Chapter 2, Problem 28T

Solution

To find: All zeros of the polynomial function. Correct Mario’s error.

The possible rational zeros of the function would be $0,\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{8},\pm 2,\pm 4,\pm 8,\pm 16$ .

Given information:

Mario made an error in listing the possible rational zero of the function: $f\left(x\right)=-8x^6+7x^5-3x^4+45x^3-1500x^2+16x$ .

Possible zeros:

  $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{4},\pm \frac{1}{2},\pm \frac{1}{16},\pm \frac{1}{8}$

Formula used:

As per Rational Zero Theorem, the possible rational zeros for any function with integer coefficients can be obtained by dividing factors of leading coefficient of highest degree term with factors of the constant term as:

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

Calculation:

First of all, factor out x from all the terms as:

  $f\left(x\right)=-x\left(8x^5+7x^4-3x^3+45x^2-1500x+16\right)$

Now apply rational zero theorem on the polynomial $8x^5+7x^4-3x^3+45x^2-1500x+16$ .

The leading coefficient is $-8$ and the constant term is $16$ .

Factors of $-8$ are $\pm 1,\pm 2,\pm 4,\pm 8$ and factors of $16$ are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16$ .

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{8},\pm \frac{2}{1},\pm \frac{2}{2},\pm \frac{2}{4},\pm \frac{2}{8},\pm \frac{4}{1},\pm \frac{4}{2},\pm \frac{4}{4},\pm \frac{4}{8},\\ \pm \frac{8}{1},\pm \frac{8}{2},\pm \frac{8}{4},\pm \frac{8}{8},\pm \frac{16}{1},\pm \frac{16}{2},\pm \frac{16}{4},\pm \frac{16}{8}\end{array}$

Now simplify and remove the duplicates as:

  $\frac{p}{q}=\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{8},\pm 2,\pm 4,\pm 8,\pm 16$

Upon comparing these zeros with Mario’s work, it can be seen that Mario use $\pm \frac{1}{16}$ instead of $\pm 16$ . Mario also forget to use zero as a solution.

Therefore, the possible rational zeros of the function would be $0,\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{8},\pm 2,\pm 4,\pm 8,\pm 16$ .