Chapter 4, Problem 35SGA

To determine

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

Answer to Problem 35SGA

  $2or0;1;-\frac{1}{2},3$

Explanation of Solution

Given information:

  $f\left(x\right)=2x^3-11x^2+12x+9$

Calculation:

Consider the given function,

  $f\left(x\right)=2x^3-11x^2+12x+9$

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

  $\begin{array}{l}f\left(x\right)=2x^3-11x^2+12x+9\\ =2-11129\\ \to \to \to \\ yesyesno\end{array}$

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,

  $\begin{array}{l}f\left(-x\right)=2{\left(-x\right)}^3-11{\left(-x\right)}^2+12\left(-x\right)+9\\ =-2x^3-11x^2-12x+9\\ \to \to \to \\ nonoyes\end{array}$

There is one change, so there are one negative real zero.

Now find the possible zeroes,

Possible value of $p:\pm 1,\pm 3\pm 9$

Possible value of $q:\pm 1,\pm 2$

Possible rational zeroes, $\frac{p}{q}:\pm 1,\pm 3\pm 9,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2}$

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

  $\begin{array}{ccccc}r& 2& -11& 12& 9\\ 1/2& 2& -10& 7& 25/2\\ -1/2& 2& -12& 18& 0\\ 1& 2& -9& 3& 12\\ -1& 2& -13& 25& -16\\ 3& 2& -5& -3& 0\\ 3& 2& -17& 63& -180\\ 9& 2& 7& 75& 684\\ -9& 2& -29& 273& -2448\end{array}$

Since, there are only two positive real zeroes and $3$ is a zero and only one negative real zero and $-\frac{1}{2}$ is a zero.

Hence the possible positive real, negative real and rational zeroes for $f\left(x\right)=2x^3-11x^2+12x+9$ is $2or0;1;-\frac{1}{2},3$