Chapter 2.6, Problem 44E

Solution

To explain: greater than three genuine zeros in a cubic function if possible. Can a cubic function have non-existent zeros.

Maximum of $3$ real roots.

At least $1$ real root.

Given information: 

The function $h\left(x\right)=x^3+x^2-x+2$ .

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

  $\frac{p}{q}=\frac{\text{ factor of constant term }{a}_0}{\text{ factor of leading coefficient }{a}_n}$

Calculation:

A cubic function can only have three real roots in total.

Let us denote the real roots by $x_1,x_2,\dots ,x_n$ .

Then each of the polynomials $x-x_1,x-x_2,\dots ,x-x_n$ is the factor of $f(x)$ .

Therefore,

  $f(x)=\left(x-x_1\right)\left(x-x_2\right)\cdots \left(x-x_n\right)\cdot g(x),$ where $g(x)$ is some polynomial (possibly?

constant polynomial).

Since the degree of

  $\left(x-x_1\right)\left(x-x_2\right)\cdots \left(x-x_n\right)\cdot g(x)$ is at least $n$ , and it must be $3$ (because the degree of $f(x)$ is $3$ !), we get

  $n⩽3$

Therefore, it is limited to having three actual roots.

It also needs to have at least one genuine root. Drawing a graph to demonstrate this shows that

If the leading coefficient is positive, then

1. $f(x)\to -\infty \text{ as }x\to -\infty$
2. $f(x)\to +\infty \text{ as }x\to +\infty$

If the leading coefficient is negative

1. $f(x)\to +\infty \text{ as }x\to -\infty$
2. $f(x)\to -\infty \text{ as }x\to +\infty$

Maximum of $3$ real roots.

At least $1$ real root.