Chapter 2.3, Problem 96E

To determine

To find the real zeroes of the polynomial function $z(x)=6x^4+33x^3-69x+30$ .

Answer to Problem 96E

  $-5,-2,\frac{1}{2},1$

Explanation of Solution

Given:

Function: $z(x)=6x^4+33x^3-69x+30$

Calculation:

The leading coefficient of given polynomial is 6 and the constant term is 30.

So, possible zeroes are: $\frac{\text{Factors of 30}}{\text{Factors of 6}}=\frac{\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30}{\pm 1,\pm 2,\pm 3,\pm 6}=\pm 1,\pm 2$

  $3\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{5}{3},\pm \frac{10}{3},\pm \frac{1}{6},\pm \frac{5}{6}$

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

  $\begin{array}{cccccc}1& 6& 33& 0& -69& 30\\ & & 6& 39& 39& -30\\ & 6& 39& 39& -30& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $z(x)$ factors as

  $z(x)=(x-1)(6x^3+39x^2+39x-30)$

Now, consider $6x^3+39x^2+39x-30.$

The leading coefficient of above polynomial is 5 and the constant term is -3.

So, possible zeroes are: $\frac{\text{Factors of}-3\text{0}}{\text{Factors of 6}}=\frac{\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30}{\pm 1,\pm 2,\pm 3,\pm 6}=\pm 1,\pm 2$

  $3\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{5}{3},\pm \frac{10}{3},\pm \frac{1}{6},\pm \frac{5}{6}$

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

  $\begin{array}{ccccc}1& 6& 39& 39& -30\\ & & 6& 45& 84\\ & 6& 45& 84& \boxed{54}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

Using synthetic c division to check whether x = -2, is a zero or not.

  $\begin{array}{ccccc}-2& 6& 39& 39& -30\\ & & -12& -54& 30\\ & 6& 27& -15& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $z(x)$ factors as

  $z(x)=(x-1)(x+2)(6x^2+27x-15)$

Now, consider $6x^2+27x-15.$

The above polynomial is a second degree polynomial.

So, zeroes of a second degree polynomial can be found using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

  $\begin{array}{l}\Rightarrow x=\frac{-27\pm \sqrt{{(27)}^2-4(6)(-15)}}{2(6)}\\ \Rightarrow x=\frac{-27\pm \sqrt{729+360}}{12}\\ \Rightarrow x=\frac{-27\pm \sqrt{1089}}{12}\\ \Rightarrow x=\frac{-27\pm 33}{12}\\ \Rightarrow x=\frac{-27+33}{12},\frac{-27-33}{12}\\ \Rightarrow x=\frac{6}{12},\frac{-60}{12}\\ \Rightarrow x=\frac{1}{2},-5\end{array}$

Conclusion:

Therefore, the zeroes of given polynomial function are $x=-5,-2,\frac{1}{2}\text{and}1.$