Chapter 2.3, Problem 93E

To determine

To find the real zeroes of the polynomial function $g(y)=2y^4+7y^3-26y^2+23y-6$ .

Answer to Problem 93E

  $1,1,\frac{1}{2}\text{,}-6$

Explanation of Solution

Given:

Function: $g(y)=2y^4+7y^3-26y^2+23y-6$

Calculation:

The leading coefficient of given polynomial is 2 and the constant term is -6.

So, possible zeroes are: $\frac{\text{Factors of}-6}{\text{Factors of 2}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1,\pm 2}=\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}$

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

  $\begin{array}{cccccc}1& 2& 7& -26& 23& -6\\ & & 2& 9& -17& 6\\ & 2& 9& -17& 6& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $g(y)$ factors as

  $g(y)=(y-1)(2y^3+9y^2-17y+6)$

Now, consider $2y^3+9y^2-17y+6.$

The leading coefficient of above polynomial is 2 and the constant term is 6.

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

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

  $\begin{array}{ccccc}1& 2& 9& -17& 6\\ & & 2& 11& -6\\ & 2& 11& -6& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

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

So, $g(y)$ factors as

  $g(y)=(y-1)(y-1)(2y^2+11y-6)$

Now, consider $2y^2+11y-6.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow y=\frac{-(11)\pm \sqrt{{(11)}^2-4(2)(-6)}}{2(2)}\\ \Rightarrow y=\frac{-11\pm \sqrt{121+48}}{4}\\ \Rightarrow y=\frac{-11\pm \sqrt{169}}{4}\\ \Rightarrow y=\frac{-11\pm 13}{4}\\ \Rightarrow y=\frac{-11+13}{4},\frac{-11-13}{4}\\ \Rightarrow y=\frac{2}{4},\frac{-24}{4}\\ \Rightarrow y=\frac{1}{2},-6\end{array}$

Conclusion:

Therefore, the zeroes of given polynomial function are $y=1,1,\frac{1}{2}\text{and}-6.$