Chapter 2.5, Problem 49E

To determine

To find a polynomial with real coefficients that has zeroes $\frac{2}{3},-1,3+\sqrt{2}i$ .

Answer to Problem 49E

  $x^4-\frac{17x^3}{3}+\frac{25x^2}{3}+\frac{23x}{3}-\frac{22}{3}$

Explanation of Solution

Given:

Zeroes of polynomial are $\frac{2}{3},-1,3+\sqrt{2}i$

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 $\frac{2}{3},-1,3+\sqrt{2}i$ . 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 $3+\sqrt{2}i$ equals to $3-\sqrt{2}i$ 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, $x=\frac{2}{3},x=-1,x=3+\sqrt{2}i,\text{ and }x=3-\sqrt{2}i$ are zeroes of the polynomial.

Thus,

  $\begin{array}{l}\left(x-\frac{2}{3}\right)\text{,}\\ \left(x-\left(-1\right)\right)=\left(x+1\right),\\ \left(x-\left(3+\sqrt{2}i\right)\right)=\left(x-3-\sqrt{2}i\right)\\ \text{, and }\left(x-\left(3-\sqrt{2}i\right)\right)=\left(x-3+\sqrt{2}i\right)\end{array}$

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

Now, to find a polynomial that has factors $\left(x-\frac{2}{3}\right)\text{,}\left(x+1\right),\left(x-3-\sqrt{2}i\right)\text{, and }\left(x-3+\sqrt{2}i\right)$ , multiply these factors, as

  $\begin{array}{c}\left(x-\frac{2}{3}\right)\left(x+1\right)\left(x-3-\sqrt{2}i\right)\left(x-3+\sqrt{2}i\right)=\left(x^2-\frac{2}{3}x+x-\frac{2}{3}\right)\left({\left(x-3\right)}^2-{\left(\sqrt{2}i\right)}^2\right)\\ =\left(x^2+\frac{1}{3}x-\frac{2}{3}\right)\left(x^2-6x+9+2\right)\\ =\left(x^2+\frac{1}{3}x-\frac{2}{3}\right)\left(x^2-6x+11\right)\\ =x^2\left(x^2-6x+11\right)+\frac{1}{3}x\left(x^2-6x+11\right)-\frac{2}{3}\left(x^2-6x+11\right)\\ =x^4-6x^3+11x^2+\frac{1}{3}{x}^3-2x^2+\frac{11}{3}x-\frac{2}{3}{x}^2+4x-\frac{22}{3}\\ =x^4+\frac{-18x^3+x^3}{3}+\frac{27x^2-2x^2}{3}+\frac{11x+12x}{3}-\frac{22}{3}\\ =x^4-\frac{17x^3}{3}+\frac{25x^2}{3}+\frac{23x}{3}-\frac{22}{3}\end{array}$

Hence, any polynomial with real coefficients that has zeroes $-1,5,3-2i,\text{ and }3+2i$ will be multiple of polynomial

  $x^4-\frac{17x^3}{3}+\frac{25x^2}{3}+\frac{23x}{3}-\frac{22}{3}$ .

Thus, a polynomial with real coefficients that has zeroes $\frac{2}{3},-1,3+\sqrt{2}i$ is $x^4-\frac{17x^3}{3}+\frac{25x^2}{3}+\frac{23x}{3}-\frac{22}{3}$ .