Chapter 2.3, Problem 100E

To determine

To find the real zeroes of the polynomial function

  $f(x)=4x^5+8x^4-15x^3-23x^2+11x+15$ .

Answer to Problem 100E

  $-\frac{5}{2},-1,-1,1,\frac{3}{2}$

Explanation of Solution

Given:

Function: $f(x)=4x^5+8x^4-15x^3-23x^2+11x+15$

Calculation:

The leading coefficient of given polynomial is 4 and the constant term is 15.

So, possible zeroes are:

  $\frac{\text{Factors of 15}}{\text{Factors of 4}}=\frac{\pm 1,\pm 3,\pm 5,\pm 15}{\pm 1,\pm 2,\pm 4}=\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{5}{4},\pm \frac{15}{4}$

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

  $\begin{array}{ccccccc}1& 4& 8& -15& -23& 11& 15\\ & & 4& 12& -3& -26& -15\\ & 4& 12& -3& -26& -15& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=(x-1)(4x^4+12x^3-3x^2-26x-15)$

Now, consider $4x^4+12x^3-3x^2-26x-15.$

The leading coefficient of above polynomial is 4 and the constant term is -15.

So, possible zeroes are:

  $\frac{\text{Factors of}-\text{15}}{\text{Factors of 4}}=\frac{\pm 1,\pm 3,\pm 5,\pm 15}{\pm 1,\pm 2,\pm 4}=\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{5}{4},\pm \frac{15}{4}$

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

  $\begin{array}{cccccc}1& 4& 12& -3& -26& -15\\ & & 4& 16& 13& -13\\ & 4& 16& 13& -13& \boxed{-28}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

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

  $\begin{array}{cccccc}-1& 4& 12& -3& -26& -15\\ & & -4& -8& 11& 15\\ & 4& 8& -11& -15& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=(x-1)(x+1)(4x^3+8x^2-11x-15)$

Now, consider $4x^3+8x^2-11x-15.$

The leading coefficient of above polynomial is 4 and the constant term is -15.

So, possible zeroes are:

  $\frac{\text{Factors of}-\text{15}}{\text{Factors of 4}}=\frac{\pm 1,\pm 3,\pm 5,\pm 15}{\pm 1,\pm 2,\pm 4}=\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{5}{4},\pm \frac{15}{4}$

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

  $\begin{array}{ccccc}-1& 4& 8& -11& -15\\ & & -4& -4& 15\\ & 4& 4& -15& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=(x-1)(x+1)(x+1)(4x^2+4x-15)$

Now, consider $4x^2+4x-15.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-4\pm \sqrt{{(4)}^2-4(4)(-15)}}{2(4)}\\ \Rightarrow x=\frac{-4\pm \sqrt{16+240}}{8}\\ \Rightarrow x=\frac{-4\pm \sqrt{256}}{8}\\ \Rightarrow x=\frac{-4\pm 16}{8}\\ \Rightarrow x=\frac{-4+16}{8},\frac{-4-16}{8}\\ \Rightarrow x=\frac{12}{8},\frac{-20}{8}\\ \Rightarrow x=\frac{3}{2},-\frac{5}{2}\end{array}$

Conclusion:

Therefore, the zeroes of given polynomial function are $x=-\frac{5}{2},-1,-1,1\text{and }\frac{3}{2}.$