# For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root

For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root.

how can i show that above statement is true or false.can anyone help me please

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You can show (for example using the derivative) that the function $x \mapsto x^3 + x + c$ is increasing for any real $c$.

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strictly increasing even. – lhf Nov 23 '12 at 13:14

Here’s yet another argument, not using calculus. As @xen remarks, a strictly increasing real-valued function will have at most one root. Letting $f$ be our function, we find that $f(x+\varepsilon)-f(x)=\varepsilon\lbrack 3x^2+3x\varepsilon+\varepsilon^2+1\rbrack$, and the quantity in brackets is a sum of squares, namely $3x^2/4+(3x/2+\varepsilon)^2+1$. So the computed difference has the same sign as $\varepsilon$, the function is strictly increasing.

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The easiest way is to show that $x \mapsto x^3+x+c$ is increasing or use Rolle's theorem. Another way is to use Descartes' rule of signs.
If $c>0$ it has a negative root while if $c\leq0$ it has a nonnegative root.

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See wikipedia for a algebric condition in wikipedia section "Roots of a cubic function". Let $$(1) \quad ax^3+bx^2+cx+d=0.$$ Every cubic equation $(1)$ with real coefficients has at least one solution ''x'' among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminante $\Delta$ of $(1)$,

$$\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2.$$ The following cases need to be considered:

If $\Delta > 0$, then the equation has three distinct real roots.

If $\Delta = 0$, then the equation has multiple root and all its roots are real.

If $\Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.

For reference see the book

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For a complet dedution of $\Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped. – MathOverview Nov 23 '12 at 13:34

$x^3 + x + c = 0$
$x^3 + x = y = -c$
$x(x^2 + 1) = y$

$(x^2 + 1)$ is always positive and greater than zero, and $x$ takes both negative and positive values. So it is possible to sweep the entire real numbers, including $d$, by giving $x$ values from $-\infty$ to $+\infty$. In other words, $y=x(x^2 + 1)$ is a smooth function from $\mathbb{R}$ to $\mathbb{R}$.

So, $x^3 + x + c = 0$ must have at least real root, since $y = x(x^2 + 1)$ always has a solution $x_0$ for any given $y_0$.

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