Chapter 2.2, Problem 73E

To determine

To find: The polynomial function which has given zeros.

Answer to Problem 73E

The polynomial function whose zeros are 2, $2+\sqrt{5}$ and $2-\sqrt{5}$ is $f\left(x\right)=x^3-6x^2+7x+2$

Explanation of Solution

Given:

The given zeros are 2, $2+\sqrt{5}$ and $2-\sqrt{5}$ .

Concept used:

If $x={\alpha }_1,x={\alpha }_2,x={\alpha }_3,\dots x={\alpha }_n,$ are the zeros of a polynomial function $f\left(x\right)$ (say) then $\left(x-{\alpha }_1\right)$ , $\left(x-{\alpha }_2\right)$ , $,\left(x-{\alpha }_3\right),\dots ,\left(x-{\alpha }_n\right)$ are the factors of $f\left(x\right)$ that is polynomial function of degree n given by $f\left(x\right)=\left(x-{\alpha }_1\right)\left(x-{\alpha }_2\right)\left(x-{\alpha }_3\right)\dots \left(x-{\alpha }_n\right)$ .

Calculations:

The polynomial function $f\left(x\right)$ having zeros $x={\alpha }_1$ , $x={\alpha }_2$ and $x={\alpha }_3$ is

  $f\left(x\right)=\left(x-{\alpha }_1\right)\left(x-{\alpha }_2\right)\left(x-{\alpha }_3\right)$ ..............(1)

Given zeros are 2, $2+\sqrt{5}$ and $2-\sqrt{5}$ , sosubstituting ${\alpha }_1=2$ , ${\alpha }_2=2+\sqrt{5}$ and ${\alpha }_3=2-\sqrt{5}$ in equation (1), the polynomial function whose zeros are 2, $2+\sqrt{5}$ and $2-\sqrt{5}$ is

  $f\left(x\right)=\left(x-2\right)\left(x-( 2+\sqrt{5})\right)\left(x-( 2-\sqrt{5})\right)$

  $\Rightarrow f\left(x\right)=\left(x-2\right)\left((x-2)-\sqrt{5}\right)\left((x-2)+\sqrt{5}\right)$

  $\Rightarrow f\left(x\right)=\left(x-2\right)\left({(x-2)}^2-{(\sqrt{5})}^2\right)$$\because \left(a+b\right)\left(a-b\right)=a^2-b^2$

  $\Rightarrow f\left(x\right)=\left(x-2\right)\left(x^2-4x+4-5\right)$

  $\Rightarrow f\left(x\right)=\left(x-2\right)\left(x^2-4x-1\right)$

  $\Rightarrow f\left(x\right)=x^3-6x^2+7x+2$

Conclusion:

The polynomial function whose zeros are 2, $2+\sqrt{5}$ and $2-\sqrt{5}$ is $f\left(x\right)=x^3-6x^2+7x+2$