Chapter 2.6, Problem 33E

Solution

To find: all real zeros of the function.

The real zeros of the function $\frac{5}{2},-2,1,3$ .

Given information: 

The function $f\left(x\right)=2x^4-9x^3+37x-30$ .

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

  $f\left(x\right)=2x^4-9x^3+37x-30$

Since the leading coefficient is $2$ and the constant term is $-30$ , the possible rational zeros are:

  $\begin{array}{l}\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{5}{1},\pm \frac{6}{1},\pm \frac{10}{1},\pm \frac{15}{1},\pm \frac{30}{1},\\ \pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{6}{2},\pm \frac{10}{2},\pm \frac{15}{2},\pm \frac{30}{2}\end{array}$ i.e.,

  $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2}$

Test these potential zeros by the synthetic division until it finds 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 $0$ .

Let's test $1$ first:

  $/$ polyhornerscheme $[x=1]\left\{2x^4-9x^3+37x-30\right\}$

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 $2$ :

  $/$ polyhornerscheme $[x=2]\left\{2x^3-7x^2-7x+30\right\}$

And $2$ is not a zero.

Let's test $3$

  $/$ polyhornerscheme $[x=3]\left\{2x^3-7x^2-7x+30\right\}$

And $3$ 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, $2x^2-x-10=0$ .

Solve this by factoring:

  $\begin{array}{c}2x^2-x+10=0\\ \left(2x-5\right)\left(x+2\right)=0\\ x=\frac{5}{2}\end{array}$

or $x=-2$

Therefore, the only real zeros are $\frac{5}{2},-2,1$ and 3.