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).
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?