Factorisation

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

The first three terms are a perfect square because \(4x^2=(2x)^2\), \(y^2=y^2\), and the middle term is \(4xy=2(2x)(y)\). Thus \(4x^2+4xy+y^2=(2x+y)^2\). The last term, \(9z^2\), is \((3z)^2\), so the expression is a difference of squares.

Using \(A^2-B^2=(A-B)(A+B)\), with \(A=2x+y\) and \(B=3z\), gives \((2x+y-3z)(2x+y+3z)\). This is the complete factorisation, so option A is correct. Option C stops before splitting the difference of squares and is only a partial form. Option B changes the sign of the \(4xy\) term, and option D uses the wrong square root of \(9z^2\).

The direct answer is A: \(x-2y-5)(x-2y+5)\). First, \(x^2-4xy+4y^2=(x-2y)^2\), because the middle term is \(-2(x)(2y)=-4xy\). Therefore the full expression is \((x-2y)^2-25=(x-2y)^2-5^2\). The difference-of-squares identity gives option A. Option B uses \(x+2y\), which would produce a positive \(4xy\). Option C is only an intermediate form, not fully factorised into two factors. Option D does not have equal square roots and expands incorrectly. Thus A is the complete correct factorisation. Find the perfect-square trinomial before factoring the final difference.

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

Grouping works by arranging terms so that the same bracket appears in both groups. For \(2xy-3x+4y-6\), group the first two terms and the last two terms: \(x(2y-3)+2(2y-3)\). Both groups contain the identical factor \(2y-3\), so it can be taken outside the bracket.

Taking out this common factor gives \( (x+2)(2y-3)\), which is option A. Expanding confirms the result: \(x(2y-3)+2(2y-3)=2xy-3x+4y-6\). Option B changes the signs and would produce different terms. Option C groups the expression incorrectly, and option D has factors whose product does not reproduce the original four terms. The repeated bracket is the key recognition step.

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.

The expression is a perfect square because it matches \(u^2-2uv+v^2=(u-v)^2\), with \(u=x^2\) and \(v=y^2\). Thus it first becomes \((x^2-y^2)^2\). Since \(x^2-y^2=(x-y)(x+y)\), the complete factorisation is \((x-y)^2(x+y)^2\), so option B is correct.

The check is direct: \((x-y)^2(x+y)^2=[(x-y)(x+y)]^2=(x^2-y^2)^2\), which expands to \(x^4-2x^2y^2+y^4\). Option A is an equivalent intermediate form but is not fully factorised. Option C has the wrong sign in the middle term, and D omits one copy of each factor. Continue factoring whenever a difference of squares remains.

(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.

The grouping gives (x^2(x-4)-1(x-4)), and (x^2-1) factors. In exams look for both common bracket and difference of squares.

The grouping gives (x^2(2x+3)-4(2x+3)), and (x^2-4) factors. In exams factorise the remaining binomial after grouping.

The grouping gives (x^2(3x-5)+4(3x-5)). In exams do not factor (x^2+4) into real linear factors.

Grouping ((a^3-b^3)+ab(a-b)) gives ((a-b)(a^2+ab+b^2+ab)). In exams check whether the final bracket forms a perfect square.

\((p+3q)^2=p^2+2(p)(3q)+(3q)^2=p^2+6pq+9q^2\), so A is correct. In C, the middle term is negative, so it matches \((a-b)^2\), not \((a+b)^2\). Exam tip: compare the middle term with \(2ab\).

In \(p^2+14p+49\), \(49=7^2\) and the middle term \(14p=2\times p\times7\). Hence it is \((p+7)^2\). In exams, check whether the middle term equals \(2ab\).

By the Factor Theorem, \(x+2\) is a factor only if the polynomial becomes 0 at \(x=-2\). For option A, \(-8+12-4=0\). Option B gives \(-4\), so it is not divisible by \(x+2\). Exam tip: substitute the zero of the linear factor.

First (4x^2-12xy+9y^2=(2x-3y)^2), then apply difference of squares. In exams take the square root of the group correctly.

First (9a^2+12ab+4b^2=(3a+2b)^2), then use (16c^2=(4c)^2). In exams apply difference of squares after forming the perfect square group.

(x^3+8y^3=(x+2y)(x^2-2xy+4y^2)) and (x^2y+2xy^2=xy(x+2y)). In exams simplify inside terms after a common binomial appears.