Question 587930
{{{x2+x-(a+2)(a+1)=0}}}
If you had {{{x2+(b+c)x+bc=0}}} (it would be confusing if I used a),
you would factor to solve and have
{{{(x+b)(x+c)=0}}} with solutions {{{x=-b}}} and {{{x=-c}}}
The constant (independent term) in a quadratic (degree 2) polynomial is the product of the constants (independent terms) of the factors.
The coefficient of the term is x is the sum of those same factors.
The trick to factoring is to find the pair of factors for the constant that add to the coefficient of the term is x.
The constant in your equation is {{{ -(a+2)(a+1)}}}
The coefficient of x in your equation is 1, and
{{{1=(a+2)-(a+1)}}}
So your factors are {{{a+2}}} and {{{-(a+1)}}}.
Your quadratic polynomial factors like this:
{{{x2+x-(a+2)(a+1)=(x+(a+2))(x-(a+1))}}}
Your equation is equivalent to
{{{(x+(a+2))(x-(a+1))=0}}}
and the solutions are
{{{x=a+1}}} and {{{x=-(a+2)=-a-2}}}