Algebraic identities

The correct identity is \(p^2-q^2=(p+q)(p-q)\). Expanding gives \((p+q)(p-q)=p^2-pq+pq-q^2\), so the middle terms cancel. Option C misses the \(-2pq\) term. Exam tip: difference of squares factors into a sum and a difference.

Option A is correct. Expanding \((x+y)(x-y)\) gives \(x^2-xy+xy-y^2=x^2-y^2\), as the middle terms cancel. Exam tip: check that the binomial signs are opposite.

Apply the identity \((a+b)^2-(a-b)^2=4ab\). Here, \(a=5x\) and \(b=1\). Therefore, the value is \(4\times 5x\times 1=20x\). The expression \(25x^2-1\) is the value of \((5x+1)(5x-1)\), not the difference of the two squares. Exam tip: Recognise the \(4ab\) identity to avoid unnecessary expansion.

\(a^2-b^2\) is a difference of squares, so its standard factorisation is \((a+b)(a-b)\). On multiplying, the middle terms cancel and give \(a^2-b^2\). Exam tip: look for conjugate binomials when you see a difference of squares.

The difference of two squares identity is \(a^2-b^2=(a-b)(a+b)\), so option A is correct. Options B and C are perfect-square identities. Exam tip: for a difference, first check that both terms are squares.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Therefore, \((a+b)^2-(a^2+b^2)=a^2+2ab+b^2-a^2-b^2=2ab\). Hence, the correct answer is \(2ab\). The expression \(a^2-b^2\) belongs to the different identity \((a+b)(a-b)\), so it is not applicable here. Exam tip: do not miss the middle term \(2ab\) while expanding a square.

Use the identity \((a-b)^2=a^2-2ab+b^2\). Thus, \((a-b)^2-(a^2+b^2)=(a^2-2ab+b^2)-a^2-b^2=-2ab\). Hence, the correct answer is \(-2ab\). The option \(0\) is obtained only in the special case \(ab=0\), not as the general simplified form. Exam tip: when subtracting an expression in brackets, distribute the negative sign to every term.

In \((a-b)^2\), the terms have opposite signs, so the middle term is \(-2ab\). Option B has \(+2ab\) because both terms are added. Exam tip: the last term in either binomial-square identity is always positive.

The identity \(p^2-q^2=(p+q)(p-q)\) factorises the difference of two squares. Option D would expand to \((p-q)^2=p^2-2pq+q^2\), not \(p^2-q^2\). Exam tip: the two factors have opposite signs.

When the sum and difference of the same terms are multiplied, the middle terms cancel, giving \((a+b)(a-b)=a^2-b^2\). Option A is the square of a sum, not a product. Exam tip: spot opposite signs first.

To expand \((x+a)(x+b)\), multiply each term in the first bracket by each term in the second bracket. This gives \(x\times x+x\times b+a imes x+a imes b\), or \(x^2+bx+ax+ab\). The middle terms have the common factor \(x\), so \(bx+ax=(a+b)x\). Therefore the general expansion is \(x^2+(a+b)x+ab\), which is option D.

The expression contains a square term from multiplying x by x, a linear term from the two cross-products, and a constant term from multiplying a by b. Option B has \(a-b\), although the cross-products are added, not subtracted. Option C omits the linear term, and option A has an incorrect product and constant arrangement. Hence option D follows directly from distribution.

Expanding \((2x+1)(2x+7)\) gives \(4x^2+14x+2x+7=4x^2+16x+7\). Therefore, the coefficient of \(x\) is \(16\). The value \(14\) comes only from \(2x\times 7\); the term \(1\times 2x=2x\) must also be included. Exam tip: combine all like \(x\)-terms before identifying the coefficient.

Expand the expression: \((3a-2)(3a+8)=9a^2+24a-6a-16=9a^2+18a-16\). Therefore, the coefficient of \(a\) is \(18\). The value \(24\) comes only from \(3a\times 8\); the term \(-2\times 3a=-6a\) must also be included. Exam tip: write all four products before combining like terms.

After removing square terms from the three-term square, (2ab+2bc+2ca) remains. Exam tip: remember pairwise terms.

In the square of a trinomial, the product of each pair of distinct terms appears twice. From \(x\) and \(-3z\), we get \(2\times x\times(-3z)=-6xz\). Hence, the \(xz\)-term is \(-6xz\). The option \(-3xz\) misses the required factor of 2. Exam tip: while finding cross terms, include both \(2ab\) and the sign of each term.

Twice the product of (2p) and (-3q) is (2\cdot2p\cdot(-3q)=-12pq). Exam tip: check both pair and sign.

\(64x^2-1=(8x)^2-1^2\). Applying the identity \(a^2-b^2=(a+b)(a-b)\) gives \((8x+1)(8x-1)\). Option B would produce a middle term when expanded, so it is not correct. Exam tip: first rewrite both terms as perfect squares and then check for the difference-of-squares identity.

The expression is a difference of two squares: \(121=11^2\) and \(4y^2=(2y)^2\). Using \(a^2-b^2=(a+b)(a-b)\), with \(a=11\) and \(b=2y\), we get \(121-4y^2=(11+2y)(11-2y)\). Option A is the square of a binomial, so it is not the factorisation of a difference of squares. Exam tip: Rewrite both terms as perfect squares before selecting an identity.

Using the identity \((a+b)(a-b)=a^2-b^2\), \((x+6)(x-6)=x^2-36\). Therefore, \((x+6)(x-6)+36=x^2-36+36=x^2\). Option A is only the product and misses the outside \(+36\). Exam tip: When two binomials have opposite signs, look for the difference-of-squares identity first.

Use the identity \((a+b)^2-(a-b)^2=4ab\). Here, \(a=3x\) and \(b=4\). Therefore, the expression becomes \(4\times 3x\times 4=48x\). The expression \(9x^2-16\) is the product \((3x+4)(3x-4)\), not the difference of the two squares given here. Exam tip: Recognising the identity is usually faster than expanding both squares.