Chapter 4, Problem 17RE

To determine

To calculate: The potential rational zeros of $f(x)=4x^3+4x^2-7x+2$ . Also, factor f over the real numbers.

Answer to Problem 17RE

There are 8 potential rational zeros of f with factors -2 and $\frac{1}{2}$ .

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 $rs=c$ .

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 ,

  $f(x)=4x^3+4x^2-7x+2$

where

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

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 $(4x^2-4x+1)=0$ .

  $=x=-2$

Solving equation $(4x^2-4x+1)$ with discriminate we get,

  $\begin{array}{l}a=4\\ b=-4\\ c=1\end{array}$

  $\begin{array}{l}={(}^{-}-4(4)(1)\\ =16-16\\ =0\end{array}$

The quadratic equation has one real solution

  $\begin{array}{l}x=\frac{-(-4)\pm \sqrt{{(-4)}^2-4(-4)(1)}}{2(4)}\\ x=\frac{-(-4)\pm \sqrt{16+16}}{8}\\ x=\frac{4}{8}\\ x=\frac{1}{2}\end{array}$

Therefore,

  $f(x)=x^3-3x^2-6x+8$

  $=(x+2)(4x^2-4x+1)$

  $x=-2,\frac{1}{2}$

The zeros of f are $-2$ and $\frac{1}{2}$ .

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