Chapter 3, Problem 22CLT

To determine

To find a polynomial with real coefficients that has zeroes $0,-3,\text{ and }1+\sqrt{5}i$ .

Answer to Problem 22CLT

  $x^4+x^3+12x$

Explanation of Solution

Given:

Zeroes of polynomial are $0,-3,\text{ and }1+\sqrt{5}i$

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 $0,-3,\text{ and }1+\sqrt{5}i$ . 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, conjugate of $1+\sqrt{5}i$ equals to $1-\sqrt{5}i$ 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, $x=0,x=-3,x=1+\sqrt{5}i,\text{ and }x=1-\sqrt{5}i$ are zeroes of the polynomial.

Thus,

  $\begin{array}{l}\left(x-0\right)=x\text{,}\\ \left(x-\left(-3\right)\right)=\left(x+3\right),\\ \left(x-\left(1+\sqrt{5}i\right)\right)=\left(x-1-\sqrt{5}i\right)\\ \text{, and }\left(x-\left(1-\sqrt{5}i\right)\right)=\left(x-1+\sqrt{5}i\right)\end{array}$

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

Now, to find a polynomial that has factors $x\text{,}\left(x+3\right),\left(x-1-\sqrt{5}i\right)\text{, and }\left(x-1+\sqrt{5}i\right)$ , multiply these factors, as

  $\begin{array}{c}x\left(x+3\right)\left(x-1-\sqrt{5}i\right)\left(x-1+\sqrt{5}i\right)=\left(x^2+3x\right)\left({\left(x-1\right)}^2-{\left(\sqrt{5}i\right)}^2\right)\\ =\left(x^2+3x\right)\left(x^2-2x+1+5\right)\\ =\left(x^2+3x\right)\left(x^2-2x+6\right)\\ =x^2\left(x^2-2x+6\right)+3x\left(x^2-2x+4\right)\\ =x^4-2x^3+6x^2+3x^3-6x^2+12x\\ =x^4+x^3+12x\end{array}$

Hence, any polynomial with real coefficients that has zeroes $x=0,x=-3,x=1+\sqrt{5}i,\text{ and }x=1-\sqrt{5}i$ will be multiple of polynomial $x^4+x^3+12x$ .

Thus, a polynomial with real coefficients that has zeroes $0,-3,1+\sqrt{5}i$ is $x^4+x^3+12x$ .