$x^3+b^2x^2+2x+3=0$,Find several integer values of b such that the equation has roots.

Question

$x^3+b^2x^2+2x+3=0$, Find several integer values of $b$ such that the equation has roots.

My solution: I use the rational root theorem. $-3,3$ can be its rational roots.

$$P(x)=x^3+b^2x^2+2x+3$$

$$P(3)=9b^2+36=0$$ which is not possible.

$$P(-3)=9b^2-30=0$$ which gives $b=-\frac{\sqrt{10}}{3},\frac{\sqrt{10}}{3}$

But these two values of $b$ are not integers.

Can you solve this with another approach? Are there other values of $b$?

EDIT: $x=-1,1$ can also be its rational roots.

$P(1)=0$ is not possible and $P(-1)=0$ gives $b=0$

EDIT: Please also kindly discuss the cases of the existence of two real roots and three real roots.I only discuss the rational roots because I do not know how to handle these other two cases.

Thanks

Answer 1

From your responses it is clear you want to gain a fuller understanding of the cubic solution. Study the Cardano's solution. Next, if you are willing to use Mathematica, Manipulate command graphs all transitions or possibilities you want to see. I have included B < 0 also, so you gain broad understanding. As B is varied, cubic curve moves. Single cutting of red line is at the real root ( other two are two complex). Triple cutting for three real roots. One double real root with an isolated real root can be also seen approximately in tangential situation.

Manipulate[ Plot[x^3 + B x ^2 + 2 x + 3, {x, -10, 10}, PlotRange -> 10, GridLines -> Automatic, PlotStyle -> {Red, Thick}], {B, -5, 5, 0.1}]

Answer 2

If $x>-1$, then the function $f(x)=x^3+b^2x^2+2x+3$ gets positive values, thus the range of solutions is $x\in(-\infty,-1]$ and $b \in \mathbb{Z}$.

As mentioned $x=-1$ leads to the option $b=0$.

Now we need to look on the range $(-\infty,-1)$.

By Sturm's theorem: $$p_0(x)=f(x)=x^3+b^2x^2+2x+3\\ p_1(x)=f'(x)=3x^2+2b^2x+2 \\ p_2(x)=-\mathrm{rem\,}(p_0,p_1)=\frac{2b^4-12}{9}x+\frac{2b^2-27}{9} \\ p_3(x)=-\mathrm{rem\,}(p_1,p_2)=\frac{(4b^6-30b^2+81)(2b^2-27)}{(2b^4-12)^2}-2$$

We have two cases:

1. $p_3(x)>0$, then when evaluating these polynomials at $-\infty$ we get the sequence of signs $-+-+$, thus there are three sign changes. When evaluating at $+\infty$ we get the sequence of signs $++++$, thus there are no sign changes. By the theorem the number of real roots is the difference between the number of changes when evaluated at $-\infty$ to the number of changes when evaluated at $+\infty$, hence the number of real roots is $3-0=3$.
2. $p_3(x)<0$ ,then when evaluating these polynomials at $-\infty$ we get the sequence of signs $-+--$, thus there are two sign changes. When evaluating at $+\infty$ we get the sequence of signs $+++-$, thus there is one sign change. By the theorem the number of real roots is the difference between the number of changes when evaluated at $-\infty$ to the number of changes when evaluated at $+\infty$, hence the number of real roots is $2-1=1$.

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Now we need to find for what values of $b$ we get $p_3(x)>0$ and for what values $p_3(x)<0$.

We will solve for the first case. $$\frac{(4b^6-30b^2+81)(2b^2-27)}{(2b^4-12)^2}-2>0 \rightarrow \frac{(4b^6-30b^2+81)(2b^2-27)-2(2b^4-12)^2}{(2b^4-12)^2}>0$$ The denomirator is positive for $b \ne \pm\sqrt[4]{6}$ (we are interested in integral solutions for b, so we don't have to check this two). We need to check when the numerator is positive, hence $$(4b^6-30b^2+81)(2b^2-27)-2(2b^4-12)^2>0 \rightarrow 0>108b^6-36b^4-972^2+2475 \\ 0>12b^6-4b^4-108b^2+275$$ Define $b^2=t$ we get the cubic inequation $0>12t^3-4t^2-108t+275$.

We shall look on the equation $12t^3-4t^2-108t+275=0$ and check how many solutions it has. Using Sturm's theorem (We can use Wolfram Widget) $$p_0(t)=12t^3-4t^2-108t+275 \\ p_1(t)=36t^2-8t-108 \\ p_2(t)=\frac{1952}{27}x-271 \\ p_3(t)=-\frac{350402625}{952576}$$

Evaluating these polynomials at $-\infty$ we get the sequence of signs $-++-$, thus there are two sign changes. Evaluating these polynomials at $+\infty$ we get the sequence of signs $+++-$, thus there is one sign change. By the theorem we get that there is one real solution (you can find it here).

Now, $p_0(0)>0 \ , \ p_0(-100)<0$, thus by IVT the solution $t_1$ is in $(-100,0)$.

We defined $t=b^2$, hence the equation $12b^6-4b^4+108b^2-275=0$ has no real solution, so there is no such $b$ such that the equation $x^3+b^2x^2+2x+3=0$ has three real solutions.

This result implies that $\forall b: \ p_3(x)<0$, so there is only one real solution.

This result also implies that for any $b\in\mathbb{Z}$ there is a solution.

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