((a+b)(a-b)=a^2-b^2\).

In \((a+b)^2\), the first and last terms are \(a^2\) and \(b^2\), while the middle term is \(2ab\); hence A is correct. Option B represents \((a-b)^2\). Exam tip: always check the sign and coefficient 2 of the middle term.

This matches the identity \((a+b)(a-b)=a^2-b^2\). Taking \(a=m\) and \(b=3\), we get \((m+3)(m-3)=m^2-3^2=m^2-9\). \(m^2+9\) is incorrect because the signs are opposite here, so the squared terms are subtracted. Exam tip: Whenever you see \((a+b)(a-b)\), use the difference-of-squares identity directly.

Using the identity \((a+b)^2=a^2+2ab+b^2\), take \(a=2x\) and \(b=1\). Then \((2x+1)^2=(2x)^2+2(2x)(1)+1^2=4x^2+4x+1\). Hence, option D is correct. In option C, the middle term is written as \(2x\), but it should be \(4x\). Exam tip: calculate \(2ab\) separately to avoid errors in the middle term.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect square binomial. In option B, the constant should be \(5^2=25\), not 20. Exam tip: match the middle term with \(2ab\).

Using the distributive property, \((x+2)(x+5)=x(x+5)+2(x+5)=x^2+5x+2x+10=x^2+7x+10\). Therefore, option B is correct. In \(x^2+10x+7\), both the middle term and the constant term are incorrect. Exam tip: in \((x+a)(x+b)\), the middle term is \((a+b)x\) and the constant term is \(ab\).