Chapter 4, Problem 30RE

To determine

To calculate: The complex zeros of polynomial function $f(x)=2x^4+2x^3-11x^2+x-6$ . Also, writefin factored form.

Answer to Problem 30RE

The complex zeros are $2,-3,\pm i\sqrt{\frac{1}{2}}$ .

The factored form is $f(x)=(x-2)(x+3)(2x^2+1)$

Explanation of Solution

Given information:

The polynomial $f(x)=2x^4+2x^3-11x^2+x-6$.

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 factorthe 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 ,

  $f(x)=2x^4+2x^3-11x^2+x-6$

Since , the degree is 4 . So, there will be 4 complex zeros. The coefficients are :-

  $\begin{array}{l}{a}_0=-6\\ a_1=1\\ a_2=-11\\ a_3=2\\ a_4=2\end{array}$

Let coefficient of the constant term be$p=-6$andcoefficient of the leading termbe$q=2$ .

  $=p=\pm 1,\pm 2,\pm 3,\pm 6$ (Factors of the constant term )

  $=q=\pm 1,\pm 2$(Factors of the leading term )

Now, the possible ratios are :-

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

If fhasrational zeros, it will one of ratio mentioned in the list, which contain 10 possibilities.

  

Figure 1.

Using graphing utility, the graph is drawn. The figure interpret that fhas two zeroes: one is between 0 and -5 and second is between 0 and 5.

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

  $f(x)=2x^4+2x^3-11x^2+x-6$and $g(x)=x-2$ .

  $2x^3$

  $x-2\overline{)\begin{array}{l}2x^4+2x^3-11x^2+x-6\\ 2x^4-4x^3\end{array}}$

  $x-2\overline{)\begin{array}{l}2x^4+2x^3-11x^2+x-6\\ -2x^4+4x^3\end{array}}$

  $x-2\overline{)6x^3-11x^2+x-6}$

  $6x^2$

  $x-2\overline{)\begin{array}{l}6x^3-11x^2+x-6\\ 6x^3-12x^2\end{array}}$

  $x-2\overline{)\begin{array}{l}6x^3-11x^2+x-6\\ -6x^3+12x^2\end{array}}$

  $x-2\overline{)x^2+x-6}$

  $x$

  $x-2\overline{)\begin{array}{l}{x}^2+x-6\\ x^2-2x\end{array}}$

  $x-2\overline{)\begin{array}{l}{x}^2+x-6\\ -x^2+2x\end{array}}$

  $x-2\overline{)3x-6}$

  $3$

  $x-2\overline{)\begin{array}{l}3x-6\\ 3x-6\end{array}}$

  $x-2\overline{)\begin{array}{l}3x-6\\ -3x+6\end{array}}$

The remainder is zero and the quotient $2x^3+6x^2+x+3$ .

After division we get,

  $f(x)=2x^4+2x^3-11x^2+x-6$

  $=(x-2)(2x^3+6x^2+x+3)$

This implies either

$=x-2=0$ or $(2x^3+6x^2+x+3)=0$ .

  $=x=2$

Now , solving $2x^3+6x^2+x+3$ we get:-

$\begin{array}{l}=2x^3+6x^2+x+3\\ =2x^2(x+3)+1(x+3)\end{array}$ (Taking $2x^2$ common )

  $=(2x^2+1)(x+3)$(by grouping )

This implies either

$=2x^2+1=0$ or $x+3=0$ .

  $\begin{array}{l}=2x^2=-1\\ =x^2=-\frac{1}{2}\\ =x=\pm \sqrt{-\frac{1}{2}}\\ =x=\pm \sqrt{\frac{1}{2}{i}^2}\\ =x=\pm \sqrt{\frac{1}{2}}i\end{array}$

  $x=-3$

Therefore, the complex zeros are :-

  $\begin{array}{l}x=2+0i,-3+0i,0\pm i\sqrt{\frac{1}{2}}\\ x=2,-3,\pm i\sqrt{\frac{1}{2}}\end{array}$

The factored for of f is :-

  $f(x)=2x^4+2x^3-11x^2+x-6$

  $=(x-2)(2x^2+1)(x+3)$