Chapter 2, Problem 18T

Solution

To find: All zeros of the polynomial function.

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

Given information:

A polynomial function: $f\left(x\right)=4x^4-8x^3-19x^2+23x-6$ .

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:

The leading coefficient is 4 and the constant term is $-6$ .

Factors of 4 are $\pm 1,\pm 2,\pm 4$ and factors of $-6$ are $\pm 1,\pm 2,\pm 3,\pm 6$ .

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{2}{1},\pm \frac{2}{2},\pm \frac{2}{4},\pm \frac{3}{1},\pm \frac{3}{2},\pm \frac{3}{4},\pm \frac{6}{1},\pm \frac{6}{2},\pm \frac{6}{4}\\ =\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6\end{array}$

By checking possible zeros in the given function, it can be seen that $x=3$ is a zero of the function. Now, use synthetic division as shown below:

  $\begin{array}{l}3\overline{)\begin{array}{l}4-8-1923-6\\ \downarrow 1212-216\end{array}}\\ 44-720\end{array}$

Because $x=3$ is a zero of the function, so the function can be written $f\left(x\right)=\left(x-3\right)\left(4x^3+4x^2-7x+2\right)$ .

By checking possible zeros in the given function, it can be seen that $x=-2$ is a zero of the function. Now, use synthetic division as shown below:

  $\begin{array}{l}-2\overline{)\begin{array}{l}44-72\\ \downarrow -88-2\end{array}}\\ 4-410\end{array}$

Because $x=-2$ is a zero of the function, so the function can be written $f\left(x\right)=\left(x-3\right)\left(x+2\right)\left(4x^2-4x+1\right)$ .

Solve for the quadratic equation as:

  $\begin{array}{c}\left(4x^2-4x+1\right)=0\\ 4x^2-2x-2x+1=0\\ 2x\left(2x-1\right)-1\left(2x-1\right)=0\\ \left(2x-1\right)\left(2x-1\right)=0\\ {\left(2x-1\right)}^2=0\\ x=\frac{1}{2}\end{array}$

Therefore, the zeros of the function would be $-2,\frac{1}{2},3$ .