“FOIL”
\((x+a)(x+b)\)
\(=x^2+(b+a)x +ab\)
“Difference of Squares”
\(a^2-b^2= (a+b)(a-b)\)
\(a^2+2ab+b^2=(a+b)(a+b) \)
\(a^2-2ab+b^2=(a-b)(a-b)\)
“Reverse FOIL”
\(x^2+(b+a)x+ab=\) \((x+a)(x+b)\)
You can use Reverse FOIL to factor a polynomial by thinking about two numbers a and b which add to the number in front of the x, and which multiply to give the constant. For example, to factor \(x^2+5x+6\), the numbers add to 5 and multiply to 6, i.e.:
\(a=2\) and \(b=3\), so that \(x^2+5x+6=(x+2)(x+3)\).
To solve a quadratic such as \(x^2+bx+c=0\), first factor the left side to get \((x+a)(x+b)=0\), then set each part in parentheses equal to zero.
For example, \(x^2+4x+3= (x+3)(x+1)=0\) so that \(x=-3\) or \(x=-1\).
To solve two linear equations in x and y: use the first equation to substitute for a variable in the second. E.g., suppose \(x+y=3\) and \(4x-y=2\). The first equation gives y=3-x, so the second equation becomes \(4x-(3-x)=2 ⇒ 5x-3=2\) \(⇒ x=1,y=2\).