Chapter 5, Problem 42E

Solution

To calculate : The possible rational roots of the given function using Rational Root Theorem.

The given polynomial function has no rational roots.

Given information :

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

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

Find the possible rational roots.

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

  $p=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6\text{ and}\pm 12$

Find the integer factor of the leading coefficient $4$ .

  $q=\pm 1,\pm 2\text{ and }\pm 4$

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

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

Apply the synthetic division method to test the roots.

  $\left\{\begin{array}{l}\overline{)1}\underset{\downarrow }{4}-\text{2 }10-12\\ 4233\\ \overline{\text{ 4 }23\text{ 3 }\overline{)-9}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-1}\underset{\downarrow }{4}-\text{2 }10-12\\ -46-77\\ \overline{\text{ 4 }-67-7\overline{)-5}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)2}\underset{\downarrow }{4}-\text{2 }10-12\\ 8122652\\ \overline{\text{ 4 }6\text{ 1}3\text{ 26 }\overline{)40}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-2}\underset{\downarrow }{4}-\text{2 }10-12\\ -8\text{ 20 }-42\text{ 84}\\ \overline{\text{ 4 }-1021-42\overline{)72}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)3}\underset{\downarrow }{4}-\text{2 }10-12\\ 123093279\\ \overline{\text{ 4 }1031\text{ 93 }\overline{)267}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-3}\underset{\downarrow }{4}-\text{2 }10-12\\ -1242-129387\\ \overline{4-1443-129\overline{)375}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)4}\underset{\downarrow }{4}-\text{2 }10-12\\ 1656228912\\ \overline{41457228\overline{)900}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-4}\underset{\downarrow }{4}-\text{2 }10-12\\ -1672-2921168\\ \overline{4-1873-292\overline{)1156}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)6}\underset{\downarrow }{4}-\text{2 }10-12\\ 241327984788\\ \overline{422133798\overline{)4776}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-6}\underset{\downarrow }{4}-\text{2 }10-12\\ -24156-9425652\\ \overline{\text{ 4 }-26157-942\overline{)5640}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)\frac{1}{2}}\underset{\downarrow }{4}-\text{2 }10-12\\ 20\frac{1}{2}\frac{1}{4}\\ \overline{\text{ 4 }01\frac{1}{2}\overline{)-\frac{47}{4}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{2}}\underset{\downarrow }{4}-\text{2 }10-12\\ -22-\frac{3}{2}\frac{3}{4}\\ \overline{\text{ 4 }-43-\frac{3}{2}\overline{)-\frac{45}{4}}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)\frac{3}{2}}\underset{\downarrow }{4}-\text{2 }10-12\\ 66\frac{21}{2}\frac{63}{4}\\ \overline{\text{ 4 }47\frac{21}{2}\overline{)-\frac{15}{4}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{3}{2}}\underset{\downarrow }{4}-\text{2 }10-12\\ -6\text{ 1}2-\frac{39}{2}\frac{117}{4}\\ \overline{\text{ 4 }-8\text{ 1}3-\frac{39}{2}\overline{)\frac{69}{4}}}\end{array}\right\}$

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

  $\left\{\begin{array}{l}\overline{)\frac{1}{4}}\underset{\downarrow }{4}-\text{2 }10-12\\ 1-\frac{1}{4}\frac{3}{16}\frac{63}{4}\\ \overline{\text{ 4 }-1\frac{3}{4}\frac{3}{16}\overline{)-\frac{765}{64}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{1}{4}}\underset{\downarrow }{4}-\text{2 }10-12\\ -1\frac{3}{4}-\frac{7}{16}\frac{7}{64}\\ \overline{\text{ 4 }-3\frac{7}{4}-\frac{7}{16}\overline{)-\frac{761}{64}}}\end{array}\right\}$

Here, the remainders are not equal to zero, so $x\ne \pm \frac{1}{4}$ .

  $\left\{\begin{array}{l}\overline{)\frac{3}{4}}\underset{\downarrow }{4}-\text{2 }10-12\\ 3\frac{3}{4}\frac{21}{16}\frac{63}{64}\\ \overline{\text{ 4 }1\frac{7}{4}\frac{21}{16}\overline{)-\frac{705}{64}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{3}{4}}\underset{\downarrow }{4}-\text{2 }10-12\\ -3\frac{15}{4}-\frac{57}{16}\frac{171}{64}\\ \overline{\text{ 4 }-5\frac{19}{4}-\frac{57}{16}\overline{)-\frac{597}{64}}}\end{array}\right\}$

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

Hence, the given polynomial function has no rational roots.