Algebraic identities

Using identity ( (a+b)^2=a^2+2ab+b^2 ) gives the answer. Exam tip: do not forget the middle term (2ab).

By ( (a-b)^2=a^2-2ab+b^2 ) the middle term is negative. Exam tip: note that (2\times y\times4=8y).

This is the form ( (a+b)(a-b)=a^2-b^2 ). Exam tip: identify difference of squares in opposite-sign factors.

The middle term is (2ab) so (2\times m\times2=4m). Exam tip: distinguish squares from the middle term.

The identity \\(p+q)^2=p^2+2pq+q^2\\) helps expand a square of two terms. Here, the first term is \\(3a\\) and the second term is \\(b\\). Therefore, the expansion must contain the square of \\(3a\\), a middle term involving both terms, and the square of \\(b\\). This structure is important because the plus sign between the terms makes the middle term positive.

Substituting gives \\((3a+b)^2=(3a)^2+2(3a)(b)+b^2\\). Now, \\((3a)^2=9a^2\\) and \\(2(3a)(b)=6ab\\). Hence the result is \\(9a^2+6ab+b^2\\), which is option A. Option C misses the middle term, while option D has the wrong sign.

\(x^2-16=x^2-4^2\) is a difference of two perfect squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-4)(x+4)\). \(x^2+16\) is a sum of squares. Exam tip: check for square terms separated by a minus sign.

Multiplying \((a+b)(a+b)\) gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\), so A is an identity. Option B misses the middle term \(2ab\). Exam tip: always check the middle term and its sign.

Option B is correct because in \((a-b)(a-b)\), the two middle products \(-ab\) and \(-ab\) combine to give \(-2ab\). \(a^2-b^2\) is the difference of squares, not the square of a difference. Exam tip: check the sign of the middle term.

Expanding \((a+b)^2\) gives \(a^2+2ab+b^2\), so it is the required perfect-square trinomial. In \((a-b)^2\), the middle term is \(-2ab\). Exam tip: always check the sign of the middle term.

Write 98 as \(100-2\). Using \((a-b)^2=a^2-2ab+b^2\), \((100-2)^2=100^2-2\times100\times2+2^2=10000-400+4=9604\). Therefore, 9604 is correct. In 9804, the middle-term subtraction has not been handled correctly. Exam tip: in \((a-b)^2\), the middle term is \(-2ab\), so keep its negative sign.

\((a+b)^2=a^2+2ab+b^2\). Here, \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\). Option C represents \((x-5)^2\). Exam tip: match the middle term with \(2ab\).

On expansion (a^2+12a+36-a^2+12a-36=24a). Exam tip: apply the minus sign to the whole bracket.

This uses the identity \((a+b)(a-b)=a^2-b^2\). Taking \(a=r\) and \(b=9\), \((r+9)(r-9)=r^2-9^2=r^2-81\). \(r^2+81\) is incorrect because conjugate binomials give a difference of squares. Exam tip: the middle terms cancel in this identity.

Write \(72\times68\) as \((70+2)(70-2)\). Using \((a+b)(a-b)=a^2-b^2\), we get \(70^2-2^2=4900-4=4896\). Hence, 4896 is correct. 4900 is only \(70^2\); subtracting \(2^2\) is necessary. Exam tip: use the difference-of-squares identity when two numbers are equally distant from a common number.

The expression is a difference of two squares. Instead of calculating both large squares separately, use the identity \\(a^2-b^2=(a+b)(a-b)\\). Here, take \\(a=64\\) and \\(b=36\\). Then \\(64+36=100\\), and \\(64-36=28\\). Multiplying these simpler numbers gives \\(100\\times28=2800\\). This method is quicker and also reduces the chance of an arithmetic error.

Therefore, the value is 2800, so option B is correct. Direct checking gives \\(64^2=4096\\) and \\(36^2=1296\\); their difference is \\(4096-1296=2800\\). The other options do not match this result. The important skill is recognising the pattern \\(a^2-b^2\\) before expanding or calculating separately.

A trinomial is a perfect square when it follows the pattern \\(x^2+2xy+y^2=(x+y)^2\\). The first term, \\(x^2\\), shows that one part of the bracket is \\(x\\). The last term, \\(36\\), is \\(6^2\\), so the other part is 6. The middle term must then be twice their product. Since \\(2\times x\times6=12x\\), the given expression matches the plus form.

Thus, \\(x^2+12x+36=x^2+2(x)(6)+6^2=(x+6)^2\\). Therefore option C is correct. The choices with 12 use the last number itself instead of its square root. The expression \\((x-6)^2\\) would produce \\(x^2-12x+36\\), so it cannot be correct because the middle term here is positive.

This trinomial has the pattern of a perfect square. The identity is \\(a^2-2ab+b^2=(a-b)^2\\). In \\(t^2-14t+49\\), the first term is \\(t^2\\), and the last term is \\(49=7^2\\). The middle term is \\(-14t=-2\\times t\\times7\\), so the required form is \\(t^2-2\\times t\\times7+7^2\\). Therefore the expression factors as \\((t-7)^2\\).

Option B is correct. Expanding it confirms the result: \\((t-7)^2=t^2-14t+49\\). Option A would produce a positive middle term, \\(t^2+14t+49\\), so it cannot match. The constants 14 and 49 should be checked through the square identity rather than guessed from the final term alone.

This expression is a difference of squares: \(x^2-64=x^2-8^2\). Using \(a^2-b^2=(a+b)(a-b)\), we get \((x+8)(x-8)\). In contrast, \((x-8)^2\) expands to \(x^2-16x+64\), so it is not correct. Exam tip: rewrite the constant term as a perfect square before applying the difference-of-squares identity.

The expression is a difference of squares: \(4x^2=(2x)^2\) and \(25=5^2\). Using \(a^2-b^2=(a+b)(a-b)\), with \(a=2x\) and \(b=5\), gives \((2x+5)(2x-5)\). On expansion, option B gives \(4x^2-15x-25\), so it is not correct. Exam tip: first check whether both terms are perfect squares separated by a minus sign.

Using the distributive property, \((x+3)(x-4)=x^2-4x+3x-12=x^2-x-12\). Therefore, option B is correct. In option A, \(-4x+3x\) has incorrectly been simplified as \(+x\). Exam tip: check the signs carefully before combining like terms.