Factorisation

Both terms of \(2x+6\) have 2 as a common factor. Dividing each term by 2 gives \(2x\div2=x\) and \(6\div2=3\), so \(2x+6=2(x+3)\). Expanding \(2(x+6)\) gives \(2x+12\), so it is incorrect. Exam tip: expand the brackets to check a factorised expression.

The greatest common factor of (15) and (20) is (5). Exam tip: first find the HCF of the numbers.

The highest common factor of \(15p\) and \(20q\) is \(5\). Factoring out \(5\) gives \(15p+20q=5(3p+4q)\). Expanding option B gives \(30p+20q\), so it is incorrect. Exam tip: check a factorisation by expanding the brackets and comparing it with the original expression.

In \(x^2+6x+9\), \(x^2=(x)^2\), \(9=3^2\), and the middle term is \(6x=2\times x\times3\). Therefore, \(x^2+6x+9=(x+3)^2=(x+3)(x+3)\), so option A is correct. In option B, \(8\) is not \(3^2\), so it cannot equal \((x+3)^2\). Exam tip: for a perfect-square trinomial, check whether the middle term is twice the product of the square roots of the first and last terms.

\(x^2+8x+16=x^2+2\cdot x\cdot4+4^2=(x+4)^2=(x+4)(x+4)\). Hence, it is the product of two identical binomials. In option C, since \(16=4^2\), the middle term should be \(2\cdot x\cdot4=8x\), not \(6x\). Exam tip: use \(a^2+2ab+b^2=(a+b)^2\) to check the middle term.

(4x^2=(2x)^2), (9=3^2), and (12x=2\cdot2x\cdot3). Exam tip: match the roots of the first and last terms.

In option A, both \(3p^2\) and \(12p\) are divisible by \(3p\): \(3p^2+12p=3p(p+4)\). In option B, 12 has no factor \(p\), so \(3p\) cannot be a common factor of all the terms. Exam tip: test a common factor by checking it separately in every term.

\(x^2-16=x^2-4^2\) is a difference of two perfect squares. Using \(a^2-b^2=(a-b)(a+b)\), we get \(x^2-16=(x-4)(x+4)\); hence it is a product of two binomials. In contrast, \(x^2+8x+17=(x+4)^2+1\), so it cannot be factorised into two binomials with real coefficients. Exam tip: for a difference of squares, check that both terms are perfect squares and that there is a minus sign between them.

The factor \(a\) is common to every term in \(ab+ac+ad\). Taking \(a\) out gives \(ab+ac+ad=a(b+c+d)\). In contrast, \(b(a+c+d)\) expands to \(ab+bc+bd\), so it is not equivalent to the given expression. Exam tip: verify factorisation by expanding the bracketed expression.

\(y^2-49=y^2-7^2\) is a difference of two perfect squares, so it factorises as \((y-7)(y+7)\). \(y^2+49\) is a sum of squares. In exams, check for perfect squares separated by a minus sign.

\((x+3)(x+4)=x^2+4x+3x+12=x^2+7x+12\), so the statement is correct. The numbers 3 and 4 add to 7 and multiply to 12. In exams, expand the factors once to verify.

\(x^2-49=x^2-7^2\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-7)(x+7)\). In contrast, \((x-7)^2\) contains the middle term \(-14x\). Exam tip: check for two squared terms with subtraction.

\(x^2-9=x^2-3^2\), so it has the form \(a^2-b^2\). Therefore, it factorises as \((x+3)(x-3)\). \(x^2+6x+9=(x+3)^2\) is a perfect-square trinomial, while \(x^2+9\) has a plus sign between the terms. Exam tip: look for two square terms separated by a minus sign before applying the difference-of-squares identity.

The identity for the difference of two squares is \(a^2-b^2=(a-b)(a+b)\). Therefore, one factor has a minus sign and the other has a plus sign. Using two plus signs, \((a+b)(a+b)\), gives \(a^2+2ab+b^2\), so it is not correct. Exam tip: when you see a difference of squares, write \((a-b)(a+b)\).

This expression is a difference of two perfect squares: \(16a^2=(4a)^2\) and \(25b^2=(5b)^2\). Applying \(x^2-y^2=(x-y)(x+y)\) gives \(16a^2-25b^2=(4a-5b)(4a+5b)\). Options A and B are squares of binomials and produce a middle term when expanded, so they are incorrect. Exam tip: first check whether both terms are perfect squares separated by a minus sign.

\(p^2-q^2\) is a difference of two perfect squares. Applying \(a^2-b^2=(a-b)(a+b)\), it factorises as \((p-q)(p+q)\). In \(p^2+q^2\), the squares are added, not subtracted, so it does not fit this identity. Also, \(p^2\pm2pq+q^2\) are perfect-square trinomials: \((p+q)^2\) and \((p-q)^2\). Exam tip: when two square terms are separated by a minus sign, check for \((a-b)(a+b)\).

Using the identity \(a^2-2ab+b^2=(a-b)^2\), with \(a=x\) and \(b=y\), we get \(x^2-2xy+y^2=(x-y)^2\). Therefore, option B is correct. Expanding option A gives a middle term of \(+2xy\), whereas the given expression has \(-2xy\). Exam tip: for a perfect square trinomial, take the square roots of the first and last terms and check the sign of the middle term.

(3) is common in all three terms. Exam tip: taking out the common factor first makes further factorisation easier.

First take (3) common and write (x^2+2x+1=(x+1)^2). Exam tip: continue until complete factorisation.

Both terms of \(a^2+5a\) have \(a\) as a common factor. Factoring out \(a\) gives \(a(a+5)\), because \(a(a+5)=a^2+5a\). Option \((a+5)^2\) expands to \(a^2+10a+25\), so it is incorrect. Exam tip: multiply the factors to check your factorisation.