How does my prof's logic work? Find four roots of $p(x)$.

Question

Below is the solution to a problem where you have to find four intervals of length $1/16$ that contain roots of $p(x)$.

Apparently,

Since the cubic and quadratic terms have large negative coefficients, we suspect that the other two roots are close to x=1.

My problem with this is, if you make some roots within $0.0001$ "distance" of each other, there's no way on an exam you can use IVT and bisection method (the problem statement said we should use them) to find those two intervals of length $1/16$. Or even 0.1, it leaves you guessing all over the place.

Also, In Desmos I graphed the same function, but with even more negative coefficients for the cube and square and lo and behold, the roots were not around 1 anymore, unlike in the solution statement.

Is the solution's argument absurd, or am I missing something important? How would one solve this painlessly in a calculator-less, exam situation?

Answer 1

$$225 x^4-210 x^3-812 x^2+706 x+315=0$$ $$(3 x-5) (3 x+1) (5 x-7) (5 x+9)=0$$

$$x_1=-\frac{9}{5},x=_2-\frac{1}{3},x_3=\frac{7}{5},x_4=\frac{5}{3}$$

Hope this helps

Answer 2

For a quartic with positive fourth power term to have four roots it comes down from $(-\infty, +\infty)$ crosses the $x$ axis four times and heads off to $(+\infty,+\infty)$ With the tabluated values in hand, it looks like the fourth power term is already dominant by $-3$ and $+3$, so the roots need to be between those. The sign changes locate two of them. The fact that $f(1)$ is the smallest tabulated value gives a hint that the roots are nearby. You are correct that if you make the second and third power terms more negative you can move the roots well away from $1$, but that is likely to depress $f(2)$ and make it closer to $0$ than $f(1)$. I don't think the "large and negative" claim holds water, but the fact that $f(1)$ looks like a local minimum is more telling.