Factorise: x2 + 5x + 6. A. (x+1)(x+6) B. (x+2)(x+3) C. (x+1)(x+5) D. (x+3)(x+2)

The expression is in the form $ax^2 + bx + c$, where $a=1$, $b=5$, and $c=6$.

We are looking for two numbers, let's call them $p$ and $q$, such that:
$p \times q = 6$
$p + q = 5$

• Factors of 6: $(1, 6)$, $(-1, -6)$, $(2, 3)$, $(-2, -3)$.
• Sums of factors: $1 + 6 = 7$ (not 5) $-1 + (-6) = -7$ (not 5) $2 + 3 = 5$ (This is the correct sum) $-2 + (-3) = -5$ (not 5)

The two numbers are 2 and 3. So, the factored form of the expression is $(x+2)(x+3)$.

A. $(x+1)(x+6)$
B. $(x+2)(x+3)$
C. $(x+1)(x+5)$
D. $(x+3)(x+2)$

Both options B and D are equivalent to $(x+2)(x+3)$ due to the commutative property of multiplication. We will select option B.

The final answer is $\boxed{\text{B. } (x+2)(x+3)}$.