Algebraic identities

Using \((a+b)^2=a^2+2ab+b^2\), take \(a=100\) and \(b=3\). Then \(103^2=(100+3)^2=10000+600+9=10609\). Hence, 10609 is correct. The option 10600 misses the \(3^2=9\) term. Exam tip: while applying the square identity, include both the middle term \(2ab\) and the final term \(b^2\).

\(x^2-49=x^2-7^2\) is a difference of two squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-7)(x+7)\). \((x-7)^2\) would produce a middle term \(-14x\). Exam tip: recognise \(49=7^2\).

The difference of two squares, \(a^2-b^2\), factors as \((a-b)(a+b)\). On multiplying, the middle \(ab\) terms cancel. Option B gives \(a^2-2ab+b^2\), not the required expression. Exam tip: spot two squared terms joined by a minus sign.

Use the identity \(a^2-b^2=(a+b)(a-b)\). Thus, \(54^2-46^2=(54+46)(54-46)=100\times8=800\). Therefore, the correct answer is 800. A value such as 900 can result from an error in finding the difference or multiplying. Exam tip: for a difference of squares, find the sum and the difference first to simplify the calculation.

Using \((x-y)^2=x^2-2xy+y^2\), take \(x=5a\) and \(y=2b\). Thus, \((5a-2b)^2=(5a)^2-2(5a)(2b)+(2b)^2=25a^2-20ab+4b^2\). Option B has a positive middle term, which belongs to \((x+y)^2\). Exam tip: square the complete first and last terms, and include \(-2\) in the middle term.

The expression uses the identity \((a+b)(a-b)=a^2-b^2\), with \(a=7m\) and \(b=3n\). Therefore, \((7m+3n)(7m-3n)=(7m)^2-(3n)^2=49m^2-9n^2\). Option D resembles the expansion of \((a-b)^2\), but this expression is a product of conjugate binomials. Exam tip: in \((a+b)(a-b)\), the middle terms always cancel.

Use the distributive property: \((x-4)(x+9)=x^2+9x-4x-36=x^2+5x-36\). Therefore, option B is correct. In option A, the middle terms have incorrectly been treated as \(9x+4x\) instead of \(9x-4x\). Exam tip: while combining like terms, check the signs carefully.

Due to opposite signs, the middle term cancels. Exam tip: ( (a+b)(a-b) ) has no middle term.

This uses the (4ab) identity with second term (2b), so (4\cdot a\cdot2b=8ab). Exam tip: take the full second term.

\(x^2-49=x^2-7^2\), so it is a difference of two squares and factorises as \((x+7)(x-7)\). \(x^2+49\) is a sum of squares. Exam tip: check for perfect squares with a minus sign between them.

Use the identity \(a^2-b^2=(a-b)(a+b)\). Since \(49=7^2\), \(x^2-49=x^2-7^2=(x-7)(x+7)\). The close distractor \((x-7)^2\) expands to \(x^2-14x+49\), so it is not equal to the given expression. Exam tip: find the square root of the constant term before applying the difference-of-squares identity.

The expression is a difference of two squares: \(121=11^2\) and \(4y^2=(2y)^2\). Using \(a^2-b^2=(a+b)(a-b)\), we get \(121-4y^2=(11+2y)(11-2y)\). The option \((11-2y)^2\) would also produce a middle term \(-44y\), so it is not correct. Exam tip: identify the square roots of both terms before applying the difference-of-squares identity.

\(a^2+2ab+b^2=(a+b)^2\), so it is a perfect-square trinomial. In option B, the middle term is not \(2ab\). Exam tip: take square roots of the first and last terms, then verify the middle term.

One is a perfect square and the other is difference of squares. Exam tip: distinguish bracket square from difference of squares.

In the square of three terms, the double product of each pair is added. Exam tip: remember (2ab+2bc+2ca).

\((a+b)(a-b)=a^2-b^2\) is the difference-of-squares identity because the \(+ab\) and \(-ab\) terms cancel on multiplication. Option A is the identity for the square of a sum. Exam tip: look for binomials with opposite signs.

Using \((a+b)^2=a^2+2ab+b^2\), put \(a=x\) and \(b=5\). The middle term is \(2\times x\times5=10x\), and the last term is \(25\). Hence A is a perfect-square trinomial. Exam tip: check whether the middle coefficient is twice the product of the terms.

Expanding \((a-b)(a-b)\) gives \(a^2-ab-ab+b^2=a^2-2ab+b^2\). Option B is the identity for the square of a sum. Exam tip: always check the sign of the middle term.

\((x+6)(x-1)=x^2-x+6x-6=x^2+5x-6\). Hence, the term without \(x\), namely \(-6\), is the constant term. The number \(5\) is the coefficient of the middle term \(5x\), not the constant term. Exam tip: in a product of two binomials, multiply the constant terms to find the constant term.

To find the coefficient of \\(x\\), expand the product and collect the terms containing exactly one \\(x\\). In \\((x+6)(x-1)\\), multiplying each term gives \\(x\cdot x+x\cdot(-1)+6\cdot x+6\cdot(-1)\\). The two terms containing \\(x\\) are \\(-x\\) and \\(6x\\). Their sum is \\(5x\\), so the coefficient of \\(x\\) is 5.

Equivalently, for \\((x+r)(x+s)\\), the coefficient of \\(x\\) is \\(r+s\\). Here, \\(r=6\\) and \\(s=-1\\), so \\(r+s=6-1=5\\). Therefore option A is correct. The number 6 is only one constant from the first bracket, and \\(-6\\) is the constant term after multiplication, not the coefficient of \\(x\\).