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I've been trying to simplify the problem but I can't. I tried long division factoring and perfect cube but I can't still solve it. My calculator shows $x=3$ and $x=-1$ but wait how? this is for the numerator

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  • $\begingroup$ What do you mean $x=3$ and $x=-1$? The function is $\geq 0$ in between those values? $\endgroup$
    – Jack M
    Feb 18, 2015 at 6:44
  • $\begingroup$ the value of the x's I just entered it into my calculator with EQN mode $\endgroup$
    – Mickey
    Feb 18, 2015 at 6:45
  • $\begingroup$ Those are the values at which the function equals zero. Is that what you're interested in? $\endgroup$
    – Jack M
    Feb 18, 2015 at 6:46
  • $\begingroup$ I want to know how to solve it. and solve the whole inequality, I solved some rationals but those problems are factorable unlike this one which i don't know how to start. $\endgroup$
    – Mickey
    Feb 18, 2015 at 6:49
  • $\begingroup$ I was hoping if there's a way to simplify the problem $\endgroup$
    – Mickey
    Feb 18, 2015 at 6:50

3 Answers 3

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Use the rational root theorem and a little arithmetic to find that $3$ or $-1$ is a root of the numerator. Let's say that you find $3$. Then you know that $x-3$ is a factor. Divide it out to get $x^2+2x+1$; this easily factors as $(x+1)^2$, so the numerator is $(x+1)^2(x-3)$.

The denominator has an obvious factor of $x$; pull it out to leave $x^2-x-15$. This doesn't have a factorization with integer coefficients, but the quadratic formula tells you that its roots are $\frac12(1\pm\sqrt{61})$, so the denominator is $x\left(x-\frac12(1-\sqrt{61})\right)\left(x-\frac12(1+\sqrt{61})\right)$.

You now know exactly where the fraction can change sign: at $x=-1,0,3$, and $\frac12(1\pm\sqrt{61})$. Thus, you need to test values in the intervals defined by these breakpoints.

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  • $\begingroup$ The product of the roots of that quadratic is $-15$. The product of those 2 fractions is not an integer. $\endgroup$
    – Mike
    Feb 18, 2015 at 8:18
  • $\begingroup$ @Mike: I really shouldn't do this on a Kindle: too little screen, too easy to make stupid careless mistakes. sigh Thanks for catching it. $\endgroup$ Feb 18, 2015 at 8:49
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it seems you can factor $x^3-x^2 - 5x - 3 = (x+1)^2(x-3)$ and $x^3 - x^2-15x=x(x^2-x-15)=x(x -a)(x-b)$ where $a = \dfrac{1-\sqrt{61}}2, b = \dfrac{1+\sqrt{61}}2$

we need to solve $$y = \dfrac{(x+1)^2(x-3)}{(x-a)x(x-b)} \ge 0$$ you can draw a line and put the numbers $a, -1, 0, 3$ and $b$ and do a sign analysis or you can see that $y = 1+ \cdots$ for large $x$ and switches signs across $a, 0, 3$ and $b$ but at $x = 1$ it touches the $x$-axis from below. you will see that $$\dfrac{x^3-x^2 - 5x - 3}{x^3 - x^2 - 15x} \ge 0\text{ on }-\infty < x < a, 0 < x \le 3\text{ and }b < x < \infty.$$

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I have solved. So, you can transform your inequality to $1+\frac{10 x-3}{x(x^2-x-15)} \geq 0$ which is equivalent to $\frac{10 x - 3}{x(x^2-x-15}$ which is equivalent, if $x(x^2-x-15) \geq 0$ with $10 x - 3 \geq x(-x^2+x+15)$ which gives us inequality : $-x^3+x^2+5 x+ 3 \leq 0$ take now the free element in a polynomial which is 3. So, you can check that 3 satisfies it and when you divide polinomial with x - 3 you get $-x^2-2*x-1$ which is -$(x+1)^2$. I have omited really high school mathematics.

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