## College Algebra (6th Edition)

Solution set: $(-\displaystyle \infty, -\frac{1}{2})\cup(\frac{1}{3},\infty)$
Follow the "Procedure for Solving Polynomial lnequalities",\ p.412: 1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$ where $f$ is a polynomial function. $6x^{2}+x >1$ $6x^{2}+x-1>0$ $f(x)= 6x^{2}+x-1$ factor the trinomial... find factors of $6(-1)=-6$ that add to $+1:$ ($3$ and $-2$ ) $6x^{2}+x-1=6x^{2}+3x-2x-1= \quad$ ... in pairs ... $=3x(2x+1)-(2x+1)=(3x-1)(2x+1)$ $f(x)=(3x-1)(2x+1)>0$ 2. Solve the equation $f(x)=0$. The real solutions are the boundary points. $(3x-1)(2x+1)=0$ $x=-\displaystyle \frac{1}{2} x=\displaystyle \frac{1}{3}$ 3. Locate these boundary points on a number line, thereby dividing the number line into intervals. 4. Test each interval's sign of $f(x)$ with a test value, $\begin{array}{llll} Intervals: & (-\infty, -\frac{1}{2}) & (-\frac{1}{2},\frac{1}{3}) & (\frac{1}{3},\infty)\\ a=test.val. & -1 & 0 & 1\\ f(a) & (-4)(-1) & (-1)(1) & (2)(3)\\ f(a) > 0 ? & T & F & T \end{array}$ 5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points. Solution set: $(-\displaystyle \infty, -\frac{1}{2})\cup(\frac{1}{3},\infty)$