Factorisation

First (x) is common and (2x^2+7x+3=(2x+1)(x+3)). In exams, factor the trinomial after taking the common factor.

First take (a) common and then factor (4a^2-9b^2) by difference of squares. In exams, factorise completely.

Let \(x^2+px+q=(x+a)^2\). Comparing coefficients gives \(p=2a\) and \(q=a^2\), hence \(p^2=(2a)^2=4q\). The condition \(p^2=q\) does not match the middle-term coefficient. Exam tip: square the coefficient of the middle term and compare it with four times the constant term.

The expression can be written as \((a+b)^2-4c^2=(a+b)^2-(2c)^2\). Applying \(x^2-y^2=(x-y)(x+y)\), where \(x=a+b\) and \(y=2c\), gives \((a+b-2c)(a+b+2c)\). Option B would give \((4c)^2=16c^2\), so it is incorrect. Exam tip: treat a complete bracket such as \((a+b)\) as one term when applying an identity.

Write it as (x^2-4y^2+2(x+2y)), so ((x+2y)) is common. In exams, group after difference of squares.

\(x^2-49=x^2-7^2\), which matches \(a^2-b^2=(a-b)(a+b)\). Therefore, it factorises as \((x-7)(x+7)\). Option C is a perfect-square trinomial, \((x+7)^2\), not a difference of squares. Exam tip: check that both terms are perfect squares with a minus sign between them.

Direct answer: Option B, (p-q)(p+q-5). First use p²-q²=(p-q)(p+q). The remaining terms are -5p+5q=-5(p-q), because factoring out -5 from -5p+5q leaves p-q. Thus the expression becomes (p-q)(p+q)-5(p-q). Taking the common factor (p-q) gives (p-q)(p+q-5). Option A has p+q as the common factor, but the linear terms are not divisible by p+q in this way. Option B reproduces p²-q²-5p+5q when expanded. Option C gives +5p-5q after expansion, so both signs are wrong. Option D is algebraically the same product as A, and is therefore also wrong. Memory cue: first factor the difference of squares, then factor the common (p-q) from the last two terms.

The expression simplifies to (x^2+4x+3), so the factors are ((x+1)(x+3)). First simplify the given terms carefully.

For \(x^2+x+1\), the discriminant is \(b^2-4ac = 1-4 = -3\), not a perfect square, so it has no linear factors with integer coefficients. The others factor as a difference of squares, \((x+2)(x+3)\), and \((x-2)^2\). Exam tip: check the discriminant first.

For \(x^2+7x+10\), the discriminant is \(b^2-4ac=49-40=9\), a perfect square. Its roots are \(-5\) and \(-2\), so it factors as \((x+5)(x+2)\). In exams, check whether the discriminant is a perfect square.

A perfect-square trinomial has the form \(a^2-2ab+b^2=(a-b)^2\). Here, \(9p^2=(3p)^2\), \(4q^2=(2q)^2\), and \(-12pq=-2(3p)(2q)\). Therefore, \(9p^2-12pq+4q^2=(3p-2q)^2\). In option B, the last term is \(2q^2\), not \(4q^2\), so it is not a perfect square. Exam tip: take the square roots of the first and last terms, then check whether the middle term is \(\pm2ab\).

First take (4) common and (3x^2-5xy+2y^2=(3x-2y)(x-y)). In exams, factor the trinomial after the common factor.

Treat it as ((x^2-y^2)+5(x-y)+6) and use (u=x-y). In exams, keep the remaining (x+y) term carefully.

In \(ax+ay+bx+by=a(x+y)+b(x+y)\), both groups contain the common binomial \((x+y)\), giving \((a+b)(x+y)\). In option B, the binomials are different. Exam tip: group terms in pairs and check for an identical factor.

The direct answer is option A: \(4xy\). Use the difference-of-squares identity with \(A=x+y\) and \(B=x-y\): \(A^2-B^2=(A-B)(A+B)\). Now \(A-B=(x+y)-(x-y)=2y\), and \(A+B=(x+y)+(x-y)=2x\). Their product is \(2y\cdot2x=4xy\). Option A is therefore correct. Option B is too small by a factor of two. Option C, \(x^2-y^2\), is the result of a different expression and does not account for both given squares. Option D gives the sum-like result \(2x^2+2y^2\), because it incorrectly treats the cross terms as if they had the same sign. Memory cue: subtracting the conjugate squares leaves four times the product.

A perfect-square trinomial has the form \(a^2+2ab+b^2\). In option A, \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect square. In option B, the middle term is \(10x\), but the constant term is \(20\) rather than \(25\), so it is not a perfect square. Exam tip: take the square roots of the first and last terms and check whether the middle term is twice their product.

Difference of squares directly gives (A-B) and (A+B). In exams, you may simplify it to ((2a-3b)(4a-b)).

The direct answer is option A: \(x+2y-3z)(x+2y+3z)\). Rearrange the first three terms as \(x^2+4xy+4y^2=(x+2y)^2\). The last term is \(-9z^2=-(3z)^2\). Thus the expression is \( (x+2y)^2-(3z)^2\). Applying \(A^2-B^2=(A-B)(A+B)\) gives option A. Option B uses \(x-2y\), whose square has \(-4xy\), but the expression has \(+4xy\). Option C adds \(9z^2\) instead of subtracting it, so it has the wrong sign. Option D is not arranged as the difference-of-squares product and does not give the same factorisation. Memory cue: check the middle sign before choosing the perfect square.

The first three terms form a perfect square. Since \((4a)^2=16a^2\), \((3b)^2=9b^2\), and the middle term of \((4a−3b)^2\) is \(−2(4a)(3b)=−24ab\), we have \(16a^2−24ab+9b^2=(4a−3b)^2\). Also, \(25c^2=(5c)^2\). The expression is therefore \((4a−3b)^2−(5c)^2\). Applying the difference-of-squares identity gives \((4a−3b−5c)(4a−3b+5c)\), option A.

Multiplying these factors confirms the answer because they are conjugates: their product is \((4a−3b)^2−25c^2\), exactly the original expression. The plus sign between 4a and 3b would produce a positive \(24ab\) term, so option B cannot work. Option D lacks the subtraction of \(25c^2\), and option C does not follow from either a common factor or a standard identity.

Writing it as ((x+y)^2+4(x+y)-4z^2) does not directly factor as option (D). Verify by expansion carefully.

In \(x^2+10x+25\), \(25=5^2\) and the middle term is \(2\cdot x\cdot5=10x\). Hence it is \((x+5)^2\). The expression ending in 20 does not fit this pattern. Exam tip: verify \(b^2=4ac\).

The direct answer is option A: \((3a+b)(a-2b)\). Multiply term by term: \(3a\cdot a=3a^2\), \(3a\cdot(-2b)=-6ab\), \(b\cdot a=ab\), and \(b\cdot(-2b)=-2b^2\). Combining the middle terms gives \(-6ab+ab=-5ab\), so the result is \(3a^2-5ab-2b^2\). Option A is exactly correct. Option B gives \(3a^2+6ab-ab-2b^2=3a^2+5ab-2b^2\), so the middle sign is wrong. Option C gives \(3a^2-2ab-b^2\), with both the middle and last coefficients wrong. Option D gives \(3a^2+3ab-2ab-2b^2=3a^2+ab-2b^2\), not the required middle term. Memory cue: always multiply all four pairs and combine only like terms.

In option A, \(49a^2=(7a)^2\) and \(64b^2=(8b)^2\). Thus, \(49a^2-64b^2=(7a)^2-(8b)^2\), which is a difference of two perfect squares. Its factorisation is \((7a-8b)(7a+8b)\). Option B is a sum of squares, while in options C and D the second term is not a perfect square. Exam tip: before using \(x^2-y^2=(x-y)(x+y)\), verify that both terms are perfect squares.

The direct answer is A: \((2x+y)(2x+y-3)\). Group the first three terms: \(4x^2+4xy+y^2=(2x+y)^2\). The remaining terms are \(-6x-3y=-3(2x+y)\). Taking the common factor \(2x+y\) gives \((2x+y)^2-3(2x+y)=(2x+y)(2x+y-3)\). Option A follows exactly. Option B changes the sign of \(y\), so its first product is not the given square. Option C is a square but expands differently and does not represent the expression. Option D gives different quadratic and linear terms. The useful method is to notice the same binomial in the square and in the remaining terms.

First (x^2-2xy+y^2=(x-y)^2) and the remaining part is (-4(x-y)). In exams, treat the common binomial as one term.