Chapter 2.2, Problem 52E

Solution

To find the value of $f\left(-5\right)$ .

  $f\left(-5\right)=-480$

Given:

The leading coefficient of a cubic polynomial function $f$ is 2 and the constant term is $-5$ and also $f\left(1\right)=0$ and $f\left(2\right)=3$ .

Calculation:

Let the cubic polynomial be,

  $f\left(x\right)=a{x}^3+b{x}^2+cx+d$

Since the leading coefficient is 2 and the constant term is $-5$ , therefore

  $a=2\&d=-5$

So the above equation becomes,

  $f\left(x\right)=2x^3+b{x}^2+cx-5\to \left(1\right)$

Now consider,

  $\begin{array}{c}f\left(1\right)=0\\ 2{\left(1\right)}^3+b{\left(1\right)}^2+c\left(1\right)-5=0\\ b+c=3\to \left(2\right)\end{array}$

Now,

  $\begin{array}{c}f\left(2\right)=3\\ 2{\left(2\right)}^3+b{\left(2\right)}^2+c\left(2\right)-5=3\\ 4b+2c=-8\to \left(3\right)\end{array}$

Now,

  $\left[2\times \text{Equation}\left(2\right)\right]-\left[\text{Equation}\left(3\right)\right]$

It follows that

  $\begin{array}{c}2\left(b+c\right)-\left(4b+2c\right)=2\left(3\right)-\left(-8\right)\\ 2b+2c-4b-2c=6+8\\ -2b=14\\ b=-7\end{array}$

Now put $b=-7$ in equation (2), it follows that

  $\begin{array}{c}-7+c=3\\ c=10\end{array}$

Now put $b=-7$ and $c=10$ in equation (1), it follows that

  $f\left(x\right)=2x^3-7x^2+10x-5$

Now put $x=-5$ in the above equation, it follows that

  $\begin{array}{c}f\left(-5\right)=2{\left(-5\right)}^3-7{\left(-5\right)}^2+10\left(-5\right)-5\\ =-250-175-50-5\\ =-480\end{array}$

Process of solving:

First consider the standard cubic polynomial function, and then substitute the given values of leading coefficient and the constant term, and then use the given function values to find the other constant terms.

Finally substitute $x=-5$ in the obtained polynomial to get the value of $f\left(-5\right)$ .