Chapter 5.5, Problem 34PPSE

Solution

To calculate : The rational roots of the given function.

The roots of the given polynomial function are $-\frac{1}{3},\text{ and }\frac{5}{2}$ .

Given information :

The function is $P\left(x\right)=6x^4-13x^3+13x^2-39x-15$ .

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: $P\left(x\right)=6x^4-13x^3+13x^2-39x-15$

Find the possible rational roots.

Determine the integer factor of the constant term $15$ .

  $p=\text{±1,±3,±5,}\pm 15$

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 $\left(\frac{p}{q}\right)$ .

  $\left(\frac{p}{q}\right)=\text{±1,±3,±5,}\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},\pm \frac{1}{3},\pm \frac{5}{3},\pm \frac{1}{6},\pm \frac{5}{6}$

Apply the synthetic division method to test the roots.

  $\left\{\begin{array}{l}\overline{)1}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 6 }-76-33\\ \overline{\text{ 6 }-76-33\overline{)-48}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-1}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -619-3271\\ \overline{\text{ 6 }-1932-71\overline{)56}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)3}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 18 }1584135\\ \overline{\text{ 6 }52845\overline{)120}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-3}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -1893-318\text{ 10}71\\ \overline{\text{ 6 }-31106-357\overline{)1056}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm 3$ .

  $\left\{\begin{array}{l}\overline{)5}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 30 }854902255\\ \overline{\text{ 6 }1798451\overline{)2240}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-5}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -30215-11405895\\ \overline{\text{ 6 }-43228-1179\overline{)5880}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm 5$ .

  $\left\{\begin{array}{l}\overline{)15}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 90 }115517520262215\\ \overline{\text{ 6 }77116817481\overline{)262200}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-15}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -901545-23370351135\\ \overline{\text{ 6 }-1031558-23409\overline{)351120}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm 15$ .

  $\left\{\begin{array}{l}\overline{)\frac{1}{2}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 3 }-54-\frac{35}{2}\\ \overline{\text{ 6 }-108-35\overline{)-\frac{65}{2}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{2}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -38-\frac{21}{2}\frac{99}{4}\\ \overline{\text{ 6 }-1621-\frac{99}{2}\overline{)\frac{39}{4}}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{1}{2}$ .

  $\left\{\begin{array}{l}\overline{)\frac{3}{2}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 9 }-6\frac{21}{2}-\frac{171}{4}\\ \overline{\text{ 6 }-47-\frac{57}{2}\overline{)-\frac{231}{4}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{3}{2}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -933-69162\\ \overline{\text{ 6 }-2246-108\overline{)147}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{3}{2}$ .

  $\left\{\begin{array}{l}\overline{)\frac{5}{2}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 15 }54515\\ \overline{\text{ 6 }2186\overline{)0}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{5}{2}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -1570-\frac{415}{2}\frac{2465}{4}\\ \overline{\text{ 6 }-2883-\frac{493}{2}\overline{)\frac{1225}{2}}}\end{array}\right\}$

Here, the remainder is zero, so $x=\frac{5}{2}$ but $x\ne -\frac{5}{2}$ as the remainder is not zero.

  $\left\{\begin{array}{l}\overline{)\frac{15}{2}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 45 }240\frac{3795}{2}\frac{55755}{4}\\ \overline{\text{ 6 3}2253\frac{3717}{2}\overline{)\frac{55695}{4}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{15}{2}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -45435-3360\frac{50985}{2}\\ \overline{\text{ 6 }-58448-3399\overline{)\frac{50955}{2}}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{15}{2}$ .

  $\left\{\begin{array}{l}\overline{)\frac{1}{3}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 2 }-\frac{11}{3}\frac{28}{9}-\frac{323}{27}\\ \overline{\text{ 6 }-11\frac{28}{3}-\frac{323}{9}\overline{)-\frac{728}{27}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{3}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -25-615\\ \overline{\text{ 6 }-1518-45\overline{)0}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)\frac{5}{3}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 10 }-5\frac{40}{3}-\frac{385}{9}\\ \overline{\text{ 6 }-38-\frac{77}{3}\overline{)-\frac{520}{9}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{5}{3}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -10\frac{115}{3}-\frac{770}{9}\frac{5605}{27}\\ \overline{\text{ 6 }-23\frac{154}{3}-\frac{1121}{9}\overline{)\frac{5200}{27}}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{5}{3}$ .

  $\left\{\begin{array}{l}\overline{)\frac{1}{6}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 1 }-2\frac{11}{6}-\frac{223}{36}\\ \overline{\text{ 6 }-12\text{ 11 }-\frac{223}{6}\overline{)-\frac{763}{36}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{6}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -1\frac{7}{3}-\frac{23}{9}\frac{187}{27}\\ \overline{\text{ 6 }-14\frac{46}{3}-\frac{374}{9}\overline{)-\frac{187}{27}}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)\frac{5}{6}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ \text{ 5 }-\frac{20}{3}\frac{95}{18}-\frac{3035}{108}\\ \overline{\text{ 6 }-8\frac{19}{3}-\frac{607}{18}\overline{)-\frac{4655}{108}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{5}{6}}\underset{\downarrow }{6}-13\text{ 13 }-39-15\\ -515-\frac{70}{3}\frac{935}{18}\\ \overline{\text{ 6 }-1828-\frac{187}{3}\overline{)\frac{665}{18}}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{5}{6}$ .

Hence, the roots of the given polynomial function are $-\frac{1}{3},\text{ and }\frac{5}{2}$ .