Chapter 2.2, Problem 79E

To determine

To find: The polynomial function with given zeros, multiplicity, and degree. Also, it rises to the left and falls to right.

Answer to Problem 79E

The polynomial function with zeros are -1 of multiplicity 2, and 2 of multiplicity 1,, and of degree 3 that rises to the left and falls to the right is $f\left(x\right)=-\left(x^3+4x^2+5x+2\right)$ .

Explanation of Solution

Given data:

The given zeros of polynomial function are -1 of multiplicity2, and -2 of multiplicity1, and of degree3. The function rises to the left and falls to the right.

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) of multiplicities $k_1,k_2,k_3,\dots ,k_n$ then ${\left(x-{\alpha }_1\right)}^{k_1}$ , ${\left(x-{\alpha }_2\right)}^{k_2}$ , ${\left(x-{\alpha }_3\right)}^{k_3}$$\dots ,{\left(x-{\alpha }_n\right)}^{k_n}$ are the factors of $f\left(x\right)$ that is a polynomial function of degree n given by $f\left(x\right)={\left(x-{\alpha }_1\right)}^{k_1}{\left(x-{\alpha }_2\right)}^{k_2}{\left(x-{\alpha }_3\right)}^{k_3}\dots {\left(x-{\alpha }_n\right)}^{k_n}$ .

The polynomial function $f\left(x\right)$ rise on left and falls on right if it is of odd degree and has leading coefficient negative.

Calculations:

Let $f\left(x\right)$ be the polynomial function having zeros $x={\alpha }_1$ of multiplicity $k_1$ and $x={\alpha }_2$ of multiplicity $k_1$ then $f\left(x\right)$ is polynomial of degree $\left(k_1+k_2\right)$ given by

  $f\left(x\right)={\left(x-{\alpha }_1\right)}^{k_1}{\left(x-{\alpha }_2\right)}^{k_2}$ ..............(1)

Given zeros are -1of multiplicity 2, and 2of multiplicity 1, thus we have ${\alpha }_1=-1$ , $k_1=2$ , ${\alpha }_2=-2$ , $k_2=1$ .

Now, substituting these values in equation (1), the polynomial function $f\left(x\right)$ is given by $f\left(x\right)={\left(x-(-1)\right)}^2{\left(x-(-2)\right)}^1$

  $\Rightarrow f\left(x\right)={\left(x+1\right)}^2{\left(x+2\right)}^1$

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

The function $f\left(x\right)$ rises to the left and falls to the right if the leading coefficient is negative. $\Rightarrow f\left(x\right)=-\left(x^2+2x+1\right)\left(x+2\right)$

  $\Rightarrow f\left(x\right)=-\left(x^3+2x^2+x+x^2+2x+1\right)$

  $\Rightarrow f\left(x\right)=-\left(x^3+4x^2+5x+2\right)$ .

Conclusion:

The polynomial function with zeros are -1 of multiplicity 2, and 2 of multiplicity 1,, and of degree 3that rises to the left and falls to the right is $f\left(x\right)=-\left(x^3+4x^2+5x+2\right)$ .