Chapter 2.3, Problem 77E

To determine

To calculate: the rational zeros of the polynomial function $P(x)=x^4-\frac{13}{4}{x}^2+\frac{9}{4}=\frac{1}{4}\left(4x^4-13x^2+9\right)$

Answer to Problem 77E

  $-1,1,\frac{-3}{2},\frac{3}{2}$ .

Explanation of Solution

Given information: Given function is

  $P(x)=x^4-\frac{13}{4}{x}^2+\frac{9}{4}=\frac{1}{4}\left(4x^4-13x^2+9\right)$

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

Calculation:- Given that $P(x)=\frac{1}{4}\left(4x^4-13x^2+9\right)$

Here, constant term = 9 and leading coefficient = 4

Possible factors of constant terms are $\pm 1,\pm 3,\pm 9$

Possible factors of leading coefficient = 1, 2, 4

Possible rational roots are =

  $\begin{array}{l}\frac{\pm 9}{4},\frac{\pm 9}{2},\frac{\pm 9}{1},\\ \frac{\pm 3}{4},\frac{\pm 3}{2},\frac{\pm 3}{1},\\ \frac{\pm 1}{4},\frac{\pm 1}{2},\frac{\pm 1}{1}\end{array}$

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.

At $x=\frac{\pm 9}{4}$

  $\begin{array}{l}P(\frac{\pm 9}{4})=\frac{1}{4}\left(4{\left(\frac{\pm 9}{4}\right)}^4-13{\left(\frac{\pm 9}{4}\right)}^2+9\right)=\frac{1}{4}\left(\frac{9^4}{4^3}-13\frac{81}{16}+9\right)\\ P(\frac{\pm 9}{4})=\frac{1}{4}\left(\frac{6561-4212+576}{4^3}\right)=\frac{2925}{256}\\ P(\frac{\pm 9}{4})\ne 0\end{array}$

At $x=\frac{\pm 9}{2}$

  $\begin{array}{l}P(\frac{\pm 9}{2})=\frac{1}{4}\left(4{\left(\frac{\pm 9}{2}\right)}^4-13{\left(\frac{\pm 9}{2}\right)}^2+9\right)=\frac{1}{4}\left(\frac{9^4}{4}-13\frac{81}{4}+9\right)\\ P(\frac{\pm 9}{2})=\frac{1}{4}\left(\frac{6561-1053+36}{4}\right)=\frac{5544}{16}\\ P(\frac{\pm 9}{2})\ne 0\end{array}$

At $x=\pm 9$

  $\begin{array}{l}P(\pm 9)=\frac{1}{4}\left(4\times 9^4-13\times 9^2+9\right)=\frac{1}{4}\left(26244-1053+9\right)\\ P(\pm 9)=\frac{25200}{4}\\ P(\pm 9)\ne 0\end{array}$

At $x=\frac{\pm 3}{4}$

  $\begin{array}{l}P(\frac{\pm 3}{4})=\frac{1}{4}\left(4{\left(\frac{\pm 3}{4}\right)}^4-13{\left(\frac{\pm 3}{4}\right)}^2+9\right)=\frac{1}{4}\left(\frac{3^4}{4^3}-13\frac{9}{16}+9\right)\\ P(\frac{\pm 3}{4})=\frac{1}{4}\left(\frac{81-468+576}{4^3}\right)=\frac{189}{256}\\ P(\frac{\pm 3}{4})\ne 0\end{array}$

At $x=\frac{\pm 3}{2}$

  $\begin{array}{l}P(\frac{\pm 3}{2})=\frac{1}{4}\left(4{\left(\frac{\pm 3}{2}\right)}^4-13{\left(\frac{\pm 3}{2}\right)}^2+9\right)=\frac{1}{4}\left(\frac{81}{4}-\frac{117}{4}+9\right)\\ P(\frac{\pm 3}{2})=\frac{1}{4}\left(\frac{81-117+36}{4}\right)=\frac{117-117}{16}\\ P(\frac{\pm 3}{2})=0\end{array}$

At $x=\pm 3$

  $\begin{array}{l}P(\pm 3)=\frac{1}{4}\left(4{\left(\pm 3\right)}^4-13{\left(\pm 3\right)}^2+9\right)=\frac{1}{4}\left(324-117+9\right)\\ P(\pm 3)=\frac{216}{4}=54\\ P(\pm 3)\ne 0\end{array}$

At $x=\frac{\pm 1}{4}$

  $\begin{array}{l}P(\frac{\pm 1}{4})=\frac{1}{4}\left(4{\left(\frac{\pm 1}{4}\right)}^4-13{\left(\frac{\pm 1}{4}\right)}^2+9\right)=\frac{1}{4}\left(\frac{1}{4^3}-13\frac{1}{16}+9\right)\\ P(\frac{\pm 1}{4})=\frac{1}{4}\left(\frac{1-52+576}{4^3}\right)=\frac{525}{256}\\ P(\frac{\pm 1}{4})\ne 0\end{array}$

  $\begin{array}{l}Atx=\frac{\pm 1}{2}\\ P(\frac{\pm 1}{2})=\frac{1}{4}\left(4{\left(\frac{\pm 1}{2}\right)}^4-13{\left(\frac{\pm 1}{2}\right)}^2+9\right)=\frac{1}{4}\left(\frac{1}{4}-\frac{13}{4}+9\right)\\ P(\frac{\pm 1}{2})=\frac{1}{4}\left(\frac{1-13+36}{4}\right)=\frac{24}{16}\\ P(\frac{\pm 1}{2})\ne 0\end{array}$

  $\begin{array}{l}Atx=\pm 1\\ P(\pm 1)=\frac{1}{4}\left(4{\left(\pm 1\right)}^4-13{\left(\pm 1\right)}^2+9\right)=\frac{1}{4}\left(4-13+9\right)\\ P(\pm 1)=\frac{1}{4}\left(0\right)\\ P(\pm 1)=0\end{array}$

Hence, x = $\frac{\pm 3}{2},\pm 1$or $\left(\frac{-3}{2},\frac{-3}{2},-1,+1\right)$