Chapter 2.3, Problem 90E

To determine

To find the real zeroes of the polynomial function $f(x)=4x^4-11x^3-22x^2+8x$ .

Answer to Problem 90E

  $0,4,\frac{-5+\sqrt{57}}{8}\text{,}\frac{-5-\sqrt{57}}{8}$

Explanation of Solution

Given:

Function: $f(x)=4x^4-11x^3-22x^2+8x$

Calculation:

The leading coefficient of given polynomial is 4 and the constant term is 0.

So, possible zeroes are: $\frac{\text{Factors of 0}}{\text{Factors of 4}}=\frac{\pm \text{0}}{\pm 1,\pm 2,\pm 4}=0$

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

  $\begin{array}{cccccc}0& 4& -11& -22& 8& 0\\ & & 0& 0& 0& 0\\ & 4& -11& -22& 8& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=(x-0)(4x^3-11x^2-22x+8)$

Now, consider $4x^3-11x^2-22x+8.$

The leading coefficient of above polynomial is 4 and the constant term is 8.

So, possible zeroes are: $\frac{\text{Factors of }8}{\text{Factors of 4}}=\frac{\pm 1,\pm 2,\pm 4,\pm 8}{\pm 1,\pm 2,\pm 4}=\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2},\pm \frac{1}{4}$

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

  $\begin{array}{ccccc}1& 4& -11& -22& 8\\ & & 4& -7& -29\\ & 4& -7& -29& \boxed{-21}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

Using synthetic c division to check whether x = 4, is a zero or not.

  $\begin{array}{ccccc}4& 4& -11& -22& 8\\ & & 16& 20& -8\\ & 4& 5& -2& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = 4 is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=(x-0)(x-4)(4x^2+5x-2)$

Now, consider $4x^2+5x-2.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(5)\pm \sqrt{{(5)}^2-4(4)(-2)}}{2(4)}\\ \Rightarrow x=\frac{-5\pm \sqrt{25+32}}{8}\\ \Rightarrow x=\frac{-5\pm \sqrt{57}}{8}\end{array}$

Conclusion:

Therefore, the zeroes of given polynomial function are $x=0,4,\frac{-5+\sqrt{57}}{8}\text{and}\frac{-5-\sqrt{57}}{8}.$