Factorisation

\(x^2-49\) is a difference of squares: \(x^2-7^2=(x+7)(x-7)\). Since one side is \(x+7\), the other is \(x-7\). Choosing \(x+7\) gives the wrong area. Exam tip: use \(a^2-b^2=(a+b)(a-b)\).

The greatest common factor of 3a and 6 is 3. Taking 3 outside gives 3a+6 = 3(a+2), because expanding the bracket gives 3a+6. Option 6(a+3) expands to 6a+18, so it is not correct. Exam tip: expand your factorised answer to check it.

In \(p^2-q^2\), both \(p^2\) and \(q^2\) are perfect squares and they are separated by subtraction, so it is a difference of squares. \(p^2+q^2\) has addition. Exam tip: look for “square minus square.”

In \(x^2+8x+16\), the first and last terms are \(x^2\) and \(4^2\), and the middle term is \(2\times x\times4=8x\). Hence it is \((x+4)^2\). Option B does not have 16 as the last term. Exam tip: match the middle term with \(2ab\).

Since (9=3^2) and the middle term is (-6x), the form is ((x-3)^2). Exam tip: watch the negative sign.

Both terms have common factor (2x). Exam tip: check the common numerical and variable parts.

(3+4=7) and (3\times4=12). Exam tip: find two numbers whose sum is the middle coefficient.

Since (9=3^2), (m^2-9) is a difference of squares. Exam tip: identify difference of squares quickly.

\((x+p)(x+q)=x^2+(p+q)x+pq\). Hence, the sum of the two numbers must give the coefficient of \(x\), and their product must give the constant term. Exam tip: check product first, then sum.

To factorise \(x^2+9x+20\), choose two numbers whose sum is \(9\) and product is \(20\). Since \(4+5=9\) and \(4\times5=20\), \((x+4)(x+5)=x^2+9x+20\). In option A, \(2+10=12\), so its middle term would be \(12x\), not \(9x\). Exam tip: check that the sum of the constant terms in the factors equals the coefficient of the middle term.

To take out a common factor, find a quantity present in every term. In 5p+15q, both numerical coefficients 5 and 15 are divisible by 5, while p and q are not common variables. Dividing each term by 5 gives p+3q. Therefore the expression becomes 5(p+3q).

Option A is correct. Multiplying back verifies the result: 5(p+3q)=5p+15q, exactly the original expression. Option B leaves 15q unchanged inside and would give 5p+75q. Option C incorrectly introduces p as a common factor, and option D uses 15 even though 15 is not a factor of 5.

In \(7m(m+2)\), \(7m\) has been taken out as the common factor, and no further common factor remains inside the bracket. In C and D, \(7\) or \(m\) can still be extracted. Exam tip: always check the bracket for a remaining common factor.

Use the identity \(a^2-b^2=(a-b)(a+b)\). Since \(49=7^2\), \(x^2-49=x^2-7^2=(x-7)(x+7)\). Option B gives \((x-7)^2=x^2-14x+49\), which is not the given expression. Exam tip: rewrite the constant term as a perfect square before applying the difference-of-squares identity.

(16=4^2) and (8y=2\times4\times y), so it is ((y+4)^2). Exam tip: identify the perfect square.

Using the identity \(x^2-2xy+y^2=(x-y)^2\), with \(x=a\) and \(y=b\), we get \(a^2-2ab+b^2=(a-b)^2\). The option \((a+b)^2\) has the middle term \(+2ab\), so it is not correct. Exam tip: use the sign of the middle term to distinguish between \((a-b)^2\) and \((a+b)^2\).

The greatest common factor of \(10x^2\) and \(-15x\) is \(5x\). Taking it out gives \(10x^2-15x=5x(2x-3)\). Option A has a positive sign, so it would produce \(10x^2+15x\). Exam tip: multiply the factors back to quickly verify the expression.

In grouping, a common binomial is factored from both groups: \(x(a+b)+y(a+b)=(x+y)(a+b)\). Equal exponents alone do not ensure grouping works. Exam tip: find the common factor in each group first.

Since \(25=5^2\), we get \(a^2-25=a^2-5^2\), a difference of two squares. Hence \(a^2-5^2=(a-5)(a+5)\). The square identities apply to three-term expressions. Exam tip: look for two perfect squares separated by a minus sign.

In 7xy + 14x, 7x is the common factor of both terms. Taking 7x outside gives 7x(y + 2), since 7x × y = 7xy and 7x × 2 = 14x. Expanding 14x(y + 7) gives 14xy + 98x, so it does not match the given expression. Exam tip: expand the factorised form once to verify your answer.

(1+2=3) and (1\times2=2), so these are correct factors. Exam tip: start with small numbers.