Algebraic identities

The correct identity is \(a^2-b^2=(a+b)(a-b)\), because multiplying the factors cancels the middle terms. Option C misses the \(-2ab\) term. Exam tip: recognise a difference of squares and factor it into sum and difference.

Use the identity \((a-b)(a+b)=a^2-b^2\). Here, \((x-3)(x+3)=x^2-3^2=x^2-9\). Therefore, \((x^2-9)+9=x^2\), so option A is correct. \(x^2-9\) is only the product and does not include the outside \(+9\). Exam tip: when you see conjugate binomials \((a-b)(a+b)\), apply the difference-of-squares identity directly.

The difference-of-squares identity is \(a^2-b^2=(a-b)(a+b)\). Expanding gives \(a^2+ab-ab-b^2=a^2-b^2\). Option B introduces an extra \(-2ab\) term. Exam tip: use this identity when two squared terms are separated by a minus sign.

When the same algebraic factor is multiplied by itself, the product is its square: \\(A\cdot A=A^2\\). Here the repeated factor is \\(x+7\\), so \\((x+7)(x+7)=(x+7)^2\\). Expanding it would give \\(x^2+14x+49\\), which is also consistent with the identity \\( (x+y)^2=x^2+2xy+y^2\\) for \\(y=7\\).

Therefore option A is correct. Option B changes the plus sign to a minus sign and represents a different expression. Options C and D correspond to the difference of squares, \\((x+7)(x-7)=x^2-49\\), not to two identical factors. The essential recognition is that repeated multiplication of one expression is written as a power of 2.

( (a-b)^2=a^2-2ab+b^2 ) while (a^2-b^2=(a+b)(a-b)). Exam tip: keep square of bracket and difference of squares separate.

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.

Since (98=100-2), ( (100-2)^2 ) makes calculation easy. Exam tip: choose a nearby round number.

\((x+3)(x+8)=x^2+8x+3x+24=x^2+11x+24\). 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.

(100=10^2) and (-20y=-2\cdot y\cdot10). Exam tip: identify the minus square from the negative middle term.

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.