Chapter 2.5, Problem 41E

To determine

To find a polynomial with real coefficients that has zeroes $1,5i$ .

Answer to Problem 41E

  $x^3-x^2+5x-5$

Explanation of Solution

Given:

Zeroes of polynomial $1,5i$

Concept Used:

• If $x=a$ is a zero of a polynomial, then $\left(x-a\right)$ is a factor of that polynomial.
• Complex zeroes of a polynomial with real coefficients always with their conjugates, that is, if $a+ib$ is a zero of polynomial $P\left(x\right)$ then its conjugate $a-ib$ will also be a zero of this polynomial.

Calculation:

To find a polynomial with real coefficients that has zeroes $1,5i$ . First use the concept that if $a+ib$ is a zero of polynomial with real coefficients then $a-ib$ will also be a zero of that polynomial.

Thus, $-5i$ is also a zero of this polynomial.

Now, if $x=a$ is a zero of a polynomial, then $\left(x-a\right)$ is a factor of that polynomial.

Since, $x=1,x=5i,x=-5i$ are zeroes of the polynomial thus

  $\left(x-1\right)\text{,}\left(x-5i\right)\text{, and }\left(x+5i\right)$

are the factors corresponding to the zeros of the polynomial respectively.

Now, to find a polynomial that has factors $\left(x-1\right)\text{,}\left(x-5i\right)\text{, and }\left(x+5i\right)$ , multiply these factors, as

  $\begin{array}{c}\left(x-1\right)\left(x-5i\right)\left(x+5i\right)=\left(x-1\right)\left(x^2-{\left(5i\right)}^2\right)\\ =\left(x-1\right)\left(x^2+5\right)\\ =x\left(x^2+5\right)-\left(x^2+5\right)\\ =x^3+5x-x^2-5\\ =x^3-x^2+5x-5\end{array}$

Hence, any polynomial with real coefficients that has zeroes $1,5i$ will be multiple of polynomial $x^3-x^2+5x-5$ .

Thus, a polynomial with real coefficients that has zeroes $1,5i$ is $x^3-x^2+5x-5$ .