Chapter 6.6, Problem 41HP

To determine

To find: the cubic function that has remainder of 8 for $f(2)$and a remainder of $-5$ for $f(3)$ .

Answer to Problem 41HP

The cubic function is $f(x)=x^3-4x^2-12x+40$ .

Explanation of Solution

Given information:

The cubic function has remainder of 8 for $f(2)$and a remainder of $-5$ for $f(3)$ .

Formula used:

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $\begin{array}{l}f(2)=8\text{ (i)}\\ f(3)=-5\text{ (ii)}\end{array}$.

Now , the general form of cubic function is

  $f(x)=a{x}^3+b{x}^2+cx+d$

We get ,

  $\begin{array}{}$ f(2)=a{(2)}^3+b{(2)}^2+c(2)+d$$ \text{ = }8a+4b+2c+d\text{ (iii)}$$ f(3)=a{(3)}^3+b{(3)}^2+c(3)+d$$ \text{ = 27}a+9b+3c+d\text{ (iv)}$\end{array}$

Using equations (i) ,(ii), (iii) and (iv) we get,

  $\begin{array}{l}=8a+4b+2c+d=8\text{ (v)}\\ =27a+9b+3c+d=-5\text{ (vi)}\end{array}$

Let $a=1\text{ and }b=-4$ .

Putting values in equation (v) and (vi) and solving for c and d we get,

  $\begin{array}{l}=8a+4b+2c+d=8\\ =8(1)+4(-4)+2c+d=8\\ =8-16+2c+d=8\\ =-8+2c+d=8\\ =2c+d=8+8\\ =2c+d=16\text{ (vii)}\end{array}$

  $\begin{array}{l}=27a+9b+3c+d=-5\\ =27(1)+9(-4)+3c+d=-5\\ =27-36+3c+d=-5\\ =-9+3c+d=-5\\ =3c+d=-5+9\\ =3c+d=4\text{ (viii)}\end{array}$

Now, subtracting equations (vii) and (viii) we get,

  $\begin{array}{l}=2c+d-3c-d=16-4\\ =-c=12\\ =c=-12\end{array}$

Putting value of c in equation (viii).

  $\begin{array}{l}=3(-12)+d=4\\ =-36+d=4\\ =d=4+36\\ =d=40\end{array}$

Therefore, the cubic function is $f(x)=x^3-4x^2-12x+40$