Factorisation

This is a perfect square trinomial because (25=5^2) and (10x=2\cdot5\cdot x). Exam tip: identify the perfect square first.

This is a difference of squares where (9a^2=(3a)^2) and (25b^2=(5b)^2). Exam tip: take square roots of both terms.

Since (3+4=7) and (3\cdot4=12), the correct factors are ((x+3)(x+4)). Exam tip: match both sum and product.

The first and last terms are perfect squares and the middle term is (2\cdot2p\cdot3q). Exam tip: confirm with the (2ab) term.

For a monic quadratic \(y^2+by+c\), find two numbers whose sum is \(b\) and whose product is \(c\). Here the middle coefficient is \(-11\) and the constant term is \(30\). The numbers \(-5\) and \(-6\) have sum \(-5-6=-11\) and product \((-5)(-6)=30\). Since both are negative, their product is positive but their sum is negative.

Therefore \(y^2-11y+30=(y-5)(y-6)\). Expanding gives \(y^2-6y-5y+30=y^2-11y+30\), so option A is correct. Option B has a positive middle coefficient. Option C uses 3 and 10, whose sum is 13 rather than 11, while option D does not provide the required pair of factors.

The direct answer is B: \((4m-5n)^2\). Use the identity \((p-q)^2=p^2-2pq+q^2\). Here \(p=4m\) and \(q=5n\). Thus \((4m-5n)^2=(4m)^2-2(4m)(5n)+(5n)^2=16m^2-40mn+25n^2\). Option B matches every coefficient and sign. Option A, \((4m+5n)^2\), would have a positive middle term \(+40mn\), so it is wrong. Option C has incorrect square roots: squaring \(16m-25n\) produces much larger coefficients and a different mixed term. Option D expands to \(16m^2-96mn-25n^2\), so it also fails. The negative middle term tells us that the two terms inside the square must have opposite signs. Memory cue: for a perfect square, check the first square, last square, and twice their product.

The formula for difference of cubes is (a^3-b^3=(a-b)(a^2+ab+b^2)). Exam tip: keep all signs positive in the second bracket.

In sum of cubes, the first factor is ((a+b)) and the second is (a^2-ab+b^2). Exam tip: remember the negative middle sign.

In option A, the first and last terms are \((2p)^2\) and \((5q)^2\). Its middle term is \(-2(2p)(5q)=-20pq\), so it equals \((2p-5q)^2\). Option B has an incorrect middle-term coefficient. Exam tip: for a perfect square trinomial, always check whether the middle term is \(\pm2ab\).

For option A, \(b^2=(-12)^2=144\) and \(4ac=4\times9\times4=144\), so it has equal linear factors: \((3x-2)^2\). Option B fails this test. Exam tip: check \(b^2=4ac\) for a perfect-square trinomial.

The direct answer is A: \((3a-2b)(4a-3b)\). Multiply the leading terms: \(3a\cdot4a=12a^2\), and the constants give \((-2b)(-3b)=6b^2\). The cross terms are \(-9ab\) and \(-8ab\), whose sum is \(-17ab\). Thus the product is exactly \(12a^2-17ab+6b^2\). Option A is correct. Option B gives leading coefficient \(12a^2\) but cross terms \(-36ab-2ab=-38ab\), so it fails. Option C expands to \(12a^2-18ab+6b^2\), not the required middle term. Option D gives positive cross terms, so the sign is wrong. Check the product, sum, and signs of the split middle terms.

The direct answer is option B: (2x+5)(y+3). Group the terms as (2xy+6x)+(5y+15). From the first group take 2x common: 2x(y+3). From the second group take 5 common: 5(y+3). The repeated bracket is (y+3), so factor it out: (2x+5)(y+3). Option B expands to 2xy+6x+5y+15, exactly the original expression. Option A expands to 2xy+3x+10y+15, so its middle terms do not match. Option C expands to 2xy+10x+3y+15, again with the coefficients interchanged. Option D expands to 2xy+5x+6y+15, so it is also different. The method is called factorisation by grouping: arrange terms so each pair produces the same bracket. Memory cue: after factoring each pair, look for one identical binomial; take it out only once.

First take (a^2+2ab+b^2=(a+b)^2), then apply difference of squares. Exam tip: apply combined identities stepwise.

The direct answer is option C: \((5x-3y)^2\). Use \(A^2-2AB+B^2\) with \(A=5x\) and \(B=3y\). Then \(A^2=25x^2\), \(2AB=2\cdot5x\cdot3y=30xy\), and \(B^2=9y^2\). Hence \(25x^2-30xy+9y^2=(5x-3y)^2\). Option A has a plus sign and would give \(+30xy\). Option B squares \(25x-9y\), producing wrong degrees and coefficients. Option C gives all three required terms exactly. Option D expands to \(25x^2-40xy-9y^2\), so it is wrong. Memory cue: a negative middle term means the two parts of the perfect square are subtracted.

This is (x^3+2^3), so use the sum of cubes formula. Exam tip: keep the middle term negative in the second bracket.

\(x^3+8=x^3+2^3\), so it is a sum of two cubes and \(a^3+b^3=(a+b)(a^2-ab+b^2)\) applies. \(x^3-8\) is a difference of cubes. Exam tip: recognise \(8=2^3\).

First take common factor (5), then factor (x^2-4). Exam tip: apply identities after taking the common factor.

Look for a factor common to every term. In \(3a^2b+6ab^2-9ab\), each term contains \(3ab\). Dividing each term by it leaves \(a\), \(2b\), and \(-3\), respectively. Therefore the expression factors as \(3ab(a+2b-3)\), which is option A.

Verification by distribution gives \(3ab\cdot a=3a^2b\), \(3ab\cdot2b=6ab^2\), and \(3ab\cdot(-3)=-9ab\). Option B leaves an extra factor of \(b\) missing from the first term. C places the original factors incorrectly inside the bracket, and D does not preserve the terms after expansion. Extracting the greatest common factor is usually the first step in factorisation.

The first three terms are the perfect square \(x^2+2xy+y^2=(x+y)^2\). The expression becomes \((x+y)^2-z^2\), which is a difference of squares. Using \(u^2-v^2=(u-v)(u+v)\), with \(u=x+y\) and \(v=z\), gives \((x+y-z)(x+y+z)\). Therefore option A is correct.

Expanding confirms it: \((x+y-z)(x+y+z)=(x+y)^2-z^2=x^2+2xy+y^2-z^2\). Option C is an equivalent intermediate form but not the complete factorisation. Option B changes \(x+y\) to \(x-y\), altering the middle term, and D does not reproduce the expression. Recognising a perfect square before applying difference of squares makes the process short and reliable.

\(9a^2=(3a)^2\) and \(4b^2=(2b)^2\). For a perfect-square trinomial, the middle term must be \(2\times 3a\times 2b=12ab\). Hence, \(9a^2+12ab+4b^2=(3a+2b)^2=(3a+2b)(3a+2b)\). In option B, the last term is not \(4b^2\), while option C does not have the required middle term \(12ab\). Exam tip: take the square roots of the first and last terms, then check the middle term using \(2pq\).

The direct answer is option A: \((2x+5)(x+2)\). For \(2x^2+9x+10\), multiply the first and last coefficients: \(2\times10=20\). We need two numbers with product 20 and sum 9; 4 and 5 work. Split the middle term: \(9x=4x+5x\). Then \(2x^2+4x+5x+10 = 2x(x+2)+5(x+2)\). Taking the common bracket gives \((2x+5)(x+2)\). Option A expands as \(2x^2+4x+5x+10=2x^2+9x+10\). Option B expands to \(2x^2+10x+2x+10=2x^2+12x+10\), so its middle coefficient is wrong. Option C gives \(x^2+9x+20\), with the leading and constant terms wrong. Option D gives \(2x^2+12x+10\), same incorrect middle coefficient as B. Always verify by multiplication. Memory cue: multiply the outside coefficients, split the middle term, then group.

In \(a^2-b^2\), both terms are perfect squares joined by subtraction, so it becomes \((a-b)(a+b)\). A sum of squares does not use this identity. Exam tip: check the sign between the squared terms first.

Write it as ((x+y)(x-y)+2(x+y)), then take ((x+y)) common. Exam tip: identify the difference of squares first.

Grouping gives (m(n+p)+q(n+p)), so the factor is ((m+q)(n+p)). Exam tip: look for the common binomial bracket.

The expression is \(8x^3+27y^3=(2x)^3+(3y)^3\). Applying the sum-of-cubes identity \(a^3+b^3=(a+b)(a^2-ab+b^2)\) gives \((2x+3y)(4x^2-6xy+9y^2)\). Hence, option A is correct. Option B is associated with the difference-of-cubes identity \(a^3-b^3\), so it does not apply here. Exam tip: in a sum of cubes, the middle term of the second factor is negative.