Chapter 4.3, Problem 45E

To determine

To find: the value of $a,b\text{}\text{ and }c$ for the given function.

Answer to Problem 45E

  $a=1,b=-6,c=25.$

Explanation of Solution

Given:

The polynomial equation as:

  $\begin{array}{l}P\left(x\right)=a{x}^2+bx+c.\\ P\left(3+4i\right)=0\text{ and }P\left(3-4i\right)=0.\end{array}$

Concept used:

Factor theorem:

The binomial $x-r$ is a factor of the polynomial $P\left(x\right)$ if and only if $P\left(r\right)=0$ .

Calculation:

Consider the polynomial:

  $P\left(x\right)=a{x}^2+bx+c$ .

Factor theorem:

The binomial $x-r$ is a factor of the polynomial $P\left(x\right)$ if and only if $P\left(r\right)=0$ .

Now,

By factor theorem,

  $P\left(3+4i\right)=0$ if and only if $x-\left(3+4i\right)$ is a factor of the polynomial:

  $P\left(x\right)=a{x}^2+bx+c$ .

And by factor theorem:

  $P\left(3-4i\right)=0$ if and only if $x-\left(3+4i\right)$ is a factor of the polynomial:

  $P\left(x\right)=a{x}^2+bx+c$ .

Since, the polynomial $P\left(x\right)=a{x}^2+bx+c$ is of degree two, it has two factors.

But $x-\left(3+4i\right)$ and $x-\left(3-4i\right)$ are factors of $P\left(x\right)=a{x}^2+bx+c$ .

Now, the polynomial will be:

The polynomial will be:

  $\begin{array}{l}a{x}^2+bx+c=\left(x-\left(3+4i\right)\right)\left(x-\left(3-4i\right)\right).\\ \Rightarrow \left(x-3-4i\right)\left(x-3+4i\right).\\ \Rightarrow {\left(x-3\right)}^2-{\left(4i\right)}^2.\\ \Rightarrow x^2-6x+25.\end{array}$

Comparing with the $a{x}^2+bx+c$ :

Hence, $a=1,b=-6,c=25.$