Chapter 2.3, Problem 89E

To determine

To find the real zeroes of the polynomial function $f(x)=5x^4+9x^3-19x^2-3x$ .

Answer to Problem 89E

  $0,-3,\frac{3+\sqrt{14}}{5}\text{,}\frac{3-\sqrt{14}}{5}$

Explanation of Solution

Given:

Function: $f(x)=5x^4+9x^3-19x^2-3x$

Calculation:

The leading coefficient of given polynomial is 5 and the constant term is 0.

So, possible zeroes are: $\frac{\text{Factors of 0}}{\text{Factors of 5}}=\frac{\pm \text{0}}{\pm 1,\pm 5}=0$

Using synthetic division to check whether x = 0, is a zero or not.

  $\begin{array}{cccccc}0& 5& 9& -19& -3& 0\\ & & 0& 0& 0& 0\\ & 5& 9& -19& -3& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=(x-0)(5x^3+9x^2-19x-3)$

Now, consider $5x^3+9x^2-19x-3.$

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

So, possible zeroes are: $\frac{\text{Factors of}-\text{3}}{\text{Factors of 5}}=\frac{\pm 1,\pm 3}{\pm 1,\pm 5}=\pm 1,\pm 3,\pm \frac{3}{5},\pm \frac{1}{5}$

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

  $\begin{array}{ccccc}-1& 5& 9& -19& -3\\ & & -5& -4& 23\\ & 5& 4& -23& \boxed{20}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

Using synthetic c division to check whether x = -3, is a zero or not.

  $\begin{array}{ccccc}-3& 5& 9& -19& -3\\ & & -15& 18& 3\\ & 5& -6& -1& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = -3 is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=(x-0)(x+3)(5x^2-6x-1)$

Now, consider $5x^2-6x-1.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(-6)\pm \sqrt{{(-6)}^2-4(5)(-1)}}{2(5)}\\ \Rightarrow x=\frac{6\pm \sqrt{36+20}}{10}\\ \Rightarrow x=\frac{6\pm \sqrt{56}}{10}\\ \Rightarrow x=\frac{6\pm 2\sqrt{14}}{10}\\ \Rightarrow x=\frac{3\pm \sqrt{14}}{5}\end{array}$

Conclusion:

Therefore, the zeroes of given polynomial function are $x=0,-3,\frac{3+\sqrt{14}}{5}\text{and}\frac{3-\sqrt{14}}{5}.$