Solve the inequality 6p5 - 15p4 + 10p3 is less than p.

Step-by-step solution

1. 1

   Move everything to one side and factor out $p$.

   $6p^5-15p^4+10p^3-p<0\quad\Rightarrow\quad p\left(6p^4-15p^3+10p^2-1\right)<0$

   Working with a single expression compared to zero is what makes a sign chart possible.

2. 2

   Find the rational roots of the quartic. Testing the candidates from the rational root theorem: $p=1$ gives $6-15+10-1=0$, and $p=\tfrac12$ gives $\tfrac38-\tfrac{15}8+\tfrac52-1=0$. So $(p-1)$ and $(2p-1)$ are factors. Dividing them out,

   $6p^4-15p^3+10p^2-1=(p-1)(2p-1)\left(3p^2-3p-1\right)$

   (Leading coefficients $1\cdot 2\cdot 3=6$ ✓ and constants $(-1)(-1)(-1)=-1$ ✓.)

3. 3

   Complete the factorization and list the critical points.

   $p(p-1)(2p-1)\left(3p^2-3p-1\right)<0$

   The last factor has roots $p=\frac{3\pm\sqrt{21}}{6}$, i.e. $-0.26376$ and $1.26376$. Sorted, the five simple roots are

   $-0.26376,\quad 0,\quad 0.5,\quad 1,\quad 1.26376$

4. 4

   Build the sign chart by testing one point per interval. Substituting into the original inequality as $\text{LHS}-\text{RHS}$:

   $p=-1:\ -30\ (<0)\quad p=-0.2:\ +0.094\quad p=0.25:\ -0.1465\ (<0)$
   $p=0.75:\ +0.1465\quad p=1.1:\ -0.0884\ (<0)\quad p=2:\ +30$

   All five roots are simple, so the sign alternates — and the leading coefficient $6>0$ with odd degree $5$ forces the expression positive far to the right, which pins the whole pattern.

5. 5

   Read off the solution. The expression is negative on three intervals:

   $p<\frac{3-\sqrt{21}}{6},\qquad 0<p<\frac12,\qquad 1<p<\frac{3+\sqrt{21}}{6}$

   Beware: a sign chart built from the wrong end gives exactly the complement of this set — an easy error to make and an easy one to catch, since $p=2$ must fail ($192-240+80=32$, and $32<2$ is false) while $p=0.25$ must hold ($0.1035<0.25$ ✓).

6. 6

   Confirm the endpoints are excluded. At each of the five critical points the two sides are equal, and the inequality is strict, so none of $\frac{3-\sqrt{21}}{6}$, $0$, $\frac12$, $1$, $\frac{3+\sqrt{21}}{6}$ belongs to the solution set.