Chapter 5, Problem 47E

Solution

To calculate : Therational root of the given function.

The given polynomial function has four rational roots and are given by $x=1,1,-2\text{ and }-\frac{2}{3}$ .

Given information :

The function is $P\left(x\right)=3x^4+2x^3-9x^2+4$ .

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: $P\left(x\right)=3x^4+2x^3-9x^2+4$ .

Find the possible rational roots.

Determine the integer factor of the constant term $4$ .

  $p=\text{±1,}\pm \text{2 and }\pm 4$

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

  $q=\pm 1\text{ and }\pm 3$

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

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

Apply the synthetic division method to test the roots.

  $\left\{\begin{array}{l}\overline{)1}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ \text{ 3 5 }-\text{4 }-4\\ \overline{\text{ 3 5 }-\text{4 }-4\overline{)0}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-1}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ -\text{3 1 8 }-4\\ \overline{\text{ 3 }-1-8\text{ 8 }\overline{)-4}}\end{array}\right\}$

Here, the rational root is given by $x=1$ .

Solving further,

  $\left\{\begin{array}{l}\overline{)2}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ \text{ 6 16 14 28}\\ \overline{\text{ 3 8 7 }14\overline{)32}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-2}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ -6\text{ 8 2 }-4\\ \overline{\text{ 3 }-4-1\text{ 2 }\overline{)0}}\end{array}\right\}$

Here, the rational root is given by $x=-2$ .

Solving further,

  $\left\{\begin{array}{l}\overline{)4}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ \text{ 12 56 188 752}\\ \overline{\text{ 3 14 47 }188\overline{)756}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-4}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ -12\text{ 40 }-\text{124 }496\\ \overline{\text{ 3 }-10\text{ 3}1-1\text{24 }\overline{)500}}\end{array}\right\}$

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

Solving further,

  $\left\{\begin{array}{l}\overline{)\frac{2}{3}}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ \text{ 2 }\frac{8}{3}-\frac{38}{9}-\frac{76}{27}\\ \overline{\text{ 3 4 }-\frac{19}{3}-\frac{38}{9}\overline{)\frac{32}{27}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{2}{3}}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ -2\text{ 0 6 }-4\\ \overline{\text{ 3 }0-9\text{ 6 }\overline{)0}}\end{array}\right\}$

Here, the rational root is given by $x=-\frac{2}{3}$ .

Solving further,

  $\left\{\begin{array}{l}\overline{)\frac{4}{3}}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ \text{ 4 8 }-\frac{4}{3}-\frac{16}{9}\\ \overline{\text{ 3 6 }-1-\frac{4}{3}\overline{)\frac{20}{9}}}\end{array}\right\},\left\{\begin{array}{l}\overline{)-\frac{4}{3}}\underset{\downarrow }{3}\text{ 2 }-9\text{ 0 }4\\ -4\frac{8}{3}\frac{76}{9}-\frac{304}{27}\\ \overline{\text{ 3 }-2-\frac{19}{3}\frac{76}{9}\overline{)-\frac{196}{27}}}\end{array}\right\}$

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

Find the product of the factors $\left(x+2\right)\text{ and }\left(x+3\right)$ .

  $\begin{array}{l}\left(x+2\right)\left(3x+2\right)=x\left(3x+2\right)+2\left(3x+2\right)\\ =\left(3x^2+2x\right)+\left(6x+4\right)\\ =3x^2+8x+4\end{array}$

Apply long division method to divide the polynomial $P\left(x\right)=3x^4+2x^3-9x^2+4$ by $3x^2+8x+4$ as shown below:

  $3x^2+8x+4\begin{array}{c}\hfill x^2-2x+1\\ \hfill \overline{)\begin{array}{l}3x^4+2x^3-9x^2+4\\ 3x^4+8x^3+4x^2\\ \overline{\begin{array}{l}-6x^3-13x^2+4\\ -6x^3-16x^2-8x\\ \overline{\begin{array}{l}3x^2-8x+4\\ 3x^2-8x+4\\ \overline{0}\end{array}}\end{array}}\end{array}}\end{array}$

Now,

  $\begin{array}{l}{x}^2-2x+1=0\\ {\left(x-1\right)}^2=0\\ x=1,1\end{array}$

So, the other two rational roots are $x=1\text{, 1}$ .

Hence, the given polynomial function has four rational roots and are given by $x=1,1,-2\text{ and }-\frac{2}{3}$ .