Factorise x^2 + (2a + 5)x + a^2 + 5a - 6.

Step-by-step solution

1. 1

   Organise the expression in one variable. The expression is already written as a quadratic in $x$: the coefficient of $x$ is $2a+5$ and the constant term is $a^2+5a-6$. Treating $a$ as a fixed number turns this into an ordinary sum-and-product factorisation.

2. 2

   Factor the constant term. Two numbers with product $-6$ and sum $5$ are $6$ and $-1$:

   $a^{2}+5a-6=(a+6)(a-1)$

3. 3

   Test whether those two pieces add to the middle coefficient.

   $(a+6)+(a-1)=2a+5$

   This matches the coefficient of $x$ exactly, which is the signal that the quadratic factors as $(x+p)(x+q)$ with $p=a+6$ and $q=a-1$.

4. 4

   Write down the factorisation.

   $x^{2}+\big[(a+6)+(a-1)\big]x+(a+6)(a-1)=(x+a+6)(x+a-1)$

   $\boxed{(x+a+6)(x+a-1)}$

5. 5

   Expand to verify. $(x+a+6)(x+a-1)=x^2+(a-1)x+(a+6)x+(a+6)(a-1)=x^2+(2a+5)x+a^2+5a-6$ — the original expression.