Chapter 6.8, Problem 34PPS

To determine

To calculate: The zeros of the function $f\left(x\right)=2x^3-7x^2-8x+28$ .

Answer to Problem 34PPS

The zeros of the function $f\left(x\right)=2x^3-7x^2-8x+28$ are $-2,2\text{ and }\frac{7}{2}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=2x^3-7x^2-8x+28$ .

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 $f\left(x\right)=2x^3-7x^2-8x+28$ .

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 $f\left(x\right)=2x^3-7x^2-8x+28$

There is 2 or 0 positive real zeros.

Now,

  $f\left(-x\right)=-2x^3-7x^2+8x+28$

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 $f\left(-x\right)=-2x^3-7x^2+8x+28$ .

There is 1 negative real zero.

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

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

Recall that the Rational zero theorem states that provided a polynomial $P\left(x\right)$ with integral coefficients then every rational zero is of the form $\frac{p}{q}$ that is a rational number in simplest form. Here, p is a factor of the constant term and q is a factor of leading coefficient.

For the provided function leading coefficient is 2 and constant term is 28 Therefore, p is a factor of 28 and q is a factor of 2.

The possible combinations of $\frac{p}{q}$ in simplest form are,

  $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28,\pm \frac{1}{2},\pm \frac{7}{2}$

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

  $\begin{array}{ccccc}x& 2& -7& -8& 28\\ 1& 2& -5& -13& 15\\ -1& 2& -9& 1& 27\\ 2& 2& -3& -14& 0\\ -2& 2& -11& 14& 0\end{array}$

As observed two zeros are resulted at $x=-2\text{ and }x=2$ .

Now, the depressed polynomial is obtained is $2x^2-3x-14$ .

Factor the polynomial.

  $\begin{array}{c}2x^2-3x-14=0\\ 2x^2-7x+4x-14=0\\ x\left(2x-7\right)+2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x+2\right)=0\end{array}$

Either $\left(2x-7\right)=0\text{ or }\left(x+2\right)=0$

That is, $x=\frac{7}{2}\text{ or }x=-2$ .

Thus, the zeros of the function $f\left(x\right)=2x^3-7x^2-8x+28$ are $-2,2\text{ and }\frac{7}{2}$ .