3r^2 - 8r -3=0

2r^2 + r - 6=0

3r^2 - 5r - 2 = 0

Trinomials are slightly more difficult to factor when the leading coefficient is not 1. There are a couple of ways to go about doing this. Either way, you should start by factoring out the greatest common factor if you can (in these problems, the GCF is 1, so we don't have to worry about that).

One method is to use trial and error. Take \(\displaystyle 3r^2-8r-3\). The only way we can factor 3 (over the integers) is \(\displaystyle 3\cdot1\). So we look for a factorization of the form \(\displaystyle (3r+m)(r+n)\). Now our goal is find \(\displaystyle m, n\) such that \(\displaystyle mn = -3\) and \(\displaystyle m+3n=-8\). From trying a few values for \(\displaystyle m\) and \(\displaystyle n\), we can see that \(\displaystyle m=1\) and \(\displaystyle n=-3\) meets this goal. So

\(\displaystyle 3r^2-8r-3=(3r+1)(r-3)\)

There's another method that uses grouping. First multiply the leading coefficient (which we may call \(\displaystyle a\)) by the constant term (which we may call \(\displaystyle c\)): \(\displaystyle ac=3\cdot(-3)=-9\). Now look for two numbers whose product is -9 and whose sum is -8. Two such numbers are 1 and -9. Now we split up the coefficient of the linear term into these two numbers, and then we factor by grouping:

\(\displaystyle 3r^2-8r-3\)

\(\displaystyle =3r^2+r-9r-3\)

\(\displaystyle =(3r^2+r)+(-9r-3)\)

\(\displaystyle =r(3r+1)-3(3r+1)\)

\(\displaystyle =(3r+1)(r-3)\).

Another example: Factor \(\displaystyle 15x^2 + 11x + 2\).

\(\displaystyle ac = 15\cdot2=30\)

Two numbers whose product is 30 and whose sum is 11: 5 and 6

\(\displaystyle 15x^2+11x+2\)

\(\displaystyle =15x^2+5x+6x+2\)

\(\displaystyle =5x(3x+1)+2(3x+1)\)

\(\displaystyle =(3x+1)(5x+2)\).