Chapter 2.6, Problem 35E

Solution

To find: all real zeros of the function.

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

Given information: 

The function $h\left(x\right)=2x^5+5x^4-3x^3-2x^2-5x+3$ .

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

  $h\left(x\right)=2x^5+5x^4-3x^3-2x^2-5x+3$

Since the leading coefficient is $2$ and the constant term is $3$ , the possible rational zeros are:

  $\pm \frac{1}{1},\pm \frac{3}{1},\pm \frac{1}{2},\pm \frac{3}{2}$ i.e.,

  $\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

Test these zeros by the synthetic division until it finds a zero. If a test number is not a zero, the last number of the last row of the synthetic division won't be $0$ . If a test number is a zero, the last number of the last row of the synthetic division will be $0$ . Furthermore, and use the last row (without the last $0$ ) the new coefficients when testing another

number.

Let's test $1$ first:

  $\backslash$ polyhornerscheme $[x=1]\left\{2x^5+5x^4-3x^3-2x^2-5x+3\right\}$

And $1$ is a zero.

Let's test $3$ :

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

And $3$ is not a zero.

Let's test $\frac{1}{2}$ :

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

And $\frac{1}{2}$ is also a zero.

Now, let's test the negative numbers.

Let's test $-1$ first:

  $\backslash$ polyhornerscheme $[x=-1]\left\{2x^3+8x^2+8x+6\right\}$

And $-1$ is not a zero.

Let's test $-3$ :

  $\backslash$ polyhornerscheme $[x=-3]\left\{2x^3+8x^2+8x+6\right\}$

And $-3$ is a zero.

Now, use the last row to translate it back to an equation involving $x$ .

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

It divides this equation by $2$ to get $x^2+x+1=0$ .

However, this equation is not factorable. If we use the quadratic formula, it will only give us imaginary zeros.

Therefore, all real zeros: $1,-3$ and $\frac{1}{2}$ .