Exercise 56 Page 528

We have to find a counterexample to the following statement.

A polynomial equation of degree three always has three real solutions .

We need to find a polynomial of degree three that has less than three solutions. We can start by considering a quadratic equation that has no real solutions. Below we write one example equation.

x^{2} + 1 = 0

There is no real solution to the equation above. However, it is a quadratic equation. We can obtain a third degree equation by multiplying it by

x .

x (x^{2} + 1) = x (0) \Rightarrow x^{3} + x = 0

Notice that the final equation can be factored as

x (x^{2} + 1) = 0 .

Then, by the Zero Product Property we can set the following two equations.

x (x^{2} + 1) = 0 ↙ ↘ x = 0 x^{2} + 1 = 0

Since the right-hand equation has no solution, we can conclude that the unique solution to the third degree equation is

x = 0 .

Therefore, this is the counterexample we were asked for.

Polynomial Equation	Nº of Solutions
$x^{3} + x = 0$	$1$

Keep in mind that this is just an example equation and so your answer may vary.

Extra

To see why

x^{2} + 1 = 0

has no real solutions, let's solve this equation for

x .

x^{2} + 1 = 0

SubEqn

$LHS - 1 = RHS - 1$

x^{2} = - 1

SqrtEqn

$LHS = RHS$

x = \pm - 1

Since

- 1

is not a real number, we conclude that the equation

x^{2} + 1 = 0

has no real solutions.