Chapter EP, Problem 6.8.8EP

To determine

To calculate: The rational zeros of the function $f\left(x\right)=3x^4+8x^3+9x^2+32x-12$ .

Answer to Problem 6.8.8EP

The rational zeros of the function $f\left(x\right)=3x^4+8x^3+9x^2+32x-12$ are $-3\text{ and }\frac{1}{3}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=3x^4+8x^3+9x^2+32x-12$ .

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 $f\left(x\right)=3x^4+8x^3+9x^2+32x-12$ .

Observe that degree of polynomial is 4, so the functions has 4 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 $f\left(x\right)=3x^4+8x^3+9x^2+32x-12$

There are 1 sign change, so there is 1positive real zero.

Now,

  $\begin{array}{c}f\left(-x\right)=3{\left(-x\right)}^4+8{\left(-x\right)}^3+9{\left(-x\right)}^2+32\left(-x\right)-12\\ f\left(-x\right)=3x^4-8x^3+9x^2-32x-12\end{array}$

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 $f\left(-x\right)=20x^4+16x^3+11x^2+12x-3$ .

There are 3 sign changes, so there are 3or 1 negative real zeros.

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

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 1& 3& 0& 1+3+0=4\\ 1& 1& 2& 1+1+2=4\end{array}$

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 3 and constant term is $-12$ . Therefore, p is a factor of $12$ and q is a factor of 3.

Factors of $3$ are $\pm 1\text{ and }\pm \text{3}$ .

Factors of 12 are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6\text{ and }\pm 12$

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

  $\frac{p}{q}=\pm 1,\pm \frac{1}{3},\pm 2,\pm \frac{2}{3},\pm 3,\pm 4,\pm \frac{4}{3},\pm 6,\pm 12$

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

  $\begin{array}{cccccc}\frac{p}{q}& 3& 8& 9& 32& -12\\ -3& 3& -1& 12& -4& 0\\ -1& 3& 5& 4& 28& -40\\ 1& 3& 11& 20& 52& 40\\ 3& 3& 17& 60& 212& 624\end{array}$

As observed one zero is resulted at $x=-3$ . Since, only one real negative zero is obtained so use the depressed polynomial to obtain remaining zeroes.

Now, the depressed polynomial is obtained is $f\left(x\right)=3x^3-x^2+12x-4$ .

Group the terms together and factor the polynomial.

  $\begin{array}{c}3x^3-x^2+12x-4=0\\ x^2\left(3x-1\right)+4\left(3x-1\right)=0\\ \left(3x-1\right)\left(x^2+4\right)=0\end{array}$

Apply the zero product property.

Either $\left(3x-1\right)=0\text{ or }\left(x^2+4\right)=0$

That is, $x=\frac{1}{3},x=\pm i\sqrt{2}$ .

Thus, the rational zeros of the function $f\left(x\right)=3x^4+8x^3+9x^2+32x-12$ are $-3\text{ and }\frac{1}{3}$ .