Chapter 2.3, Problem 78E

To determine

To calculate: the rational zeros of the polynomial function $f(x)=x^3-\frac{3}{2}{x}^2-\frac{23}{2}x+6=\frac{1}{2}\left(2x^3-3x^2-23x+12\right)$

Answer to Problem 78E

  $x=\frac{1}{2},4,-3$ .

Explanation of Solution

Given information: Given function is

  $f(x)=x^3-\frac{3}{2}{x}^2-\frac{23}{2}x+6=\frac{1}{2}\left(2x^3-3x^2-23x+12\right)$

Concept Used:- Rational Root Test:- Possible rational roots $=\frac{\text{possible factors of the constant}}{\text{factors of the leading coefficient}}$

Calculation:- Given that $f(x)=\frac{1}{2}\left(2x^3-3x^2-23x+12\right)$

Here, constant term = 12 and leading coefficient = 2

Possible factors of constant terms are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

Possible factors of leading coefficient = 1, 2.

Possible rational roots are =

  $\begin{array}{l}\frac{\pm 12}{2},\pm 12,\frac{\pm 6}{2},\pm 6,\\ \frac{\pm 4}{2},\pm 4,\frac{\pm 3}{2},\frac{\pm 3}{1},\\ \frac{\pm 2}{2},\pm 2,\frac{\pm 1}{2},\frac{\pm 1}{1}\end{array}$

Or

  $\pm 12,\pm 6,\pm 4,\pm 3,\pm 2,\pm 1,\frac{\pm 3}{2},\frac{\pm 1}{2}$ Are the possible rational roots

Since, we know zeros of a polynomial satisfies polynomial i.e. say $f(x)$ is any polynomial and ‘a’ is zero of f(x) then $f(a)=0$ .

Now, to find roots from possible roots we just put these points in polynomial and the point, satisfies polynomial will be the root of polynomial.

  $\begin{array}{l}f(x)=\frac{1}{2}\left(2x^3-3x^2-23x+12\right)\\ Atx=12\\ f(12)=\frac{1}{2}\left(2\times {12}^3-3\times {12}^2-23\times 12+12\right)\\ f(12)=\frac{1}{2}(3456-432-276+12)=1380\\ f(12)\ne 0\\ Atx=-12\\ f(-12)=\frac{1}{2}\left(2\times {(-12)}^3-3\times {(-12)}^2-23\times (-12)+12\right)\\ f(-12)=\frac{1}{2}(-3456-432+276+12)=-1800\\ f(-12)\ne 0\end{array}$

Similarly, we can show

  $f(\pm 6)\ne 0,f(-4)\ne 0,f(\pm 2)\ne 0,f(\pm 1)\ne 0,f(-3)\ne 0,f(\frac{\pm 3}{2})\ne 0,f(\frac{-1}{2})\ne 0$

  $\begin{array}{l}Atx=-3\\ f(-3)=\frac{1}{2}\left(2\times {(-3)}^3-3\times {(-3)}^2-23\times (-3)+12\right)\\ f(-3)=\frac{1}{2}(-54-27+69+12)=\frac{1}{2}(81-81)=0\\ f(-3)=0\\ Atx=4\\ f(4)=\frac{1}{2}\left(2\times 4^3-3\times 4^2-23\times 4+12\right)\\ f(4)=\frac{1}{2}(128-48-92+12)=\frac{1}{2}(140-140)=0\\ f(4)=0\end{array}$

  $\begin{array}{l}Atx=\frac{1}{2}\\ f(\frac{1}{2})=\frac{1}{2}\left(2\times {\left(\frac{1}{2}\right)}^3-3\times {\left(\frac{1}{2}\right)}^2-23\times \frac{1}{2}+12\right)\\ f(\frac{1}{2})=\frac{1}{2}(\frac{1}{4}-\frac{3}{4}-\frac{23}{2}+12)=\frac{1}{2}(\frac{49}{4}-\frac{49}{4})=0\\ f(\frac{1}{2})=0\end{array}$

Hence, x = $\frac{1}{2}$ , 4, -3 are rational zeros of function f.