Chapter EP, Problem 6.7.15EP

To determine

To calculate: The zeros of the function $f\left(x\right)=2x^3+2x^2-34x+30$ .

Answer to Problem 6.7.15EP

The zeros of the function $f\left(x\right)=2x^3+2x^2-34x+30$ are $1,3\text{ and }-5$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=2x^3+2x^2-34x+30$ .

Formula used:

A polynomial of n degree has n zeros, which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

Calculation:

Consider the function $f\left(x\right)=2x^3+2x^2-34x+30$ .

Observe that degree of polynomial is 3, so the functions has 3 zeros which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(x\right)=2x^3+2x^2-34x+30$

The coefficients are $+2,+2,-34,+30$ .

There are 2 sign changes, so there are 2 or 0 positive real zeros.

Now,

  $\begin{array}{c}f\left(-x\right)=2{\left(-x\right)}^3+2{\left(-x\right)}^2-34\left(-x\right)+30\\ f\left(-x\right)=-2x^3+2x^2+34x+30\end{array}$

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(-x\right)=-2x^3+2x^2+34x+30$ .

The coefficients are $-2,+2,+34,+30$ .

There is 1 sign change, so there is 1 negative real zero.

Next, construct a table with possible combinations of real and imaginary zeros.

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 2& 1& 0& 2+1+0=3\\ 0& 1& 2& 0+1+2=3\end{array}$

Next, construct a table with help of synthetic substitution to compute the value of $f\left(x\right)$ for real values of $x$ .

  $\begin{array}{ccccc}x& 2& 2& -34& 30\\ -1& 2& 0& -34& 64\\ 0& 2& 2& -34& 30\\ 1& 2& 4& -30& 0\\ 2& 2& 6& -22& -14\end{array}$

As observed one zero is resulted at $x=1$ . Since, only one real positive zero is obtained so use the depressed polynomial to obtain remaining zeroes.

Now, the depressed polynomial is obtained is $f\left(x\right)=2x^2+4x-30$ .

Use the quadratic formula to obtain the remaining zeros.

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-4\pm \sqrt{16+240}}{4}\\ x=\frac{-4\pm 16}{4}\end{array}$

Either $x=\frac{-4+16}{4}\text{ or }x=\frac{-4-16}{4}$ .

That is, $x=3\text{ or }x=-5$ .

Thus, the zeros of the function $f\left(x\right)=2x^3+2x^2-34x+30$ are $1,3\text{ and }-5$ .