Chapter 5.5, Problem 37PPSE

Solution

To calculate : The rational roots of the given function.

The rational roots of the given polynomial function are $-1,\frac{1}{2},\text{ and }-\frac{1}{3}$ .

Given information :

The function is $P\left(x\right)=6x^3+5x^2-2x-1$ .

Formula Used:

Rational Root Theorem: For a given polynomial with integer coefficients, the rational roots will be in form of $\frac{p}{q}$ , where $p$ is the integer factor of the constant term and $q$ is the integer factor of the leading coefficient.

Calculation :

Given polynomial function: $P\left(x\right)=6x^3+5x^2-2x-1$

Find the possible rational roots.

Determine the integer factor of the constant term $1$ .

  $p=\text{±1}$

Find the integer factor of the leading coefficient $6$ .

  $q=\pm 1,\pm 2,\pm 3,\pm 6$

Determine the list of rational roots that will be in reduced form of $\left(\frac{p}{q}\right)$ .

  $\left(\frac{p}{q}\right)=\text{±1,}\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{6}$

Apply the synthetic division method to test the roots.

  $\left\{\begin{array}{l}\overline{)1}\underset{\downarrow }{6}5-2-1\\ \text{ 6 }119\\ \overline{\text{ 6 }119\overline{)8}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-1}\underset{\downarrow }{6}5-2-1\\ -611\\ \overline{\text{ 6 }-1-1\overline{)0}}\end{array}\right\}$

Here, the remainder is zero, so $x=-1$ but $x\ne 1$ as the remainder is not zero.

  $\left\{\begin{array}{l}\overline{)\frac{1}{2}}\underset{\downarrow }{6}5-2-1\\ \text{ 3 }41\\ \overline{\text{ 6 }82\overline{)0}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{2}}\underset{\downarrow }{6}5-2-1\\ -3-1\frac{3}{2}\\ \overline{\text{ 6 }2-3\overline{)\frac{1}{2}}}\end{array}\right\}$

Here, the remainder is zero, so $x=\frac{1}{2}$ but $x\ne -\frac{1}{2}$ as the remainder is not zero.

  $\left\{\begin{array}{l}\overline{)\frac{1}{3}}\underset{\downarrow }{6}5-2-1\\ \text{ 2 }\frac{7}{3}\frac{1}{9}\\ \overline{\text{ 6 }7\frac{1}{3}\overline{)-\frac{8}{9}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{3}}\underset{\downarrow }{6}5-2-1\\ -2-11\\ \overline{\text{ 6 }3-3\overline{)0}}\end{array}\right\}$

Here, the remainder is zero, so $x=-\frac{1}{3}$ but $x\ne \frac{1}{3}$ as the remainder is not zero.

  $\left\{\begin{array}{l}\overline{)\frac{1}{6}}\underset{\downarrow }{6}5-2-1\\ \text{ 1 }1-\frac{1}{6}\\ \overline{\text{ 6 }6-1\overline{)-\frac{7}{6}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{6}}\underset{\downarrow }{6}5-2-1\\ -1-\frac{2}{3}\frac{4}{9}\\ \overline{\text{ 6 }4-\frac{8}{3}\overline{)-\frac{5}{9}}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{1}{6}$ .

Hence, the rational roots of the given polynomial function are $-1,\frac{1}{2},\text{ and }-\frac{1}{3}$ .