Vleasekeep practising... Question 1 $$ 1.1 \quad $$ Determine the product of the following and simplify fully: $$ 1.1 .1 \quad a b^{2}\left(-2 a^{2}+4 b ight) $$ $$ 1.1 .2 \quad(x-2)\left(x^{2}+2 x+8 ight) $$ Factorise the following expressions fully: $$ 1.2 .1 \quad 2 x^{2}+7 x-4 $$ $$ 1.2 .2 \quad a^{2} x-a y-b^{2} x+b y $$ $$ 1.2 .3 \quad 125-27 m^{3} $$ Simplify the following expressions fully: $$ 1.3 \quad 25 x^{4} y^{-2} imes\left(5 x^{3} y^{0} ight)^{-2} $$ $$ 1.3 .2 \quad \frac{x^{2}+7 x+10}{x+5} \div \frac{x+2}{5} $$ $$ 1.3 .3 \quad \frac{9.3^{x+2}+5.3^{x}}{3^{x}-3^{x+1}} $$

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Vleasekeep practising... Question 1 $$ 1.1 \quad $$ Determine the product of the following and simplify fully: $$ 1.1 .1 \quad a b^{2}\left(-2 a^{2}+4 bight) $$ $$ 1.1 .2 \quad(x-2)\left(x^{2}+2 x+8ight) $$ Factorise the following expressions fully: $$ 1.2 .1 \quad 2 x^{2}+7 x-4 $$ $$ 1.2 .2 \quad a^{2} x-a y-b^{2} x+b y $$ $$ 1.2 .3 \quad 125-27 m^{3} $$ Simplify the following expressions fully: $$ 1.3 \quad 25 x^{4} y^{-2} 	imes\left(5 x^{3} y^{0}ight)^{-2} $$ $$ 1.3 .2 \quad \frac{x^{2}+7 x+10}{x+5} \div \frac{x+2}{5} $$ $$ 1.3 .3 \quad \frac{9.3^{x+2}+5.3^{x}}{3^{x}-3^{x+1}} $$

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**1.1 Determine the product and simplify fully**

**1.1.1**  
We have  
$$
a b^{2}\left(-2 a^{2}+4 b\right)
$$  
Multiply term‐by‐term:  
$$
a b^2 \times (-2a^2) = -2a^3b^2,\qquad a b^2 \times 4b = 4ab^3.
$$  
Thus, the product is  
$$
-2a^3b^2+4ab^3.
$$

**1.1.2**  
We have  
$$
(x-2)\left(x^{2}+2x+8\right).
$$  
Distribute $$ (x-2) $$ across the trinomial:  
$$
x\left(x^{2}+2x+8\right) = x^3+2x^2+8x,
$$  
$$
-2\left(x^{2}+2x+8\right) = -2x^2-4x-16.
$$  
Combine like terms:  
$$
x^3+(2x^2-2x^2)+(8x-4x)-16 = x^3+4x-16.
$$

---

**1.2 Factorise the following expressions fully**

**1.2.1**  
Factorise  
$$
2x^{2}+7x-4.
$$  
We look for two numbers that multiply to $$2 \times (-4) = -8$$ and add to $$7$$. These numbers are $$8$$ and $$-1$$. Rewrite the middle term:  
$$
2x^2+8x-x-4.
$$  
Group terms:  
$$
(2x^2+8x)+(-x-4).
$$  
Factor each group:  
$$
2x(x+4) -1(x+4).
$$  
Factor out the common binomial factor $$ (x+4) $$:  
$$
(x+4)(2x-1).
$$

**1.2.2**  
Factorise  
$$
a^{2}x - ay - b^{2}x + by.
$$  
Group the terms:  
$$
(a^{2}x - b^{2}x) + (-ay+by).
$$  
Factor out common factors:  
$$
x(a^2-b^2) + y(-a+b).
$$  
Note that $$a^2-b^2=(a-b)(a+b)$$ and $$-a+b=-(a-b)$$. Hence,  
$$
x(a-b)(a+b) - y(a-b).
$$  
Factor out the common factor $$ (a-b) $$:  
$$
(a-b)\left( x(a+b) - y\right).
$$

**1.2.3**  
Factorise  
$$
125-27m^{3}.
$$  
Recognize that  
$$
125=(5)^3,\qquad 27m^3=(3m)^3.
$$  
This is a difference of cubes. Recall the identity  
$$
a^3-b^3=(a-b)(a^2+ab+b^2).
$$  
Here, $$a=5$$ and $$b=3m$$. Thus,  
$$
125-27m^3=(5-3m)(5^2+5(3m)+(3m)^2).
$$  
Simplify inside the second factor:  
$$
5^2=25,\quad 5(3m)=15m,\quad (3m)^2=9m^2.
$$  
So, the factorised form is  
$$
(5-3m)(25+15m+9m^2).
$$

---

**1.3 Simplify the following expressions fully**

**1.3.1**  
Simplify  
$$
25x^{4}y^{-2}\times\left(5x^{3}y^{0}\right)^{-2}.
$$  
Since $$ y^0 = 1 $$, we have  
$$
\left(5x^{3}\right)^{-2} = 5^{-2}x^{-6} = \frac{1}{25}x^{-6}.
$$  
Then,  
$$
25x^4y^{-2} \times \frac{1}{25}x^{-6} = x^{4-6}y^{-2} = x^{-2}y^{-2}.
$$  
Writing with positive exponents:  
$$
\frac{1}{x^2y^2}.
$$

**1.3.2**  
Simplify  
$$
\frac{x^{2}+7x+10}{x+5}\div \frac{x+2}{5}.
$$  
First, factorise the quadratic in the numerator:  
$$
x^2+7x+10=(x+5)(x+2).
$$  
So the expression becomes  
$$
\frac{(x+5)(x+2)}{x+5}\div \frac{x+2}{5}.
$$  
Cancel the common factor $$x+5$$ (provided $$x\neq -5$$):  
$$
x+2 \div \frac{x+2}{5}.
$$  
Then, divide by a fraction by multiplying by its reciprocal, cancelling $$x+2$$ (provided $$x\neq -2$$):  
$$
(x+2)\times\frac{5}{x+2}=5.
$$

**1.3.3**  
Simplify  
$$
\frac{9\cdot3^{x+2}+5\cdot3^{x}}{3^{x}-3^{x+1}}.
$$  
Rewrite the terms in the numerator:  
$$
9\cdot3^{x+2} = 9\cdot3^{2}\cdot3^x = 9\cdot9\cdot3^x = 81\cdot3^x.
$$
Thus, the numerator becomes:  
$$
81\cdot3^x+5\cdot3^x = (81+5)3^x = 86\cdot3^x.
$$
Now, factor the denominator:  
$$
3^{x}-3^{x+1} =3^x-3\cdot3^x = 3^x(1-3)= -2\cdot3^x.
$$
Cancel the common factor $$3^x$$ (since $$3^x\neq 0$$):  
$$
\frac{86\cdot3^x}{-2\cdot3^x}=\frac{86}{-2}=-43.
$$

**1.1 Determine the product and simplify fully** **1.1.1** $$ -2a^3b^2 + 4ab^3 $$ **1.1.2** $$ x^3 + 4x - 16 $$ --- **1.2 Factorise the following expressions fully** **1.2.1** $$ (x + 4)(2x - 1) $$ **1.2.2** $$ (a - b)(x(a + b) - y) $$ **1.2.3** $$ (5 - 3m)(25 + 15m + 9m^2) $$ --- **1.3 Simplify the following expressions fully** **1.3.1** $$ \frac{1}{x^2y^2} $$ **1.3.2** $$ 5 $$ **1.3.3** $$ -43 $$

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