Chapter 2, Problem 71RE

To determine

To find the zeros of the function $f(x)=6x^4-25x^3+14x^2+27x-18$ .

Answer to Problem 71RE

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

Explanation of Solution

Given:

Function: $f(x)=6x^4-25x^3+14x^2+27x-18$

Calculation:

The leading coefficient of given polynomial is 6 and the constant term is -18.

So, possible rational zeros are:

  $\frac{\text{Factors of}-18}{\text{Factors of 6}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18}{\pm 1,\pm 2,\pm 3,\pm 6}=\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm \mathrm{18.......}$

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

  $\begin{array}{cccccc}3& 6& -25& 14& 27& -18\\ & & 18& -21& -21& 18\\ & 6& -7& -7& 6& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=\left(x-3\right)(6x^3-7x^2-7x+6)$

Now, consider $6x^3-7x^2-7x+6.$

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

So, possible rational zeros are: $\frac{\text{Factors of}6}{\text{Factors of 6}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1,\pm 2,\pm 3,\pm 6}=\pm 1,\pm 2,\pm 3,\pm \mathrm{6.............}$

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

  $\begin{array}{ccccc}-1& 6& -7& -7& 6\\ & & -6& 13& -6\\ & 6& -13& 6& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=\left(x-3\right)\left(x+1\right)(6x^2-13x+6)$

Now, consider $6x^2-13x+6.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(-13)\pm \sqrt{{(-13)}^2-4(6)(6)}}{2(6)}\\ \Rightarrow x=\frac{13\pm \sqrt{169-144}}{12}\\ \Rightarrow x=\frac{13\pm \sqrt{25}}{12}\\ \Rightarrow x=\frac{13\pm 5}{12}\\ \Rightarrow x=\frac{13+5}{12},\frac{13-5}{12}\\ \Rightarrow x=\frac{18}{12},\frac{8}{12}\\ \Rightarrow x=\frac{3}{2},\frac{2}{3}\end{array}$

Conclusion:

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