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We all know, $ax^2+bx+c \geqslant 0$ is true if $a > 0$ and $4ac-b^2 \geqslant 0.$ So, what is conditon of $$ax^3+bx^2+cx+d \geqslant 0, \quad a > 0 \quad ?$$

  • You might want to look for the “cubic discriminant” on Google. – ViHdzP Dec 25 '19 at 08:21
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    A cubic polynomial (or any odd-degree polynomial) goes "up to infinity" on one end and "down to negative infinity" on the other end. (If $a>0$, the "up" end is on the right.) So, the graph must cross the $x$-axis at least once, which is to say: no condition on the coefficients will make all values non-negative. – Blue Dec 25 '19 at 08:23
  • https://images.app.goo.gl/NrzhUsBGVUTr3LKT6 can you see what must be true in order for this to have real roots? This is the formula for trinomial roots by the way – Ty Jensen Dec 25 '19 at 08:24
  • https://math.stackexchange.com/questions/1393869/conditions-for-distinct-real-roots-of-cubic-polynomials – Alessio K Dec 25 '19 at 10:19

2 Answers2

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There is no such condition since $$\lim _{x\to \infty}f(x)= \infty$$ and $$\lim _{x\to -\infty}f(x)= -\infty$$

meaning $Range(f) = \mathbb{R}$ since it is continious.

nonuser
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$$ax^3+bx^2+cx+d \geqslant 0$$

This is a false statement for real coefficients of the equation. (I can't say for imaginary ones).

Letting $$ax^3+bx^2+cx+d=y$$ (y being the range).

The graph would go on from $-\infty$ to $+\infty$ as stated by Aqua.

The graph would look somewhat like this:—

Source: Wikipedia

Finding the coefficient of Cubic equation by using Cardano's Method can be helpful.

Crocogator
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