Chapter 4, Problem 18RE

To determine

To calculate: The potential rational zeros of $f(x)=x^4-4x^3+9x^2-20x+20$ . Also, factor f over the real numbers.

Answer to Problem 18RE

There are 12 potential rational zeros of f with factors 2,2 and $\pm i\sqrt{5}$ .

Explanation of Solution

Given information:

The polynomial $f(x)=x^4-4x^3+9x^2-20x+20$ .

Formula used:

Rational Zero Theorem:

The theorem states that if the coefficients of the polynomial $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{.......}+a_{1}x+a_0$ are integers, then rational zero of the polynomial can be expressed in the form $\frac{p}{q}$ where p is the factor of constant term $a_0$ and q is a factor of the leading coefficient $a_n$ .

Factor Theorem:

The theorem states $(x-c)$ is a factor of $f(x)$ if and only if $f(c)=0$ .

Factoring Trinomials:- To factor the form $x^2+bx+c$

  $(x+r)(x+s)=x^2(r+s)x+rs$ where $r+s=b$ and $rs=c$ .

Discriminant Formula: $x=\frac{-b^2\pm \sqrt{b^2-4ac}}{2a}$ used to determine the number of roots in quadratic solution.

1. $b^2-4ac<0$ the equation has 0 real solutions
2. $b^2-4ac=0$ the equation has 1 real solution.
3. $b^2-4ac>0$ the equation has 2 real solutions.

Division Algorithm: $f(x)=q(x)g(x)+r(x)$ where $f(x)$ is dividend, $q(x)$ is quotient, $g(x)$ divisor and $r(x)$ remainder .

Calculation:

Consider ,

  $f(x)=x^4-4x^3+9x^2-20x+20$

where

  $\begin{array}{l}{a}_0=20\\ a_1=-20\\ a_2=9\\ a_3=-4\\ a_4=1\end{array}$

Let coefficient of the constant term be $p=20$ and coefficient of the leading term be $q=1$ .

  $=p=\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$ (Factors of the constant term )

  $=q=\pm 1$ (Factors of the leading term )

Now, the possible ratios are :-

  $=\frac{p}{q}:\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$

If f has rational zeros, it will one of ratio mentioned in the list, which contain 12 possibilities.

  

Figure 1.

Using graphing utility, the graph is drawn. The figure interpret that f has one zeroes: one at

2 .

Let $c=2$ such that f is divided by $(x-2)$ .

  $f(x)=x^4-4x^3+9x^2-20x+20$ and $g(x)=x-2$ .

  $x^3$

  $x-2\overline{)\begin{array}{l}{x}^4-4x^3+9x^2-20x+20\\ x^4-2x^3\end{array}}$

  $x-2\overline{)\begin{array}{l}{x}^4-4x^3+9x^2-20x+20\\ -x^4+2x^3\end{array}}$

  $x-2\overline{)-2x^3+9x^2-20x+20}$

  $-2x^2$

  $x-2\overline{)\begin{array}{l}-2x^3+9x^2-20x+20\\ -2x^3+4x^2\end{array}}$

  $x-2\overline{)\begin{array}{l}-2x^3+9x^2-20x+20\\ 2x^3-4x^2\end{array}}$

  $x-2\overline{)5x^2-20x+20}$

  $5x$

  $x-2\overline{)\begin{array}{l}5x^2-20x+20\\ 5x^2-10x\end{array}}$

  $x-2\overline{)\begin{array}{l}5x^2-20x+20\\ -5x^2+10x\end{array}}$

  $x-2\overline{)-10x+20}$

  $-10$

  $x-2\overline{)\begin{array}{l}-10x+20\\ -10x+20\end{array}}$

  $x-2\overline{)\begin{array}{l}-10x+20\\ 10x-20\end{array}}$

The remainder is zero and the quotient $x^3-2x^2+5x-10$ .

After division we get,

  $f(x)=x^4-4x^3+9x^2-20x+20$

  $=(x-2)(x^3-2x^2+5x-10)$

This implies either

  $=x-2=0$ or $(x^3-2x^2+5x-10)=0$ .

  $=x=2$

  

Figure 2.

Again ,by graphing utility, the graph is drawn. The figure interpret that f has one zeroes: one at 2 .

Let $c=2$ such that f is divided by $(x-2)$ .

  $f(x)=x^3-2x^2+5x-10$ and $g(x)=x-2$ .

  $x^2$

  $x-2\overline{)\begin{array}{l}{x}^3-2x^2+5x-10\\ x^3-2x^2\end{array}}$

  $x-2\overline{)\begin{array}{l}{x}^3-2x^2+5x-10\\ -x^3+2x^2\end{array}}$

  $x-2\overline{)5x-10}$

  $5$

  $x-2\overline{)\begin{array}{l}5x-10\\ 5x-10\end{array}}$

  $x+2\overline{)\begin{array}{l}5x-10\\ -5x+10\end{array}}$

The remainder is zero and the quotient $x^2+5$ .

After division we get,

  $f(x)=x^3-2x^2+5x-10$

  $=(x-2)(x^2+5)$

This implies either

  $=x-2=0$ or $(x^2+5)=0$ .

  $=x=2$

Solving equation $(x^2+5)$ with discriminate we get,

  $\begin{array}{l}a=1\\ b=0\\ c=5\end{array}$

  $\begin{array}{l}={(}^0-4(1)(5)\\ =0-20\\ =-20<0\end{array}$

The quadratic equation has 0 real solution

  $\begin{array}{l}x=\frac{(0)\pm \sqrt{{(0)}^2-4(1)(5)}}{2(1)}\\ x=\frac{(0)\pm \sqrt{0-20}}{2}\\ x=\pm \frac{\sqrt{-20}}{2}\\ x=\pm \frac{\sqrt{20i^2}}{2}\\ x=\pm \frac{2i\sqrt{5}}{2}\\ x=\pm i\sqrt{5}\end{array}$

Therefore,

  $f(x)=x^4-4x^3+9x^2-20x+20$

  $=(x-2)(x^3-2x^2+5x-10)$

  $=(x-2)(x-2)(x^2+5)$

  $x=2,2,\pm i\sqrt{5}$

The zeros of f are $2,2$ and $\pm i\sqrt{5}$ .

The two zeroes of f are in the list of potential rational zeros.