Algebraic identities

Using the identity \((a+b)^2=a^2+2ab+b^2\), put \(a=x\) and \(b=7\). Then \((x+7)^2=x^2+2\times x\times7+7^2=x^2+14x+49\). Hence, option A is correct. In option B, the middle term is written as \(7x\) instead of \(2ab=14x\). Exam tip: while expanding a squared binomial, always check the sign and coefficient of the \(2ab\) term.

Expanding \((a+b)^2\) gives \(a^2+2ab+b^2\), so A is correct. In \((a-b)^2\), the middle term is \(-2ab\). Exam tip: always check the sign of the middle term.

For the square of a difference, the first and last terms are the squares of the two terms, and the middle term is \(-2ab\). Hence option A is correct. Option B has \(+2ab\), so it is for a sum. Exam tip: check the sign of the middle term first.

This expression uses the identity \((a+b)(a-b)=a^2-b^2\). Taking \(a=p\) and \(b=9\), we get \((p+9)(p-9)=p^2-9^2=p^2-81\). \(p^2+81\) is a sum of squares, so it does not result from this product. Exam tip: when two binomials have the same terms with opposite signs, use the difference-of-squares identity.

To use an identity, rewrite \(98\) as \(100-2\) so that \(98^2=(100-2)^2\) matches the square of a difference: \( (a-b)^2=a^2-2ab+b^2 \). Then \((100-2)^2=100^2-2\cdot 100\cdot 2+2^2=10000-400+4=9604\).

\((x+3)(x+8)=x^2+8x+3x+24=x^2+11x+24\). Therefore, the coefficient of \(x\) is 11. Here, 24 is the constant term, while 5 is only the difference between the constants. Exam tip: in \((x+a)(x+b)\), the coefficient of \(x\) is \(a+b\).

Using the distributive property, \(m(m+2)-6(m+2)=m^2+2m-6m-12=m^2-4m-12\). Hence, option A is correct. In option B, the like terms \(2m-6m\) have been combined with the wrong sign to get \(+4m\). Exam tip: while multiplying two binomials, write all four products first and then combine like terms.

In \((a-b)^2=(a-b)(a-b)\), the middle terms give \(-ab-ab=-2ab\), so the expansion is \(a^2-2ab+b^2\). Option B is the expansion of \((a+b)^2\). Exam tip: the last squared term, \(b^2\), remains positive.

The direct answer is option C, \((y-10)^2\). Start with the identity \((a-b)^2=a^2-2ab+b^2\). Here, take \(a=y\) and \(b=10\). Then \((y-10)^2=y^2-2(y)(10)+10^2=y^2-20y+100\), exactly the given expression. Option A, \((y+10)^2\), expands to \(y^2+20y+100\), so its middle sign is wrong. Option B, \((y-20)^2\), expands to \(y^2-40y+400\), so both the middle coefficient and constant are wrong. Option C gives all three terms correctly and is therefore correct. Option D, \(y^2-100\), is a difference of squares and has no \(-20y\) term, so it is not equal. The negative middle term tells us to use subtraction inside the square. Memory cue: half of 20 is 10, and the constant is \(10^2\).

9p^2-4q^2 = (3p)^2-(2q)^2, so it is a difference of two perfect squares. Option C is a perfect-square trinomial. In exams, check for square terms separated by a minus sign.

\((a+b)^2=a^2+2ab+b^2\) and \((a-b)^2=a^2-2ab+b^2\). On adding them, \(+2ab\) and \(-2ab\) cancel, giving \(2a^2+2b^2\). \(4ab\) results from incorrectly focusing only on the middle terms. Exam tip: expand each square and combine like terms carefully.

The square of a difference is \((a-b)^2=a^2-2ab+b^2\), so its middle term is negative. \(a^2-b^2\) is a difference of squares, not a square expansion. Exam tip: use the sign of the middle term to identify the identity.

\((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\). The double product terms are \(2ab\), \(2bc\), and \(2ca\), so there are 3 such terms. \(a^2,b^2,c^2\) are square terms, not double product terms. Exam tip: make each distinct pair of variables once: \(ab\), \(bc\), and \(ca\).

In a three-term square, each pair gives a product term multiplied by (2). Exam tip: think in pairs.

In
(2x+3y+z)^2, the term containing xy comes from 2(2x)(3y). Therefore, 2 × 2x × 3y = 12xy. The value 6xy results from taking only (2x)(3y), but this product occurs twice in the expansion of a square. Exam tip: in the square of a trinomial, the mixed term for two different terms is 2ab.

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.

The direct answer is option C, \(28x\), because this term does not appear. Use \((a-b)(a+b)=a^2-b^2\), with \(a=2x\) and \(b=7\). Thus \((2x-7)(2x+7)=(2x)^2-7^2=4x^2-49\). If multiplied term by term, the two cross terms are \(+14x\) and \(-14x\), which cancel to zero. Option A, \(4x^2\), is present as the square of the first term. Option B, \(-49\), is present as the negative square of 7. Option C, \(28x\), would be a middle term, but opposite signs make the middle terms cancel; it does not appear. Option D, \((2x)^2\), is also present before simplifying it to \(4x^2\). Therefore C is correct. Exam cue: conjugate brackets with opposite signs produce only two square terms.

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.