# Inferring locations of zeroes indirectly

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

$x = - \frac{2}{\pi }$

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

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.

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Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus