Chapter 5.5, Problem 36PPSE

Solution

To calculate : The rational roots of the given function.

The given polynomial function has no rational root.

Given information :

The function is $P\left(x\right)=3x^4-7x^3+10x^2-x+12$ .

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: $P\left(x\right)=3x^4-7x^3+10x^2-x+12$

Find the possible rational roots.

Determine the integer factor of the constant term $12$ .

  $p=\text{±1,±2,±3,±4,±6,±12}$

Find the integer factor of the leading coefficient $3$ .

  $q=\pm 1,\pm 3$

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

  $\left(\frac{p}{q}\right)=\text{±1,±2,±3,±4,±6,±12,}\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3}$

Apply the synthetic division method to test the roots.

  $\left\{\begin{array}{l}\overline{)1}\underset{\downarrow }{3}-710-112\\ \text{ 3 }-465\\ \overline{\text{ 3 }-465\overline{)17}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-1}\underset{\downarrow }{3}-710-112\\ -310-2021\\ \overline{\text{ 3 }-1020-21\overline{)33}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)2}\underset{\downarrow }{3}-710-112\\ \text{ 6 }-2\text{ 1}630\\ \overline{\text{ 3 }-18\text{ 1}5\overline{)42}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-2}\underset{\downarrow }{3}-710-112\\ -626-72146\\ \overline{\text{ 3 }-1336-73\overline{)158}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)3}\underset{\downarrow }{3}-710-112\\ \text{ 9 }648141\\ \overline{\text{ 3 }21647\overline{)153}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-3}\underset{\downarrow }{3}-710-112\\ -948-174525\\ \overline{\text{ 3 }-1658-175\overline{)537}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)4}\underset{\downarrow }{3}-710-112\\ \text{ 12 }20120476\\ \overline{\text{ 3 }530119\overline{)488}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-4}\underset{\downarrow }{3}-710-112\\ -1276-3441380\\ \overline{\text{ 3 }-1986-345\overline{)1392}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm 4$ .

  $\left\{\begin{array}{l}\overline{)6}\underset{\downarrow }{3}-710-112\\ \text{ 18 }664562730\\ \overline{\text{ 3 }1176455\overline{)2718}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-6}\underset{\downarrow }{3}-710-112\\ -18150-9605766\\ \overline{\text{ 3 }-25160-961\overline{)5754}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)\frac{1}{3}}\underset{\downarrow }{3}-710-112\\ \text{ 1 }241\\ \overline{\text{ 3 }6123\overline{)13}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{3}}\underset{\downarrow }{3}-710-112\\ -1\frac{8}{3}-\frac{38}{9}\frac{47}{27}\\ \overline{\text{ 3 }-8\frac{38}{3}-\frac{47}{9}\overline{)\frac{371}{27}}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{1}{3}$ .

  $\left\{\begin{array}{l}\overline{)\frac{2}{3}}\underset{\downarrow }{3}-710-112\\ \text{ 2 }\frac{10}{3}\frac{80}{9}\frac{142}{27}\\ \overline{\text{ 3 }5\frac{40}{3}\frac{71}{9}\overline{)\frac{466}{27}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{2}{3}}\underset{\downarrow }{3}-710-112\\ -26-\frac{32}{3}\frac{70}{9}\\ \overline{\text{ 3 }-916-\frac{35}{3}\overline{)\frac{178}{9}}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)\frac{4}{3}}\underset{\downarrow }{3}-710-112\\ \text{ 4 }4\frac{56}{3}\frac{212}{9}\\ \overline{\text{ 3 3 }14\frac{53}{3}\overline{)\frac{320}{9}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{4}{3}}\underset{\downarrow }{3}-710-112\\ -4\frac{44}{3}-\frac{296}{9}\frac{1220}{27}\\ \overline{\text{ 3 }-11\frac{74}{3}-\frac{305}{9}\overline{)\frac{1544}{27}}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{4}{3}$ .

Hence, the given polynomial function has no rational root.