Factoring

[Steven G]

I was trying to show that two probabilities were equal. I completely factored one of the probabilities but the 2nd one I had no idea how to completely factor one of the factors. Since the two probabilities were in fact equal I was able to see how to completely factor this one factor.
I was hoping that someone can explain how to go about factoring \(\displaystyle rg+g^2-3r-8g+15\) as I just do not see it. If you want the answer to help see how to do it, it is \(\displaystyle (g-3)(r+g-5)\).

 

[Dr.Peterson]

I'd probably guess at the form (g + ar + b)(g + c), since r is never squared but g^2 and rg terms exist. Expanding and equating coefficients leads to 4 equations in 3 unknowns, which fortunately is consistent: a = 1, c = -3, b = -5.

 

[lookagain]

I was hoping that someone can explain how to go about factoring \(\displaystyle rg+g^2-3r-8g+15\) as I just do not see it. If you want the answer to help see how to do it, it is \(\displaystyle (g-3)(r+g-5)\).

Begin and reorder the terms:

rg + g^2 - 8g - 3r + 15 =

g(r + g - 8) - 3(r - 5) =

g(r + g - 5 - 3) - 3(r - 5) =

g(r - 5 + g - 3) - 3(r - 5) =

g(r - 5) + g(g - 3) - 3(r - 5) =

g(r - 5) - 3(r - 5) + g(g - 3) =

(r - 5)(g - 3) + g(g - 3) =

(g - 3)(r - 5 + g) =

(g - 3)(r + g - 5)