Algebraic identities

The direct answer is A: \\(x^2+10x+25\\). Use the identity \\(a+b\\)^2=a^2+2ab+b^2\\). Here a=x and b=5. First, \\(a^2=x^2\\). Next, \\(2ab=2\\times x\\times5=10x\\). Finally, \\(b^2=5^2=25\\). Combining the three parts gives \\(x^2+10x+25\\), so A is correct. Option B has 5x instead of 10x because it forgets the factor 2 in the middle term. Option C leaves out the cross term completely; it is not the square of a sum. Option D has a negative middle term, which belongs to \\(x-5\\)^2\\), not \\(x+5\\)^2\\). Direct multiplication also confirms it: \\(x+5\\)(x+5)=x^2+5x+5x+25=x^2+10x+25\\). Memory cue: in a plus square, the middle term is positive twice the product.

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

The direct answer is C: \\(p^2-49\\). The expression has the form \\(a+b\\)(a-b)=a^2-b^2\\), called the difference of squares. Take a=p and b=7. Then \\(p+7\\)(p-7)=p^2-7^2=p^2-49\\). The middle terms cancel if multiplied directly: \\(p^2-7p+7p-49=p^2-49\\). Therefore C is correct. Option A, \\(p^2+49\\), uses addition instead of subtraction and ignores the opposite signs. Option B, \\(p^2+14p+49\\), is the expansion of \\(p+7\\)^2\\), not a product with p-7. Option D, \\(p^2-14p+49\\), is the expansion of \\(p-7\\)^2\\), not the given product. Exam cue: matching opposite brackets produce a difference of squares, and the middle terms disappear.

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 \(2ab\).

The direct answer is B: \\(24a\\). There are two reliable methods. Using difference of squares, write the expression as \\([(a+6)^2-(a-6)^2]=[(a+6)-(a-6)][(a+6)+(a-6)]\\). The first bracket is 12 and the second is 2a, so the product is \\(12\\times2a=24a\\). By expansion, \\(a+6\\)^2=a^2+12a+36\\), while \\(a-6\\)^2=a^2-12a+36\\). Subtracting gives \\(a^2+12a+36-a^2+12a-36=24a\\). Thus B is correct. A, 12a, keeps only one middle contribution. C, 36a, confuses the square of 6 with the full difference. D leaves an unrelated expression and fails to cancel the a² and constant terms. Common mistake: the minus sign before the second bracket changes every sign inside it.

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.

Using the distributive property, \((x-6)(x-2)=x^2-2x-6x+12=x^2-8x+12\). Therefore, option A is correct. In option C, both the middle term and the constant term are incorrect. Exam tip: use \((x-a)(x-b)=x^2-(a+b)x+ab\) for quick expansion.

( (z+1)(z+9)=z^2+10z+9 ). Exam tip: the coefficient of (z) comes from the sum of the constants.

In the square of three terms, mixed terms are (2xy), (2yz), and (2zx). Exam tip: include all pairs.

Each pair gives a double mixed term in three terms. Exam tip: do not miss (2ab), (2bc), and (2ca).

The direct answer is B: the difference is 2x. Use the identity \\((p+q)^2=p^2+2pq+q^2\\) with p=x and q=1. Thus \\((x+1)^2=x^2+2x+1\\. Subtract the second expression: \\((x^2+2x+1)-(x^2+1)=2x\\. Option B is correct because the x² terms cancel and the constant 1 terms also cancel. Option A, x, misses the required middle-term coefficient 2. Option C, 1, ignores the middle term and is only a constant. Option D, x², fails because x² appears in both expressions and therefore cancels. This explanation treats “difference” as the first expression minus the second; if the reverse order were intended, it would be -2x. Exam cue: never forget the middle term in a square.