Vleasekeep practising... Question 1 \( 1.1 \quad \) Determine the product of the following and simplify fully: \( 1.1 .1 \quad a b^{2}\left(-2 a^{2}+4 b\right) \) \( 1.1 .2 \quad(x-2)\left(x^{2}+2 x+8\right) \) 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: \( 1.3 \quad 25 x^{4} y^{-2} \times\left(5 x^{3} y^{0}\right)^{-2} \) \( 1.3 .2 \quad \frac{x^{2}+7 x+10}{x+5} \div \frac{x+2}{5} \) \( 1.3 .3 \quad \frac{9.3^{x+2}+5.3^{x}}{3^{x}-3^{x+1}} \)

**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 \(3^x\neq 0\)): \[ \frac{86\cdot3^x}{-2\cdot3^x}=\frac{86}{-2}=-43. \]