Chapter 4, Problem 28RE

To determine

To calculate: The complex zeros of polynomial function $f(x)=4x^3+4x^2-7x+2$ . Also, write f in factored form.

Answer to Problem 28RE

The complex zeros are $-2,-\frac{1}{2},-\frac{1}{2}$ .

The factored form of polynomial is $f(x)=(x-2)\left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)$ .

Explanation of Solution

Given information:

The polynomial $f(x)=4x^3+4x^2-7x+2$ .

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 $rs=c$ .

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 ,

  $f(x)=4x^3+4x^2-7x+2$

Since the polynomial is a degree 3. It has 3 complex zeros. The coefficients are :-

  $\begin{array}{l}{a}_0=2\\ a_1=-7\\ a_2=4\\ a_3=4\end{array}$

Let coefficient of the constant term be $p=2$ and coefficient of the leading term be $q=4$ .

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

  $=q=\pm 1,\pm 2,\pm 4$ (Factors of the leading term )

Now, the possible ratios are :-

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

If f has rational zeros, it will one of ratio mentioned in the list, which contain 8 possibilities.

  

Figure 1.

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

-2 , second is between 0 and 2 .

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

  $f(x)=4x^3+4x^2-7x+2$ and $g(x)=x+2$ .

  $4x^2$

  $x+2\overline{)\begin{array}{l}4x^3+4x^2-7x+2\\ 4x^3+8x^2\end{array}}$

  $x+2\overline{)\begin{array}{l}4x^3+4x^2-7x+2\\ -4x^3-8x^2\end{array}}$

  $x+2\overline{)-4x^2-7x+2}$

  $-4x$

  $x+2\overline{)\begin{array}{l}-4x^2-7x+2\\ -4x^2-8x\end{array}}$

  $x+2\overline{)\begin{array}{l}-4x^2-7x+2\\ 4x^2+8x\end{array}}$

  $x+2\overline{)x+2}$

  $1$

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

The remainder is zero and the quotient $4x^2-4x+1$ .

After division we get,

  $f(x)=4x^3+4x^2-7x+2$

  $=(x+2)(4x^2-4x+1)$

This implies either

  $=x+2=0$ or $(4x^2-4x+1)=0$ .

  $=x=-2$

Now, solving $4x^2-4x+1$ we get,

  $\begin{array}{l}=4x^2-4x+1\\ =x^2-x+\frac{1}{4}\end{array}$

where $r=\frac{-1}{2}$ and $s=\frac{-1}{2}$

  $\begin{array}{l}r+s=-\frac{1}{2}-\frac{1}{2}=-1\\ rs=\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)=\frac{1}{4}\\ (x+r)(x+s)=\left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)\end{array}$

  $\begin{array}{l}=\left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=0\\ =-\frac{1}{2},-\frac{1}{2}\end{array}$

Therefore, the complex zeros are :-

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

The factored form of f is -

  $f(x)=x^3-3x^2-6x+8$

  $=(x+2)(x-2)(x-2)$