By deflation prove that the following equation has only one real root.

Question

The equation is $$ f(x)=x^3+2x^2+10x-20=0.$$ Can anyone tell me what the method of deflation is and how I would use it to prove the statement?

I found the root $x=1.3688$ by using Newtons method.

Any insight would be greatly appreciated.

Answer 1

The section of Factoring polynomials using the deflation method explains deflation. In summary, it says you find an initial root $r$ of the polynomial, such as by trial & error, or inspecting the graph, then

we find the other factor by dividing the polynomial by $(x − r)$. This is called deflating the polynomial.

It says to repeat this until no more real roots can be found.

In your case, you have found a root of $r \approx 1.3688$ by using Newton's method. As such, the procedure is to divide $x - 1.3688$ into your original polynomial, which gives (accurate to about $4$ decimal places for the constant term)

$$x^3 + 2x^2 + 10x − 20 \approx \left(x - 1.3688\right)\left(x^2 + 3.3688x + 14.6112\right) \tag{1}\label{eq1}$$

Next, according to the method of deflation, you should graph the function $g\left(x\right) = x^2 + 3.3688x + 14.6112$ to confirm there are no more zeros, or according to an example in the referenced Web page, you can use the "grouping" method as provided in its section 8.3.

Alternatively, you can note the discriminant of $b^2 - 4ac = 3.3688^2 - 4\left(14.6112\right) \approx -47.0959 \lt 0$, to show it has no real roots. In addition, you can just use the derivative of the original $3$rd degree polynomial, as used in the other $2$ current answers.

Answer 2

Because $$f'(x)=3x^2+4x+10>0,$$ which says that $f$ increases.

Also, $f(1)<0$, $f(2)>0$ and $f$ is a continuous function.

Answer 3

$$ 3 x^2 + 4 x + 10 $$ is always positive. $$ 3 \left(x + \frac{2}{3}\right)^2 + \frac{26}{3} $$

I have no idea what the method of deflation might be