Chapter 2.3, Problem 95E

To determine

To find the real zeroes of the polynomial function $f(x)=4x^4-55x^2-45x+36$ .

Answer to Problem 95E

  $-\frac{3}{2},-3,\frac{1}{2}\text{,}4$

Explanation of Solution

Given:

Function: $f(x)=4x^4-55x^2-45x+36$

Calculation:

The leading coefficient of given polynomial is 4 and the constant term is 36.

So, possible zeroes are: $\frac{\text{Factors of}36}{\text{Factors of 4}}=\frac{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 9,\pm 12,\pm 18,\pm 36}{\pm 1,\pm 2,\pm 4}=\pm 1,\pm 2,\pm 3,\pm 4$

  $\pm 6,\pm 9,\pm 12,\pm 18,\pm 36,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{9}{4}$

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

  $\begin{array}{cccccc}1& 4& 0& -55& -45& 36\\ & & 4& 4& -51& -96\\ & 4& 4& -51& -96& \boxed{-60}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

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

  $\begin{array}{cccccc}4& 4& 0& -55& -45& 36\\ & & 16& 64& 36& -36\\ & 4& 16& 9& -9& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=(x-4)(4x^3+16x^2+9x-9)$

Now, consider $4x^3+16x^2+9x-9.$

The leading coefficient of above polynomial is 4 and the constant term is -9.

So, possible zeroes are: $\frac{\text{Factors of}-9}{\text{Factors of 4}}=\frac{\pm 1,\pm 3,\pm 9}{\pm 1,\pm 2,\pm 4}=\pm 1,\pm 3,\pm 9,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{9}{4}$

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

  $\begin{array}{ccccc}1& 4& 16& 9& -9\\ & & 4& 20& 29\\ & 4& 20& 29& \boxed{20}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

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

  $\begin{array}{ccccc}-3& 4& 16& 9& -9\\ & & -12& -12& 9\\ & 4& 4& -3& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=(x-4)(x+3)(4x^2+4x-3)$

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

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(4)\pm \sqrt{{(4)}^2-4(4)(-3)}}{2(4)}\\ \Rightarrow x=\frac{-4\pm \sqrt{16+48}}{8}\\ \Rightarrow x=\frac{-4\pm \sqrt{64}}{8}\\ \Rightarrow x=\frac{-4\pm 8}{8}\end{array}$

  $\begin{array}{l}\Rightarrow x=\frac{-4+8}{8},\frac{-4-8}{8}\\ \Rightarrow x=\frac{4}{8},\frac{-12}{8}\\ \Rightarrow x=\frac{1}{2},-\frac{3}{2}\end{array}$

Conclusion:

Therefore, the zeroes of given polynomial function are $x=-\frac{3}{2},-3,\frac{1}{2}\text{and}4.$