1. Factorise the quadratic expression $x^2 + 5x + 6$. 2. Factorise the difference of squares $9y^2 - 16$. 3. Factorise the expression $2x^2 + 7x + 3$. 4. Factorise the expression $x^3 - 27$ using the difference of cubes formula. 5. Factorise the expression $x^2 - 4x + 4$. --- **Step 1: Factorise $x^2 + 5x + 6$** The problem is to factorise the quadratic expression $x^2 + 5x + 6$. The formula for factorising a quadratic $ax^2 + bx + c$ is to find two numbers that multiply to $ac$ and add to $b$. Here, $a=1$, $b=5$, $c=6$. We look for two numbers that multiply to $1 \times 6 = 6$ and add to $5$. These numbers are $2$ and $3$. So, we write: $$x^2 + 5x + 6 = x^2 + 2x + 3x + 6$$ Group terms: $$= (x^2 + 2x) + (3x + 6)$$ Factor each group: $$= x(x + 2) + 3(x + 2)$$ Factor out the common binomial: $$= (x + 3)(x + 2)$$ --- **Step 2: Factorise $9y^2 - 16$** This is a difference of squares: $a^2 - b^2 = (a - b)(a + b)$. Here, $9y^2 = (3y)^2$ and $16 = 4^2$. So: $$9y^2 - 16 = (3y - 4)(3y + 4)$$ --- **Step 3: Factorise $2x^2 + 7x + 3$** Find two numbers that multiply to $2 \times 3 = 6$ and add to $7$. These are $6$ and $1$. Rewrite: $$2x^2 + 6x + x + 3$$ Group: $$(2x^2 + 6x) + (x + 3)$$ Factor: $$2x(x + 3) + 1(x + 3)$$ Factor out common binomial: $$(2x + 1)(x + 3)$$ --- **Step 4: Factorise $x^3 - 27$** Use difference of cubes formula: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ Here, $a = x$, $b = 3$. So: $$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$$ --- **Step 5: Factorise $x^2 - 4x + 4$** Look for two numbers that multiply to $4$ and add to $-4$. These are $-2$ and $-2$. So: $$x^2 - 4x + 4 = (x - 2)(x - 2) = (x - 2)^2$$