Chapter 2.6, Problem 15E

Solution

To find: all real zeros of the function.

The only real zeros are $-2$ and $-1$ .

Given information: 

The function $h\left(x\right)=x^4+7x^3+26x^2+44x+24$ .

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

  $h\left(x\right)=x^4+7x^3+26x^2+44x+24$ .

Then

Since the leading coefficient is $1$ and the constant term is $24$ , the possible rational zeros are:

  $\begin{array}{l}\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{8}{1},\pm \frac{12}{1},\pm \frac{24}{1}\text{, i}\text{.e}\text{.,}\\ \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24\end{array}$

Test these potential zeros by the synthetic division until we have found a zero. If a test number is a zero, the last number of the last row of the synthetic division will be $0$ . If a test number is not a zero, the last number of the last row of the synthetic division won't be $0$ .

Here, since all the coefficients are positive, testing any positive number will not work (i.e., it won't be a zero). So, test the negative numbers, and the first one should always be $-1$ .

Let's test $-1$ first:

\polyhornerscheme $[x=-1]\left\{x^4+7x^3+26x^2+44x+24\right\}$

And $-1$ is indeed a zero.

Now, use the last row (without the last number $0$ ) of the synthetic division to test other numbers.

Let's test $-2$ first:

L polyhorner scheme $[x=-2]\left\{x^3+6x^2+20x+24\right\}$

And $-2$ is indeed a zero.

Using synthetic division, the polynomial has a degree of $4$ and two zeros.

By using the last row of the synthetic division, we have $x^2+4x+12=0$ .

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

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-4\pm \sqrt{16-48}}{2}\\ =\frac{-4\pm 4i\sqrt{2}}{2}\\ =-2\pm 2i\sqrt{2}\end{array}$

However, these two zeros are imaginary.

Therefore, the only real zeros are $-2$ and $-1$ .