Chapter EP, Problem 6.8.5EP

To determine

To calculate: The rational zeros of the function $f\left(x\right)=6x^4-31x^3-119x^2+214x+560$ .

Answer to Problem 6.8.5EP

The rational zeros of the function $f\left(x\right)=6x^4-31x^3-119x^2+214x+560$ are $-2,7,\frac{-5}{2}\text{ and }\frac{8}{3}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=6x^4-31x^3-119x^2+214x+560$ .

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 $f\left(x\right)=6x^4-31x^3-119x^2+214x+560$ .

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 $f\left(x\right)=6x^4-31x^3-119x^2+214x+560$

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

Now,

  $\begin{array}{c}f\left(-x\right)=6{\left(-x\right)}^4-31{\left(-x\right)}^3-119{\left(-x\right)}^2+214\left(-x\right)+560\\ f\left(-x\right)=6x^4+31x^3-119x^2-214x+560\end{array}$

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 $f\left(-x\right)=6x^4+31x^3-119x^2-214x+560$ .

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

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

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

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 6 and constant term is $560$ . Therefore, p is a factor of 560 and q is a factor of 6.

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

Factors of 560 are $\pm 1,\pm 2,\pm 4,\pm 5,\pm 7,\pm 8,\pm 10,\pm 14,\pm 16,\pm 20,\pm 28,\pm 35,\pm 40,\pm 56,\pm 70,\pm 80,\pm 112,\pm 140,\pm 280\text{ and }\pm 560$

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

  $\frac{p}{q}=\pm 1,\pm \frac{1}{2},\pm 2,\pm \frac{1}{3},\pm 4,\pm 5,\pm 7$

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

  $\begin{array}{cccccc}\frac{p}{q}& 6& -31& -119& 214& 560\\ -2& 6& -43& -33& 280& 0\\ -1& 6& -37& -82& 296& 264\\ 1& 6& -25& -144& 70& 630\\ 2& 6& -19& -157& -100& 360\end{array}$

As observed one zero is resulted at $x=-2$ . Since, only one real negative zero is obtained so use the depressed polynomial to obtain remaining zeroes.

Now, the depressed polynomial is obtained is $f\left(x\right)=6x^3-43x^2-33x+280$ .

Group the terms together and factor the polynomial.

  $\begin{array}{c}6x^3-43x^2-33x+280=0\\ \left(x-7\right)\left(2x+5\right)\left(3x-8\right)=0\end{array}$

Apply the zero product property.

Either $\left(x-7\right)=0\text{ or }\left(2x+5\right)=0\text{ or }\left(3x-8\right)=0$

That is, $x=7,x=\frac{-5}{2},x=\frac{8}{3}$ .

Thus, the rational zeros of the function $f\left(x\right)=6x^4-31x^3-119x^2+214x+560$ are $-2,7,\frac{-5}{2}\text{ and }\frac{8}{3}$ .