Chapter 4, Problem 36SGA

To determine

Find the number of possible positive real, negative real and rational zeroes.

Answer to Problem 36SGA

  $-3,-2,2,3$

Explanation of Solution

Given information:

  $f\left(x\right)=x^4-13x^2+36$

Calculation:

Consider the given function,

  $f\left(x\right)=x^4-13x^2+36$

To find the possible positive real zeroes, count the number of sign changes for coefficients,

  $\begin{array}{l}f\left(x\right)=x^4-13x^2+36\\ 1-1336\\ \to \to \\ yesyes\end{array}$

There are two changes, so there are two or zero positive real zeroes.

Now find the possible negative real zeroes $f\left(-x\right)$ , count the number of sign changes for coefficients,

  $\begin{array}{l}f\left(-x\right)={\left(-x\right)}^4-13{\left(-x\right)}^2+36\\ x^4-13x^2+36\\ \to \to \\ yesyes\end{array}$

There is two change, so there are two or zero negative real zeros.

Now find the possible zeroes,

Possible value of $p:\pm 1,\pm 2\pm 3\pm 4,\pm 6\pm 9\pm 12,\pm 18\pm 36$

Possible value of $q:\pm 1,$

Possible rational zeroes, $\frac{p}{q}:\pm 1,\pm 2\pm 3,\pm 4,\pm 6,\pm 9\pm 12\pm 18\pm 36$

Now test the zeroes using by synthetic division and remainder theorem,

  $\begin{array}{cccccc}r& 1& 0& -13& 0& 36\\ 1& 1& 1& -12& -12& 24\\ -1& 1& -1& -12& 12& 48\\ 2& 1& 2& -9& -18& 0\\ -2& 1& -2& -9& 18& 0\\ 3& 1& 3& -4& -12& 0\\ -3& 1& -3& -4& 12& 0\end{array}$

Since, there are only two positive real zeroes $2,3$ and $3$ is a zero and two negative real zeroes $-2,-3$ and rational zeroes are $-3,-2,2,3$

Hence the possible positive real, negative real and rational zeroes for $f\left(x\right)=x^4-13x^2+36$ is $-3,-2,2,3$