How to find solutions to high degree algebra equations

Question

I came across this math problem in Ian Bull's Extension Mathematics for years 9-10 and I am at a loss on how to solve it.

The original question is here as the image attached below.

For equation $2x^5-8x^3+7 = 0$, one solution lies between $-3$ and $-2$ and two solutions lie between $1$ and $2$. Find these solutions to two decimal places.

Answer from the book: the solutions which produce values closest to zero are: $x = -2.09, 1.07$ and $1.86$

Original Problem Full Description

These are the previous problems, the author is intending for the student to use a guess and check method but I would like to know how to find a definitive answer without trial and error. If there is an equation for that type of index.

Previous questions

Answer 1

The graph shows were are, more or less, the roots.

Consider the first one and make $x=-2+t$ to obtain $$7+64 t-112 t^2+72 t^3-20 t^4+2 t^5=0$$ So, now, the powers of $t$ become negigibl and an appoximation is given by $$7+64t=0 \implies t=-\frac 7 {64}$$ Restart the same procedure with now $x=-2-\frac 7 {64}+t$ and neglecting again, the new approximation is $$-\frac{771453991}{536870912}+\frac{764957925 t}{8388608}=0 \implies t=\frac{771453991}{48957307200}$$ SO, at this point $$x=-2-\frac 7 {64}+\frac{771453991}{48957307200}=-\frac{25624466471}{12239326800}=-2.09362$$ while the exact solution is $-2.09323$

This is more than simplistic but this is how Newton invented Newton method

Answer 2

one solution lies between $-3$ and $-2$

One way to find it is by halving the interval a.k.a. the bisection method. Start with interval $(-3,-2)$ where $f(-3) \lt 0 \lt f(-2)$. Calculate $f\left(\frac{(-3)+(-2)}{2}\right)$ and determine that $f(2.5) \lt 0$. This means the root is in $(-2.5,-2)$ so calculate $f\left(\frac{(-2.5)+(-2)}{2}\right)$ and repeat until you reach the required precision.

Another way to get a better first approximation is to look for the offset of the root with respect to one of the endpoints of the interval. For example, substitute $x=-2 + \varepsilon$ in the equation and ignore all powers $\varepsilon^k$ for $k \gt 2$ because $ |\varepsilon|\lt 1$ and $\varepsilon^k$ becomes smaller and smaller for higher $k$:

$$ \require{cancel} \begin{align} 0 &= 2(-2 + \varepsilon)^5-8(-2 + \varepsilon)^3+7 \\ &= 2(-32 + 80 \varepsilon - 80 \varepsilon^2+\dots)-8(-8 + 12 \varepsilon-6 \varepsilon^2+\dots) + 7 \\ &= 7 + 64 \varepsilon -112 \varepsilon^2 \;+ \bcancel{(\dots) \,\varepsilon^3} \end{align} $$

The quadratic $7 + 64 \varepsilon -112 \varepsilon^2$ has roots $\frac{8 \pm \sqrt{113}}{28}$, and the root closest to $0$ that you are looking for is $\varepsilon \simeq-0.094$ i.e. $x \simeq -2.094$. Then all that's left to verify is that $f(-2.10) \lt 0 \lt f(-2.09)$.