Chapter 2.6, Problem 31E

Solution

To find: all real zeros of the function.

The real zeros of the function $\text{-2,2}$ .

Given information: 

The function $h\left(x\right)=2x^4-x^3-7x^2+4x-4$ .

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

  $\frac{p}{q}=\frac{\text{ factor of constant term }{a}_0}{\text{ factor of leading coefficient }{a}_n}$

Calculation:

Consider the function

  $h\left(x\right)=2x^4-x^3-7x^2+4x-4$

Since the leading coefficient is 2 and the constant term is $-4$ , the possible rational zeros are:

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{4}{2},$ i.e.

  $\pm 1,\pm 2,\pm 4,\pm \frac{1}{2}$

Test these zeros by the synthetic division. However, since the last number of the last row of the synthetic division is not $0$ , the number we test is not a $0$ .

Let's test $1$ first:

  $\backslash$ polyhornerscheme $[x=1]\left\{2x^4-x^3-7x^2+4x-4\right\}$

And $1$ is not a zero.

Test another number from the possible rational zeros list. This time, since the last number of the last row of the synthetic division is $0$ , the number test is a $0$ .

Now let's test $2$ :

  $\backslash$ polyhornerscheme $[x=2]\left\{2x^4-x^3-7x^2+4x-4\right\}$

And $2$ is indeed a zero.

Now let's test $-2$ using the last row (without the last number $0$ ) of the synthetic division:

  $\backslash$ polyhornerscheme $[x=-2]\left\{2x^3+3x^2-x+2\right\}$

And $-2$ is indeed a zero.

The polynomial has a degree of $4$ , and used synthetic division to find two zeros.

By using the last row of the synthetic division, $2x^2-x+1=0$ .

However, this is not factorable, so, use the quadratic formula:

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{1\pm \sqrt{1-8}}{4}\\ =\frac{1\pm i\sqrt{7}}{4}\end{array}$

However, these two zeros are imaginary.

Therefore, the only real zeros are $-2$ and $2$ .