# Inferring locations of zeroes indirectly

Consider the following linear polynomial:

\(p\left( x \right):x - 2\)

Let us calculate the value of this polynomial at some specific values of *x*, as shown below:

\[\begin{array}{l}x = 1.8 & & \Rightarrow & p\left( x \right) = - 0.2\\x = 1.9 & & \Rightarrow & p\left( x \right) = - 0.1\\x = 1.99 & & \Rightarrow & p\left( x \right) = - 0.01\\x = 2.01 & & \Rightarrow & p\left( x \right) = 0.01\\x = 2.1 & & \Rightarrow & p\left( x \right) = 0.1\\x = 2.2 & & \Rightarrow & p\left( x \right) = 0.2\end{array}\]

Note that when *x* is slightly larger than 2, then the value of the polynomial is slightly larger than 0, while when *x* is slightly less than 2, then the value of the polynomial is slightly less than 0. Rephrasing this, we can say that as *x* is increased for values slightly less than 2 to values slightly greater than 2, the value of the polynomial increases from slightly less than 0 to slightly more than 0. This means that during this transition, there *must have been some point* (some value of *x*) when the value of the polynomial *would have been* *exactly* 0. Of course, that value is \(x = 2\).

The important point to understand is this: if *x* moves from one value to another (say from *a* to *b*), and correspondingly, the value of the polynomial changes from negative to positive, or positive to negative, then there must be some value of *x* between *a* and *b* for which the value of the polynomial is exactly 0.

Let’s discuss another example. Consider the polynomial

\(p\left( x \right):2 + \pi x\)

We note that

\(\begin{array}{l}p\left( 0 \right) = 2 > 0\\p\left( { - 1} \right) = 2 - \pi < 0\end{array}\)

Thus, as *x* changes from 0 to \( - 1\), the value of the polynomial changes from positive to negative. This means that there must be a value of *x* between 0 and \( - 1\) for which the value of the polynomial is exactly 0. And that value is

\[x = - \frac{2}{\pi }\]

Let’s consider a more advanced example. Consider the quadratic polynomial

\[p\left( x \right):4{x^2} - 8x + 3\]

Without actually calculating the zeroes of this polynomial, can you determine their approximate values? Let us evaluate the values of this polynomial for some specific values of *x*:

\[\begin{array}{l}p\left( 0 \right) = 3\\p\left( 1 \right) = - 1\\p\left( 2 \right) = 3\end{array}\]

Note that as *x* changed from 0 to 1, the value of the polynomial changed from 3 to \( - 1\). Thus, there must be some value of *x* between 0 and 1 for which the value of the polynomial is 0. That value is (verify this) \(x = \frac{1}{2}\).

Also, as *x* changed from 1 to 2, the value of the polynomial changed from \( - 1\) to 3. Thus, there must be some value of *x* between 1 and 2 for which the value of the polynomial is 0. In fact, that value is \(x = \frac{3}{2}\).

Thus, we see that it might be possible to infer the approximate values, or *locations*, of the zeroes of a polynomial if we know its values for some specific values of *x*.