Factorisation

In \(x^2+10x+25\), \(10x=2\times x\times5\) and \(25=5^2\), so it becomes \((x+5)^2\). In option B, the constant term is not \(5^2\). Exam tip: match the middle term with \(2ab\).

On expanding \((5x+1)(x+3)\), we get \(5x^2+15x+x+3=5x^2+16x+3\), so it is the correct factorisation. Option B gives \(5x^2+8x+3\), so its middle-term coefficient is incorrect. Exam tip: check that the sum of the outer and inner products equals the coefficient of the middle term.

The correct factorisation is \((2x+3)(3x+2)\), because expanding it gives \(6x^2+4x+9x+6=6x^2+13x+6\). In option B, the middle term becomes \(15x\), so it is incorrect. Exam tip: expand the factors and check whether the middle-term coefficients add to the required coefficient.

Expansion gives (2a^2-6a+5a-15=2a^2-a-15). In exams, use unlike signs for a negative constant term.

\(x^2-81=x^2-9^2\), so using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-9)(x+9)\). \(x^2+81\) is a sum of squares, not a difference of squares over integers. Exam tip: check for two perfect squares with a minus sign.

This is the perfect square of the difference of (2x) and (3). In exams, check (-12x=-2\cdot2x\cdot3).

This expression is a perfect-square trinomial: \(a^2+2ab+b^2=(a+b)^2\). Here, \(a=x\) and \(b=7\), so \(2ab=2\cdot x\cdot7=14x\) and \(b^2=49\). Hence, \(x^2+14x+49=(x+7)^2\). Expanding \((x-7)^2\) gives the middle term \(-14x\), so it is incorrect. Exam tip: take the square roots of the first and last terms and check whether the middle term is \(2ab\).

A perfect-square trinomial has the form \((x+a)^2=x^2+2ax+a^2\). Thus \(b=2a\) and \(c=a^2\), giving \(b^2=4c\). Option C misses the factor 4. Exam tip: halve the middle coefficient and square it.

Group the terms: \(7x+7y+2x+2y=7(x+y)+2(x+y)\). The common factor is \((x+y)\), so the expression becomes \((7+2)(x+y)=9(x+y)\). Option D leaves out the \(2x+2y\) part. Exam tip: while factorising by grouping, first identify the common binomial factor.

Group the terms: \(3a+3b+xa+xb=3(a+b)+x(a+b)\). The common factor is \((a+b)\), so the factorised form is \((a+b)(3+x)=(3+x)(a+b)\). In option B, using \(a-b\) would change the signs of the terms containing \(b\). Exam tip: for a four-term expression, group two terms at a time and look for a common bracket.

\(m^2-36n^2=m^2-(6n)^2\), so it is a difference of perfect squares and becomes \((m-6n)(m+6n)\). Option A is a sum, while C is a perfect-square trinomial. Exam tip: check for squared terms with a minus sign between them.

(100a^2=(10a)^2), and (b^2) is already a square. In exams, take square roots and write conjugate factors.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect-square trinomial. In option B, the constant term should be \(5^2=25\), not 20. Exam tip: match the middle term with \(2ab\).

Using \((a+b)^2=a^2+2ab+b^2\), option A has the required square terms and middle term \(2ab\). Option B has only \(ab\). Exam tip: check whether the middle term is twice the product of the roots of the end terms.

The correct factorisation is \((2x+5)(x+2)\), because expanding it gives \(2x^2+4x+5x+10=2x^2+9x+10\). In option B, the middle term becomes \(21x\), so it is not correct. In exams, expand the factors to check the middle term and the constant term.

This is a perfect-square trinomial. We have \(4y^2=(2y)^2\), \(1=1^2\), and the middle term is \(2\times 2y\times 1=4y\). Hence, \(4y^2+4y+1=(2y+1)^2\). The closest distractor, \((2y-1)^2\), expands to \(4y^2-4y+1\), which has a negative middle term. Exam tip: check the sign of the middle term in \(a^2+2ab+b^2=(a+b)^2\).

\(25m^2-36n^2=(5m)^2-(6n)^2\), so it is a difference of two squares and factorises as \((5m-6n)(5m+6n)\). Option B has a plus sign. Exam tip: check for two perfect squares separated by a minus sign.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so A is a perfect-square trinomial. In B, the constant should be \(25\), not \(20\). Exam tip: compare the middle term with \(2ab\).

\(x^2+10x+25=x^2+2\cdot5\cdot x+5^2=(x+5)^2\), so it is a perfect-square trinomial. Option B lacks the required constant term \(25\). Exam tip: match the middle term with \(2ab\).

(-3-7=-10) and ((-3)(-7)=21). In exams, when the constant is positive and the middle term is negative, take both factors negative.