Factorisation

A perfect-square trinomial has the form \(a^2+2ab+b^2=(a+b)^2\). Here, \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so option A is correct. In option B, the constant term would need to be \(25\), while option C is a difference of squares. Exam tip: take the square roots of the first and last terms and check whether the middle term is twice their product.

Expanding \((3a-b)(a+2b)\) gives \(3a^2+6ab-ab-2b^2=3a^2+5ab-2b^2\). Therefore, option A is correct. Expanding option B gives a middle term of \(-5ab\), whereas the given expression has \(+5ab\). Exam tip: while checking factors, add the two cross-products and verify the sign of the middle term.

The first and last terms are \((3x)^2\) and \((4y)^2\). Using \((a-b)^2=a^2-2ab+b^2\), the middle term is \(-2(3x)(4y)=-24xy\), so A is correct. In exams, always check the sign of the middle term.

Since \(4x^2=(2x)^2\) and \(25=5^2\), it is a difference of squares: \((2x-5)(2x+5)\). Option B is a perfect-square trinomial. Exam tip: check whether both terms are perfect squares first.

((3x-2)(5x-4)=15x^2-22x+8). In exams both signs can be negative when the constant is positive and the middle term is negative.

((4x-5)(5x+4)=20x^2+9x-20). In exams expand the option carefully to check the cross terms.

First (4x^2+4xy+y^2=(2x+y)^2), then apply difference of squares. In exams treat the group as one term.

First (x^2-4xy+4y^2=(x-2y)^2), then use (25=5^2). In exams identify the perfect-square trinomial.

The common binomial is (x+y), leaving coefficient (a-b+c). In exams take common binomials with signs.

The grouping gives (x(2y-3)+2(2y-3)). In exams take out the common bracket as soon as it appears.

Group the terms: \(6ab-9a+10b-15=3a(2b-3)+5(2b-3)\). Since \((2b-3)\) is the common binomial, the factorised form is \((3a+5)(2b-3)\). Expanding option B gives \(6ab+9a-10b-15\), which differs from the given expression. Exam tip: after factorising, expand once to check the signs of the middle terms.

Write \(3x^2+6x+5x+10\) as \((3x^2+6x)+(5x+10)\), giving \(3x(x+2)+5(x+2)\). The common binomial \((x+2)\) allows factorisation. In exams, first take the HCF from each group.

\((a+5b)^2=a^2+2(a)(5b)+25b^2=a^2+10ab+25b^2\), so option A is a perfect square. In exams, check whether the middle term equals twice the product of the terms.

First it is ((x^2-y^2)^2), and (x^2-y^2=(x-y)(x+y)). In exams continue until complete factorisation.

(x^4-y^4=(x^2-y^2)(x^2+y^2)), and (x^2-y^2) factors again. In exams check difference of squares twice.

First take out (3x), and the remaining (x^2+4x+4=(x+2)^2). In exams identify a perfect square after taking the common factor.

First take out (5a), and the remaining (a^2-4ab+4b^2=(a-2b)^2). In exams look for a perfect square inside the bracket.

First take out (2x), and (x^2-9) is a difference of squares. In exams check the inside identity after the common factor.

First take out (7), and the remaining (x^2-4xy+4y^2=(x-2y)^2). In exams identify the perfect square after the common factor.

The grouping gives (x^2(x+1)-1(x+1)), then (x^2-1) also factors. In exams check again after partial factorisation.