Solving Polynomial and Rational Inequalities

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Sol 1 Factoring gives $x(x^2-x-6)>0$ or $x(x-3)(x+2)>0$. Marking 0,3, and -2 on a number line, and using that

$x=1\Rightarrow y=-6<0$ and that all the exponents are odd, we get the sign chart shown below:

Therefore the solution is given by $(-2,0)\cup(3,\infty)$.

Sol 2 Factoring gives $x^2(x^2-x-6)<0$ or $x^2(x-3)(x+2)<0$. Marking 0,3, and -2 on a number line, and using that

$x=1\Rightarrow y=-6<0$ and that the sign changes at 3 and -2 but does not change at 0, we get the sign chart shown below:

Therefore the solution is given by $(-2,0)\cup(0,3)$.

Sol 3 Factoring gives

$$\frac{(x-4)(x-1)}{(x-2)(x+2)}\le0;$$

and marking 4,1,2, and -2 on a number line, and using that

$x=0\Rightarrow y=-1<0$ and that all the exponents are odd, we get the sign chart shown below:

Since the inequality is not strict, we can include the zeros of the numerator; so the solution is given by $(-2,1]\cup(2,4]$.

Sol 4 Factoring gives

$$\frac{(2x-5)(x+2)}{(4-x)(3+x)}>0;$$

and marking 5/2,-2, 4, and -3 on a number line, and using that

$x=0\Rightarrow y=-10/12=-5/6<0$ and that all the exponents are odd, we get the sign chart shown below:

Therefore the solution is given by $(-3,-2)\cup(5/2,4)$.

Sol 5 Factoring gives

$$\frac{(x^2-9)(x^2-1)}{(x^2+4)(x^2-4)}\ge0$$

or

$$\frac{(x-3)(x+3)(x-1)(x+1)}{(x^2+4)(x-2)(x+2)}\ge0.$$

Marking -3,-2,-1,1,2, and 3 on a number line, and using the facts that

$x=0\Rightarrow y=-9/16<0$ and that all the exponents are odd, we get the following sign chart:

Since the inequality is not strict, we can include the zeros of the numerator; so the solution is given by $(-\infty,-3]\cup(-2,-1]\cup[1,2)\cup[3,\infty)$.

Sol 6 Since $x^2\ge0$ for all $x$, $x^2+1\ge1$ and therefore $(x^2+1)^3>0$ for all $x$; so multiplying by $(x^2+1)^3$ gives the equivalent inequality $2x^3-6x>0$.

Factoring yields $2x(x^2-3)>0$ or $2x(x-\sqrt{3})(x+\sqrt{3})>0$; so marking $0,\sqrt{3},-\sqrt{3}$ on a number line and using that

$2(1)^3-6(1)=-4<0$ when $x=1$ and all the exponents are odd, we get the following sign chart:

Therefore the solution is given by $(-\sqrt{3},0)\cup(\sqrt{3},\infty)$.

Sol 7 Factoring gives the inequality

$$\frac{x^2(2x^2-5x+3)}{(x-4)^2(3x-1)(x+2)}\ge0$$

or

$$\frac{x^2(2x-3)(x-1)}{(x-4)^2(3x-1)(x+2)}\ge0.$$

Marking off -2,0,1/3,1,3/2, and 4, and using the facts that

$x=-1\Rightarrow y=10/(-100)=-1/10<0$ and the sign changes at 3/2,1,-2, and 1/3 and does not change at 0 or at 4, we get the following sign chart:

Since the inequality is not strict, we can include the zeros of the numerator; so the solution is given by $(-\infty,-2)\cup(1/3,1]\cup[3/2,4)\cup(4,\infty)$.
Correction:  the solution should include x=0 as well.

Sol 8 Subtracting $\frac{16}{x}$ from both sides gives $x-\frac{16}{x}<0$, so $\frac{x^2-16}{x}<0$ or $\frac{(x-4)(x+4)}{x}<0$.

Marking -4,0,and 4 on a number line, and using that

$x=1\Rightarrow y=-15<0$and that all the exponents are odd, we get the following sign chart:

Therefore $(-\infty,-4)\cup(0,4)$ is the solution.

Sol 9 Subtracting $\frac{x-5}{x+1}$ from both sides gives

$$\frac{x}{x-4}-\frac{x-5}{x+1}<0,$$

so

$$\frac{x(x+1)-(x-4)(x-5)}{(x-4)(x+1)}<0.$$

Therefore

$$\frac{x^2+x-(x^2-9x+20)}{(x-4)(x+1)}<0,$$

so

$$\frac{10x-20}{(x-4)(x+1)}<0$$

or

$$\frac{10(x-2)}{(x-4)(x+1)}<0.$$

Marking 2,4, and -1 on a number line, and using the facts that

$x=0\Rightarrow y=5>0$ and that all the exponents are odd, we get the following sign chart:

Therefore $(-\infty,-1)\cup(2,4)$ is the solution.

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