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

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**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
**5.1** $$ 3x(3x-1)-x(3x-1) = 2x(3x-1) $$ **5.2** $$ -mn^5 + mn = mn(1 - n^4) = mn(1 - n)(1 + n)(1 + n^2) $$ **5.3** $$ 30a^2 - 8ab - 18b^2 = 2(15a^2 - 4ab - 9b^2) $$ **5.4** $$ 9m^4 - 85m^2n^2 + 36n^4 = (3m - 2n)(3m + 2n)(m - 3n)(m + 3n) $$ **5.5** $$ 49x^2 + 4y^2 - 16x^2y^2 - 28xy = (7x - 2y - 4xy)(7x - 2y + 4xy) $$ **5.6** $$ 9r^4 - 243rs^3 = 9r(r^3 - 27s^3) = 9r(r - 3s)(r^2 + 3rs + 9s^2) $$

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