Chapter 4, Problem 48RE

To determine

To design:The polynomial function.

Answer to Problem 48RE

The polynomial function is $f(x)=-5{(x-b)}^3(x-c)(x^2+dx+e)$ .

Explanation of Solution

Given information:

The characteristics are: degree 6; four real zeros , one multiplicity 3; y -intercept 3; behaves like $y=-5x^6$ for large values of $\left|x\right|$ .

Formula used:

The graph of a polynomial function $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0;a_n\ne 0$

Degree of a polynomial function $f:n$

For large $\left|x\right|$ , the graph of f behaves like the graph of $y=a_n{x}^n$ .

Calculation:

We have ,

Degree: $6$

Real zeros: $4$ , one with multiplicity 3

End Behavior: $-5x^6$

y -intercept: 3

Since, the degree is 6. This implies polynomial has 4 real zeros and 2 complex zeros with one multiplicity 3.

The general form is

  $=f(x)=a{(x-b)}^3(x-c)(x^2+dx+e)$ with $d^2-4e<0$

Now,

  $=f(x)=a_6{x}^6+a_5{x}^5+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0;a_6\ne 0$ (i)

They - intercept :

  $=f(0)=3$ (ii)

Put (ii) in (i)

  $=f(x)=a_6{(0)}^6+a_5{(0)}^5+a_4{(0)}^4+a_3{(0)}^3+a_2{(0)}^2+a_1(0)+a_0=3$

  $=a_0=3$

Since the end behavior is $y=-5x^6$ this implies that leading coefficient is negative.

  $=a_6=a<0$

Therefore, we have

  $=f(x)=-5{(x-b)}^3(x-c)(x^2+dx+e)$ with $d^2-4e<0;a<0;a_0=3$

is the general form of the polynomial.

Now,

  $\begin{array}{l}=a{b}^{3}ce=3\\ =-5b^{3}ce=3\\ =b^{3}ce=-\frac{3}{5}\end{array}$

Hence, the polynomial s not unique.

If we name the real zeros then the values of b and c be given and polynomial function will still be flexible about a , d , e with the conditions.