Finding real zeros of polynomial of third degree to solve inequality

Question

\begin{align}{} & \dfrac{x^3+2 x^2}{2} < x+2 \\[18pt] & x^3 + 2x^2 < 2x+4 \\ & x^3 + 2x^2 -2x - 4 < 0 \\ \end{align}

Let $p(x) = x^3 + 2x^2 -2x - 4$. Setting $p(x) = 0$

Cauchy's bound tells me the real zeros should be in the range of $[-5, 5]$
Rational zeros theorem gives me candidates to perform polynomial division with: $\{\pm1, \pm2, \pm4 \}$
Descarte's rule of signs predicts 1 positive and either 2 or no negative real zeros.

Now, performing polynomial division of the form $p(x) \div (x - c)$, where c is each of the candidate rational zeros above, I can find $x = -2$ as a real zero, leaving polynomial $q(x) = x^2 - 2$, which is a quadratic, yielding $x = \pm \sqrt{2}$

Solving the inequality, by testing values on each interval, gives $(-\infty, -2) \cup (-\sqrt{2}, \sqrt{2})$ as the answer.

Perhaps you should check your work on testing those rational zeros. One of them should work by my calculation. — sharding4, Sep 02 '17 at 22:28
@sharding No luck with testing by evaluating the polynomial either... — pragMATHiC, Sep 02 '17 at 22:28
You've made a mistake with your polynomial. Should be $x^3+2x^2-2x-4$, shouldn't it? — sharding4, Sep 02 '17 at 22:29
Ooops... I can certainly see the error now!... thanks @sharding4 — pragMATHiC, Sep 02 '17 at 22:31
By the way, you can factor out the term $x+2$ on the left side at the beginning, this makes the further calculation easier. — Peter, Sep 02 '17 at 22:32

score 0 · Accepted Answer · answered Sep 02 '17 at 22:41

Let $p(x) = x^3 + 2x^2 -2x - 4$. Setting $p(x) = 0$

Cauchy's bound tells me the real zeros should be in the range of $[-5, 5]$
Rational zeros theorem gives me candidates to perform polynomial division with: $\{\pm1, \pm2, \pm4 \}$
Descarte's rule of signs predicts 1 positive and either 2 or no negative real zeros.

Now, performing polynomial division of the form $p(x) \div (x - c)$, where c is each of the candidate rational zeros above, I can find $x = -2$ as a real zero, leaving polynomial $q(x) = x^2 - 2$, which is a quadratic, yielding $x = \pm \sqrt{2}$

Solving the inequality, by testing values on each interval, gives $(-\infty, -2) \cup (-\sqrt{2}, \sqrt{2})$ as the answer.

Finding real zeros of polynomial of third degree to solve inequality

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