# Using factoring to solve the equation $(r^2 + 5r - 24)(r^2 - 3r + 2) = (4r - 10)(r^2 + 5r - 24)$

Solve for all values of $r$: $$(r^2 + 5r - 24)(r^2 - 3r + 2) = (4r - 10)(r^2 + 5r - 24)$$

I'm not sure how my thinking isn't really correct here. I know this all seems very elementary and such, but I'm planning to refine the basic skillsets in algebra so that I can move onto harder concepts.

What I do, is I factor both sides to get,

$(r+8)(r-3)(r-1)(r-2) = (4r-10)(r+8)(r-3)$

I then divide both sides by $(r+8)(r-3)$, giving me:

$(r-1)(r-2) = (4r-10)$

Bringing over RHS to the LHS, by factoring, we then get the roots of the quadratic and get 3 and 4.

Wolfram is giving me answers of 3, 4, and -8 though. I don't really see where the 8 came from though. Can anyone help me out and explain? Also, is my thinking/procedure correct?

Thank you! (I know, basic question sorry).

Edit:

I realize that by dividing by $(r+8)(r-3)$, I divide by a quadratic with actual roots. That means I've missed out on one of them, which is -8.

Therefore, the values that satisfy these quadratics are,

-8, 3, and 4?

I don't know. Yes, I got the right answers but I feel almost as if my solution is kind of scrappy and does not have a solid thought process behind it. Could anyone elaborate further as to show how the problem is done?

• Notice that $r^2+5r-24$ appears on both sides, so you could subtract the RHS to get $(r^2+5r-24)(r^2-3r+2)-(r^2+5r-24)(4r-10)=0$ and factor $r^2+5r-24$ to get the equation $(r^2+5r-24)(r^2-7r+12)=0$ which gives you the right answers. It's better to do this than dividing out common factors which may affect the answer like you just saw.
– Lost
Aug 23, 2014 at 17:12

$(r^2+5r-24)(r^2-3r+2-4r+10)=0$;
$(r^2+5r-24)(r^2-7r+12)=0$
$(r+8)(r-3)^2(r-4)=0$
When you divide both sides by $(r+8)(r-3)$, you need to consider the cases $r+8=0$ and $r-3=0$. Coincidentally, the latter is “caught” in your quadratic. But the former [p.s. are you sure Mathematica isn't giving you $r=-8$?] isn't handled there.