How to show an epsilon-delta relationship for this quadratic function?

Question

From Stein/Barcellos, Calculus and Analytic Geometry, 5th Ed., Sec. 2.10, 17(a):

Show that, if $0 < \delta < 1$ and $|x - 3| < \delta$, then $|x^2 + 5x - 24| < 12 \delta$. (Hint: Factor $x^2 + 5x -24$).

I can see that $|x^2 + 5x - 24| = |(x+8)(x-3)| = |x+8||x-3| < |x+8|$ (because $|x-3| < \delta < 1$). But I seem to be blind to anything after that.

Answer 1

Well, just do $$|x^2+5x-24| = |x+8||x-3| \leq (|x|+8)|x-3| <(4+8)\delta = 12\delta.$$We use that $$|x| - 3 \leq |x-3|< \delta \implies |x|<\delta+3 < 1+3 = 4.$$You might find my answer here helpful.

Answer 2

If $|x-3|<\delta<1$ then $|x+8|\leq |x-3|+11\leq 12$, hence $$ |(x+8)(x-3)|\leq 12|x-3|<12\delta$$