Chapter 2.2, Problem 80E

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 80E

The polynomial function with zeros are 1 of multiplicity 2, and 4 of multiplicity 2 and of degree 4 that falls to the left and to the right is $f\left(x\right)=-\left(x^4-10x^3+33x^2-40x+16\right)$ .

Explanation of Solution

Given data:

The given zeros of polynomial function are 1 of multiplicity2,and 4 of multiplicity, and of degree 4. The function falls to the left and tothe 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 $f\left(x\right)$ falls to the left and to the right if it is of even 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 are1of multiplicity 2, and 4of multiplicity2, thus we have ${\alpha }_1=1$ , $k_1=2$ , ${\alpha }_2=4$ , $k_2=2$ .

Now, substituting these values in equation (1), the polynomial function $f\left(x\right)$ is given by $f\left(x\right)={\left(x-1\right)}^2{\left(x-4\right)}^2$

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

  $\Rightarrow f\left(x\right)=\left(x^2-2x+1\right)\left(x^2-4x+16\right)$

The $f\left(x\right)$ falls to the right and to the left if leading coefficient of function is negative. $\Rightarrow f\left(x\right)=\left(x^2-2x+1\right)\left(x^2-4x+16\right)$

  $\Rightarrow f\left(x\right)=-\left(x^4-10x^3+33x^2-40x+16\right)$ .

Conclusion:

The polynomial function with zeros are 1 of multiplicity 2, and 4 of multiplicity 2and of degree 4that falls to the left and to the right is $f\left(x\right)=-\left(x^4-10x^3+33x^2-40x+16\right)$ .