Chapter 2.3, Problem 91E

To determine

To find the real zeroes of the polynomial function $f(z)=z^4-z^3-2z-4$ .

Answer to Problem 91E

  $-1,2$

Explanation of Solution

Given:

Function: $f(z)=z^4-z^3-2z-4$

Calculation:

The leading coefficient of given polynomial is 1 and the constant term is -4.

So, possible zeroes are: $\frac{\text{Factors of}-\text{4}}{\text{Factors of 1}}=\frac{\pm 1,\pm 2,\pm 4}{\pm 1}=\pm 1,\pm 2,\pm 4$

Using synthetic division to check whether z = 1, is a zero or not.

  $\begin{array}{cccccc}1& 1& -1& 0& -2& -4\\ & & 1& 0& 0& -2\\ & 1& 0& 0& -2& \boxed{-6}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is not 0. So, z = 1 is not a zero of the given polynomial.

Using synthetic division to check whether z = -1, is a zero or not.

  $\begin{array}{cccccc}-1& 1& -1& 0& -2& -4\\ & & -1& 2& -2& 4\\ & 1& -2& 2& -4& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, z = -1 is a zero of the given polynomial.

So, $f(z)$ factors as

  $f(z)=(z+1)(z^3-2z^2+2z-4)$

Now, consider $z^3-2z^2+2z-4.$

The leading coefficient of above polynomial is 1 and the constant term is -4.

So, possible zeroes are: $\frac{\text{Factors of}-\text{4}}{\text{Factors of 1}}=\frac{\pm 1,\pm 2,\pm 4}{\pm 1}=\pm 1,\pm 2,\pm 4$

Using synthetic division to check whether z = 1, is a zero or not.

  $\begin{array}{ccccc}1& 1& -2& 2& -4\\ & & 1& -1& 1\\ & 1& -1& 1& \boxed{-3}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is not 0. So, z = 1 is not a zero of the given polynomial.

Using synthetic division to check whether z = 2, is a zero or not.

  $\begin{array}{ccccc}2& 1& -2& 2& -4\\ & & 2& 0& 4\\ & 1& 0& 2& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, z = 2 is a zero of the given polynomial.

So, $f(z)$ factors as

  $f(z)=(z+1)(z-2)(z^2+2)$

Now, consider $z^2+2.$

The above polynomial is a second degree polynomial.

Discriminant $=b^2-4ac=0-4(2)(1)=-8<0$

Here, the value of discriminant is less than zero.

So, zeroes does not exists for above second degree polynomial.

Conclusion:

Therefore, the zeroes of given polynomial function are $x=-1\text{and}2.$