QUESTION 5 Factorise the following: \( 5.1 \quad 3 x(3 x-1)-x(3 x-1) \) \( 5.2 \quad-m n^{5}+m n \) \( 5.3 \quad 30 \mathrm{a}^{2}-8 a b-18 b^{2} \) \( 5.4 \quad 9 m^{4}-85 m^{2} n^{2}+36 n^{4} \) \( 5.5 \quad 49 x^{2}+4 y^{2}-16 x^{2} y^{2}-28 x y \) \( 5.6 \quad 9 \mathrm{r}^{4}-243 r s^{3} \) \( 5.7 \quad \frac{(\mathrm{~m}-1)^{3}+1}{3} \)

**5.1** We have \[ 3x(3x-1)-x(3x-1). \] Notice that both terms share the common factor \(x(3x-1)\). Factoring this out gives \[ x(3x-1)(3-1)=2x(3x-1). \] **5.2** The expression is \[ -mn^5 + mn. \] Factor out the common factor \(mn\): \[ mn(-n^4 + 1)=mn(1-n^4). \] Now, recognize that \(1-n^4\) is a difference of squares: \[ 1-n^4=(1-n^2)(1+n^2), \] and further, \[ 1-n^2=(1-n)(1+n). \] Thus the complete factorization is \[ mn(1-n)(1+n)(1+n^2). \] **5.3** Consider \[ 30a^2-8ab-18b^2. \] We first factor out the common factor \(2\): \[ 2(15a^2-4ab-9b^2). \] The quadratic \(15a^2-4ab-9b^2\) does not factor further over the integers. Hence the answer is \[ 2(15a^2-4ab-9b^2). \] **5.4** We wish to factor \[ 9m^4-85m^2n^2+36n^4. \] View it as a quadratic in \(m^2\) by setting \(u=m^2\). Then we have \[ 9u^2-85n^2u+36n^4. \] We look for a splitting with numbers whose product is \(9\cdot36=324\) and whose sum is \(85\) (in fact we need \(-85\), so use \(-4n^2\) and \(-81n^2\) since \(-4n^2-81n^2=-85n^2\) and \((-4n^2)(-81n^2)=324n^4\)). Write: \[ 9m^4-4m^2n^2-81m^2n^2+36n^4. \] Group the terms: \[ \bigl(9m^4-4m^2n^2\bigr)-\bigl(81m^2n^2-36n^4\bigr). \] Factor each group: \[ m^2(9m^2-4n^2)-9n^2(9m^2-4n^2). \] Then factor out the common factor \((9m^2-4n^2)\): \[ (9m^2-4n^2)(m^2-9n^2). \] Both factors are differences of squares. Factor them further: \[ 9m^2-4n^2=(3m-2n)(3m+2n), \] \[ m^2-9n^2=(m-3n)(m+3n). \] Thus, the complete factorisation is \[ (3m-2n)(3m+2n)(m-3n)(m+3n). \] **5.5** We have \[ 49x^2+4y^2-16x^2y^2-28xy. \] First, rearrange the terms: \[ 49x^2 - 16x^2y^2 - 28xy + 4y^2. \] Notice that the first three terms can be grouped as: \[ 49x^2 - 28xy + 4y^2 = (7x-2y)^2. \] So the expression becomes \[ (7x-2y)^2- (4xy)^2. \] This is a difference of two squares, which factors as \[ \bigl((7x-2y)-4xy\bigr)\bigl((7x-2y)+4xy\bigr). \] That is, \[ (7x-2y-4xy)(7x-2y+4xy). \] **5.6** The expression is \[ 9r^4-243rs^3. \] Factor out the common factor \(9r\): \[ 9r\Bigl(r^3-27s^3\Bigr