Algebraic identities

Using the identity \((x+y)^2=x^2+y^2+2xy\), we get \(x^2+y^2=(x+y)^2-2xy\). Therefore, \(x^2+y^2=9^2-2\times14=81-28=53\). Option 81 is only the value of \((x+y)^2\); subtracting \(2xy\) is necessary. Exam tip: when \(x+y\) and \(xy\) are given, apply this identity directly to find the sum of squares.

Using the identity \((p-q)^2=p^2-2pq+q^2\), we get \(p^2+q^2=(p-q)^2+2pq\). Therefore, \(p^2+q^2=5^2+2\times6=25+12=37\). Option 25 is only the value of \((p-q)^2\); the term \(2pq\) must also be added. Exam tip: when \(p-q\) and \(pq\) are given, apply \(p^2+q^2=(p-q)^2+2pq\) directly.

The difference of two squares, \(a^2-b^2\), factorises as \((a+b)(a-b)\). Options B and C are square-expansion identities, while D should contain \(-b^2\), not \(+b^2\). Exam tip: for a difference of squares, look for sum and difference factors.

Multiply the first two factors: \((x+1)(x+2)=x^2+2x+x+2=x^2+3x+2\). Hence the correct intermediate step is \((x^2+3x+2)(x+3)\). In option B, the linear terms have been combined incorrectly. Exam tip: when multiplying two binomials, multiply every term in one bracket by every term in the other bracket.

The common factor in both terms is \(x+2\): \((x+2)^2-(x+2)(x-2)=(x+2)[(x+2)-(x-2)]\). The bracket equals \(4\), so the expression becomes \(4(x+2)=4x+8\). \(4x\) is incomplete because it misses the constant term \(8\). Exam tip: In expressions involving subtraction, factor out a common factor before expanding whenever possible.

Using the identity \((a+b)^2+(a-b)^2=2a^2+2b^2\), put \(a=3x\) and \(b=2\): \(2(3x)^2+2(2)^2=18x^2+8\). Therefore, \(18x^2+8\) is correct. \(9x^2+4\) misses the doubling that occurs when both squared expressions are added. Exam tip: in such a pair, the mixed terms with opposite signs cancel out.

Use the identity \((a-b)^2=a^2-2ab+b^2\). Here, \(a=2x\) and \(b=5\), so \((2x-5)^2=4x^2-20x+25\). Therefore, the coefficient of \(x^2\) is 4. Note that 20 is associated with the middle term, \(-20x\), not with \(x^2\). Exam tip: always check the power of the variable before identifying a coefficient.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Here, \(a=x\) and \(b=3y\), so \((x+3y)^2=x^2+6xy+9y^2\). Therefore, the coefficient of \(y^2\) is 9. The number 6 is the coefficient of the middle term \(6xy\), not of \(y^2\). Exam tip: always square the numerical coefficient in the last term.

In opposite-sign product mixed terms cancel but in a square the middle term remains. Exam tip: do not mix the two forms.

\((a+b)^2=(a+b)(a+b)=a^2+2ab+b^2\), so A is correct. In B, the sign of the last term is incorrect. Exam tip: the middle term is always twice the product of the two terms.

The mixed term of (a) and (-b) is (2\times a\times(-b)=-2ab). Exam tip: keep the negative sign with the term.

Using the sum identity gives (2(3a)^2+2(4b)^2=18a^2+32b^2). Exam tip: notice cancellation of middle terms.

After removing square terms from a three-term square, only double products remain. Exam tip: remember (2xy), (2yz), (2zx).

(x^2+14x+49=(x+7)^2), so subtraction gives (0). Exam tip: identify perfect square trinomials.

Using the identity \((a+b)^2=a^2+2ab+b^2\), the expression becomes \(a^2+2ab+b^2-(a^2+b^2)\). The terms \(a^2\) and \(b^2\) cancel, leaving \(2ab\). It is not \(0\) because the middle term \(2ab\) does not cancel. Exam tip: expand the identity first, then cancel like terms carefully.

The sum of constants is (4+(-6)=-2) and the product is (-24). Exam tip: handle mixed signs carefully.

Expanding \((x+r)^2\) gives \(x^2+2rx+r^2\), so \(p=2r\) and \(q=r^2\). Hence \(p^2=4r^2=4q\). The other relations do not ensure a perfect square. Exam tip: square the middle coefficient and compare it with four times the constant term.

\(25p^2-36q^2=(5p)^2-(6q)^2\), which is a difference of two squares. Using \(a^2-b^2=(a+b)(a-b)\), we get \((5p+6q)(5p-6q)\). Options A and D are squares of a difference and a sum respectively; their expansions contain a \(pq\) term, which is absent in the given expression. Exam tip: identify the square roots first, then use \((a+b)(a-b)\) for a difference of squares.

Using \((a+b)^2=a^2+2ab+b^2\), take \(a=x\) and \(b=5y\). The middle term is \(2\times x\times5y=10xy\), and the last term is \(25y^2\). Option C represents \((x-5y)^2\). Exam tip: always check the sign of the middle term.

In \(x^2+12x+25\), the square roots of the first and last terms are \(x\) and \(5\), so the middle term should be \(2\times x\times5=10x\), not \(12x\). Exam tip: always check the middle term using \(2ab\).