A polynomial function with real coefficients is continuous everywhere; that
is, its graph has no holes or breaks. This is the basis for the following
property: If \(f(x)\) is a polynomial with real coefficients, and if \(f(a)\) and
\(f(b)\) are of opposite sign, then there is at least one real zero between \(a\)
and \(b\). This property, along with our knowledge of polynomial functions,
provides the basis for locating and approximating irrational solutions of a
polynomial equation.
Consider the equation \(x^{3}+2 x-4=0\). Applying Descartes' rule of signs, we
can determine that this equation has one positive real solution and two
nonreal complex solutions. (You may want to confirm this!) The rational root
theorem indicates that the only possible rational solutions are 1,2 , and 4 .
Using a little more compact format for synthetic division, we obtain the
following results when testing for 1 and 2 as possible solutions:
$$
\begin{array}{r|rrrr}
& 1 & 0 & 2 & -4 \\
1 & 1 & 1 & 3 & -1 \\
2 & 1 & 2 & 6 & 8
\end{array}
$$
Because \(f(1)=-1\) and \(f(2)=8\), there must be an irrational solution between 1
and 2 . Furthermore, \(-1\) is closer to 0 than is 8 , so our guess is that the
solution is closer to 1 than to 2 . Let's start looking at \(1.0,1.1,1.2\), and
so on, until we can place the solution between two numbers.
Because \(f(1.1)=-0.469\) and \(f(1.2)=0.128\), the irrational solution must be
between \(1.1\) and 1.2. Furthermore, because \(0.128\) is closer to 0 than is
\(-0.469\), our guess is that the solution is closer to \(1.2\) than to \(1.1\).
Let's start looking at \(1.15,1.16\), and so on.
$$
\begin{array}{l|rrrrr}
& 1 & 0 & 2 & -4 \\
\ 1.15 & 1 & 1.15 & 3.3225 & -0.179 \\
1.16 & 1 & 1.16 & 3.3456 & -0.119 \\
1.17 & 1 & 1.17 & 3.3689 & -0.058 \\
1.18 & 1 & 1.18 & 3.3924 & 0.003
\end{array}
$$
Because \(f(1.17)=-0.058\) and \(f(1.18)=0.003\), the irrational solution must be
between \(1.17\) and \(1.18\). Therefore we can use \(1.2\) as a rational
approximation to the nearest tenth.
For each of the following equations, (a) verify that the equation has exactly
one irrational solution, and (b) find an approximation, to the nearest tenth,
of that solution.
(a) \(x^{3}+x-6=0 \)
(b) \(x^{3}-6 x-6=0 \)
(c) \(x^{3}-27 x-60=0 \)
(d) \(x^{3}-x^{2}-x-1=0 \)
(e) \(x^{3}-2 x-10=0\)
(f) \(x^{3}-5 x^{2}-1=0 \)
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