Chapter 5.8, Problem 45PPS

To determine

To calculate: The zeros of the function $f\left(x\right)=x^5-x^4-23x^3+33x^2+126x-216$ .

Answer to Problem 45PPS

The zeros of the function $f\left(x\right)=x^5-x^4-23x^3+33x^2+126x-216$ are $-4,-3,2,3\text{ and }3$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=x^5-x^4-23x^3+33x^2+126x-216$ .

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 $f\left(x\right)=x^5-x^4-23x^3+33x^2+126x-216$ .

Observe that degree of polynomial is 5, so the functions has 5 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 $f\left(x\right)=x^5-x^4-23x^3+33x^2+126x-216$

There are 3 or 1 positive real zeros.

Now,

  $f\left(-x\right)=-x^5-x^4+23x^3+33x^2-126x-216$

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 $f\left(-x\right)=-x^5-x^4+23x^3+33x^2-126x-216$ .

There are 2 or 0 negative real zero.

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

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 3& 0& 2& 3+0+2=5\\ 3& 2& 0& 3+2+0=5\\ 1& 0& 4& 1+0+4=5\\ 1& 2& 2& 1+2+2=5\end{array}$

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 1 and constant term is 216. Therefore, p is a factor of 216 and q is a factor of 1.

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

  $\frac{p}{q}=\pm 1,\pm 2,4,\pm 8,\pm 3,\pm 9,\pm 27,\pm 6,\pm 18,\pm 54,\pm 12,\pm 36,\pm \dots$

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

  $\begin{array}{ccccccc}x& 1& -1& -23& 33& 126& -216\\ 3& 1& 2& -17& -18& 72& 0\\ -3& 1& -4& -11& 66& -72& 0\\ 2& 1& 1& -21& -9& 108& 0\\ -2& 1& -3& -17& 67& -8& -200\end{array}$

As observed three zeros are resulted at $x=-3,x=3\text{ and }x=2$ .

So, factor of the polynomial is,

  $\begin{array}{c}\left(x-3\right)\left(x+3\right)\left(x-2\right)=\left(x-2\right)\left(x^2-9\right)\\ =x^3-2x^2-9x+18\end{array}$ .

Now, divide the polynomial $f\left(x\right)=x^5-x^4-23x^3+33x^2+126x-216$ by $x^3-2x^2-9x+18$ .

  $x^5-x^4-23x^3+33x^2+126x-216=\left(x^3-2x^2-9x+18\right)\left(x^2+x-12\right)$

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

Factor the polynomial.

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

That is, $x=-4\text{ or }x=3$ .

Thus, the zeros of the function $f\left(x\right)=x^5-x^4-23x^3+33x^2+126x-216$ are $-4,-3,2,3\text{ and }3$ .