Chapter 2.5, Problem 14E

To determine

To find the number of zeroes of the polynomial function

  $h\left(t\right)={\left(t-1\right)}^2-{\left(t+1\right)}^2$

Answer to Problem 14E

1

Explanation of Solution

Given:

  $h\left(t\right)={\left(t-1\right)}^2-{\left(t+1\right)}^2$

Concept Used:

The total number of zeroes of a polynomial function of degree n is n . That is the number of zeroes of a polynomial, $P\left(x\right)=a{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\mathrm{....}+a_n;a\ne 0$ is n .

Calculation:

To determine the number of zeroes of the polynomial function

  $h\left(t\right)={\left(t-1\right)}^2-{\left(t+1\right)}^2$

Use the concept that total number of zeroes of a polynomial function of degree n is n .

First expand this polynomial, as

  $\begin{array}{c}h\left(t\right)={\left(t-1\right)}^2-{\left(t+1\right)}^2\\ =t^2-2t+1-\left(t^2+2t+1\right)\\ =t^2-2t+1-t^2-2t-1\\ =-4t\end{array}$

Now, degree of a polynomial is higher power of variable having non-zero coefficient. Thus, this polynomial $h\left(t\right)$ has degree 1.

Thus, total number of zeroes(irrespective real/complex) of given polynomial will be 1.