Algebraic identities

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.

The direct answer is option B: \\(a^2-2ab+b^2\\). Begin with the meaning of a square: \\(a-b\\)^2 means \\(a-b\\)(a-b)\\). Multiply each term in the first bracket by each term in the second: \\(a\\cdot a=a^2\\), \\(a\\cdot(-b)=-ab\\), \\((-b)\\cdot a=-ab\\), and \\((-b)\\cdot(-b)=b^2\\). Adding these gives \\(a^2-ab-ab+b^2=a^2-2ab+b^2\\). Option B is correct. Option A leaves out the two cross-products, so it is incomplete. Option C has a positive middle term and belongs to \\(a+b\\)^2\\), not \\(a-b\\)^2\\). Option D is the difference of squares, \\(a^2-b^2=(a-b)(a+b)\\), not the square of a difference. The common mistake is forgetting that two negative cross-products combine to \\(-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 \(ab\).

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

The direct answer is D: the additional term is 2rs. Start with the identity \\(r+s\\)^2=r^2+2rs+s^2\\). The square means multiplying the whole expression by itself: \\(r+s\\)(r+s). Multiplying term by term gives \\(r^2+rs+sr+s^2\\). Since \\(rs=sr\\), the two middle terms combine to \\(2rs\\). Therefore the complete expansion contains r², 2rs and s². Option D is correct. Option A, -2rs, belongs to the middle term of \\(r-s\\)^2\\), not a plus square. Option B, rs, misses one of the two equal cross-products. Option C, -rs, has the wrong sign and also the wrong coefficient. The common mistake is to write only r²+s² and forget the cross terms. Cue: plus square means positive double product.

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.