Algebraic identities

( (x+y)^2=x^2+2xy+y^2 ), so subtracting (x^2+y^2) leaves (2xy). Exam tip: identify the middle term.

( (x-y)^2=x^2-2xy+y^2 ), subtracting (x^2+y^2) gives (-2xy). Exam tip: watch the negative middle term.

In the square of three terms, the extra mixed terms are (2ab+2bc+2ca). Exam tip: remember all three pairs.

Using the distributive property, \((x+4)(x-9)=x^2-9x+4x-36=x^2-5x-36\). Therefore, option A is correct. In option B, the sign of the middle term is incorrect because \(-9x+4x=-5x\), not \(+5x\). Exam tip: multiply the first, outer, inner and last terms, then combine like terms.

((-8)+(-3)=-11), so the coefficient of (x) is (-11). Exam tip: the sum of two negative numbers remains negative.

Since \(144=12^2\), we have \(x^2-144=x^2-12^2\). Using \(a^2-b^2=(a+b)(a-b)\), its factor form is \((x+12)(x-12)\). The expansion of \((x-12)^2\) is \(x^2-24x+144\), so it is not correct. Exam tip: first check whether the constant term is a perfect square.

In \((a+b)(a-b)\), the middle terms \(-ab\) and \(+ab\) cancel, leaving \(a^2-b^2\). Options B and C are identities for squaring a binomial. Exam tip: recognise “sum × difference” as a difference of squares.

The mixed terms are (2(2a)(b)=4ab) and (2(2a)(c)=4ac). Exam tip: form mixed terms with coefficients.

The expression can be simplified by using the difference of squares. Let the two squares be formed around the same middle term. Direct expansion gives \\(x+5\\)^2=x^2+10x+25 and \\(x-5\\)^2=x^2-10x+25. Subtracting the second expression from the first cancels both \\(x^2\\) and 25, leaving \\(20x\\). The condition therefore becomes \\(20x=80\\).

Dividing both sides by 20 gives \\(x=4\\). Hence option B is correct. The same result follows from \\(a^2-b^2=(a-b)(a+b)\\): here the difference is 10 and the sum is \\(2x\\), so the product is \\(20x\\). Options A, C and D do not satisfy the equation; substituting 4 gives exactly 80. The supplied answer B is therefore verified.

The identity for the square of a sum is \((a+b)^2=a^2+2ab+b^2\). In this question, \(a=x\) and \(b=3\). The middle term is produced by twice the product of the two parts, so it is \(2\times x\times3=6x\). The constant term is the square of 3, which is 9.

Therefore, expanding the left side gives \((x+3)^2=x^2+6x+9\). Comparing this with the given expression \(x^2+kx+9\), the coefficients of the corresponding \(x\)-terms must be equal. Hence \(k=6\), so option B is correct. The value 3 is only the added number, not the coefficient of the middle term. The value 9 is the constant term, and 12 would result from an incorrect multiplication rather than the identity.

Option B follows the square-of-a-sum identity: \((x+y)^2=x^2+2xy+y^2\), so it is true for every x and y. In A, the \(2xy\) term is missing. Exam tip: expand both sides to test an identity.

Since \(49m^2=(7m)^2\) and \(25n^2=(5n)^2\), the expression is a difference of two squares. Therefore, use \(a^2-b^2=(a+b)(a-b)\). Exam tip: first check whether both terms are perfect squares.

The expansion is (16a^2-40a+25), so like terms cancel leaving (-40a). Exam tip: do not forget the negative middle term.

The difference-of-squares identity is \(a^2-b^2=(a+b)(a-b)\). On multiplying, the middle terms \(ab\) and \(-ab\) cancel. \((a-b)^2\) contains an extra \(-2ab\) term. Exam tip: check the signs before choosing an identity.

Here, \(58=50+8\) and \(42=50-8\). Using the identity \((a+b)(a-b)=a^2-b^2\), we get \((50+8)(50-8)=50^2-8^2=2500-64=2436\). Option 2500 is only \(50^2\); subtracting \(8^2\) is essential. Exam tip: when two numbers are equally distant from a common middle number, use the difference of squares identity.

A coefficient is the number or algebraic expression that multiplies a particular variable term. First expand the product using the distributive property: \\( (x+a)(x+b)=x\cdot x+x\cdot b+a\cdot x+a\cdot b\\). This gives \\(x^2+bx+ax+ab\\). Combining the two terms containing x gives \\(x^2+(a+b)x+ab\\). Therefore, the coefficient of x is \\(a+b\\), so option C is correct.

It is important to identify the term, not just any expression in the expansion. The coefficient of \\(x^2\\) is 1, and the constant term is \\(ab\\), but neither is the coefficient of x. The two contributions to the x-term are \\(bx\\) and \\(ax\\); adding them produces \\((a+b)x\\). This reasoning works for arbitrary values of a and b, including zero or negative values. Hence the third choice gives the required coefficient.

The constant term is ((-2)\times11=-22). Exam tip: multiply the constants to get the constant term.

\( (2x)(2x)=4x^2 \), \( -14x+6x=-8x \), and \(3\times(-7)=-21\). Exam tip: combine like terms.

\((a+b)^2=a^2+2ab+b^2\). Subtracting \(a^2+b^2\) gives \(2ab\). \(ab\) is only the product and misses the coefficient 2 that arises in the binomial expansion. Exam tip: whenever expanding \((a+b)^2\), check that the middle term is \(2ab\).

( (a-b)^2=a^2-2ab+b^2 ), so subtracting (a^2-b^2) gives (2b^2-2ab). Exam tip: keep both identities separate.