Chapter 2.5, Problem 46E

To determine

To find a polynomial with real coefficients that has zeroes $-\frac{5}{2},-5,1+\sqrt{3}i$ .

Answer to Problem 46E

  $x^4+\frac{11x^3}{3}-\frac{17x^2}{3}+35x+10$

Explanation of Solution

Given:

Zeroes of polynomial are $-\frac{5}{2},-5,1+\sqrt{3}i$

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 $-\frac{5}{2},-5,1+\sqrt{3}i$ . 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 $1+\sqrt{3}i$ equals to $1-\sqrt{3}i$ 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, $x=-\frac{5}{2},x=-5,x=1+\sqrt{3}i,\text{ and }x=1-\sqrt{3}i$ are zeroes of the polynomial.

Thus,

  $\begin{array}{l}\left(x+\frac{5}{2}\right)\text{,}\\ \left(x-\left(-5\right)\right)=\left(x+5\right),\\ \left(x-\left(1+\sqrt{3}i\right)\right)=\left(x-1-\sqrt{3}i\right)\\ \text{, and }\left(x-\left(1-\sqrt{3}i\right)\right)=\left(x-1+\sqrt{3}i\right)\end{array}$

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

Now, to find a polynomial that has factors $\left(x+\frac{5}{2}\right)\text{,}\left(x+5\right),\left(x-1-\sqrt{3}i\right)\text{, and }\left(x-1+\sqrt{3}i\right)$ , multiply these factors, as

  $\begin{array}{c}\left(x+\frac{5}{2}\right)\left(x+5\right)\left(x-1-\sqrt{3}i\right)\left(x-1+\sqrt{3}i\right)=\left(x^2+\frac{5}{2}x+5x+\frac{5}{2}\right)\left({\left(x-1\right)}^2-{\left(\sqrt{3}i\right)}^2\right)\\ =\left(x^2+\frac{15}{2}x+\frac{5}{2}\right)\left(x^2-2x+1+3\right)\\ =\left(x^2+\frac{15}{2}x+\frac{5}{2}\right)\left(x^2-2x+4\right)\\ =x^2\left(x^2-2x+4\right)+\frac{15}{2}x\left(x^2-2x+4\right)+\frac{5}{2}\left(x^2-2x+4\right)\\ =x^4-2x^3+4x^2+\frac{15}{2}{x}^3-15x^2+30x+\frac{5}{2}{x}^2+5x+10\\ =x^4+\frac{-4x^3+15x^3}{3}+\frac{-22x^2+5x^2}{3}+35x+10\\ =x^4+\frac{11x^3}{3}-\frac{17x^2}{3}+35x+10\end{array}$

Hence, any polynomial with real coefficients that has zeroes $-\frac{5}{2},-5,1+\sqrt{3}i$ will be multiple of polynomial $x^4+\frac{11x^3}{3}-\frac{17x^2}{3}+35x+10$ .

Thus, a polynomial with real coefficients that has zeroes $-\frac{5}{2},-5,1+\sqrt{3}i$ is $x^4+\frac{11x^3}{3}-\frac{17x^2}{3}+35x+10$ .