1 Factorise the following expressions fully: 1.1.1 4x-x3 1.1.2 x2+15x-54 1.1.3 y-xy+x-1 1.2 Simplify the following expressions fully: 1.2.1 (x+2)(x2-x+3) 1.2.2 5/(x+3) - 3/(2-x) 1.2.3 (25^x 15^(x+1))/(3^x 5^(-x)) 1.3 Determine the value of (3p+q)2 if 9p2+q2=12 and pq=-3.

Here are the solutions for Question 1 from the image (DBE NOV 18 Q1):

1.1 Factorise the following expressions fully:

1.1.1 $4x-x^3$ Step 1: Factor out the common factor $x$. $x(4-x^2)$ Step 2: Factor the difference of squares $4-x^2 = (2-x)(2+x)$. $x(2-x)(2+x)$ The fully factorised expression is $\boxed{x(2-x)(2+x)}$.

1.1.2 $x^2+15x-54$ Step 1: Find two numbers that multiply to $-54$ and add to $15$. These numbers are $18$ and $-3$. Step 2: Factor the quadratic trinomial. $(x+18)(x-3)$ The fully factorised expression is $\boxed{(x+18)(x-3)}$.

1.1.3 $y-xy+x-1$ Step 1: Group terms and factor out common factors from each group. $y(1-x) + (x-1)$ Step 2: Factor out $-1$ from the second group to get a common binomial factor. $y(1-x) - (1-x)$ Step 3: Factor out the common binomial factor $(1-x)$. $(1-x)(y-1)$ The fully factorised expression is $\boxed{(1-x)(y-1)}$.

1.2 Simplify the following expressions fully:

1.2.1 $(x+2)(x^2-x+3)$ Step 1: Expand the product by distributing each term from the first parenthesis to the second. $x(x^2-x+3) + 2(x^2-x+3)$ Step 2: Perform the multiplications. $x^3 - x^2 + 3x + 2x^2 - 2x + 6$ Step 3: Combine like terms. $x^3 + x^2 + x + 6$ The simplified expression is $\boxed{x^3+x^2+x+6}$.

1.2.2 $\frac{5}{x+3} - \frac{3}{2-x}$ Step 1: Find a common denominator, which is $(x+3)(2-x)$. $\frac{5(2-x)}{(x+3)(2-x)} - \frac{3(x+3)}{(2-x)(x+3)}$ Step 2: Combine the numerators over the common denominator. $\frac{5(2-x) - 3(x+3)}{(x+3)(2-x)}$ Step 3: Expand the numerator. $\frac{10 - 5x - 3x - 9}{(x+3)(2-x)}$ Step 4: Combine like terms in the numerator. $\frac{1 - 8x}{(x+3)(2-x)}$ The simplified expression is $\boxed{\frac{1-8x}{(x+3)(2-x)}}$.

1.2.3 $\frac{25^x \cdot 15^{x+1}}{3^x \cdot 5^x}$ Step 1: Express the bases in terms of their prime factors ($25=5^2$, $15=3 \cdot 5$). $\frac{(5^2)^x \cdot (3 \cdot 5)^{x+1}}{3^x \cdot 5^x}$ Step 2: Apply exponent rules $(a^m)^n = a^{mn}$ and $(ab)^n = a^n b^n$. $\frac{5^{2x} \cdot 3^{x+1} \cdot 5^{x+1}}{3^x \cdot 5^x}$ Step 3: Combine terms with the same base in the numerator using $a^m \cdot a^n = a^{m+n}$. $\frac{3^{x+1} \cdot 5^{2x+(x+1)}}{3^x \cdot 5^x} = \frac{3^{x+1} \cdot 5^{3x+1}}{3^x \cdot 5^x}$ Step 4: Apply the exponent rule $\frac{a^m}{a^n} = a^{m-n}$ for each base. $3^{(x+1)-x} \cdot 5^{(3x+1)-x}$ Step 5: Simplify the exponents. $3^1 \cdot 5^{2x+1}$ $3 \cdot 5^{2x+1}$ The simplified expression is $\boxed{3 \cdot 5^{2x+1}}$.

1.3 Determine the value of $(3p+q)^2$ if $9p^2+q^2=12$ and $pq=-3$. Step 1: Expand the expression $(3p+q)^2$. $(3p+q)^2 = (3p)^2 + 2(3p)(q) + q^2$ $= 9p^2 + 6pq + q^2$ Step 2: Rearrange the terms to group the given values. $= (9p^2 + q^2) + 6pq$ Step 3: Substitute the given values $9p^2+q^2=12$ and $pq=-3$. $= 12 + 6(-3)$ Step 4: Perform the arithmetic. $= 12 - 18$ $= -6$ The value of $(3p+q)^2$ is $\boxed{-6}$.