Chapter 2.5, Problem 91E

To determine

To find : all the zeros of the function

Answer to Problem 91E

The zeros of the function are: $\boxed{\boxed{x=-0.5;0.5;0.75}}$

Explanation of Solution

Given information: $f\left(x\right)=16x^3-12x^2-4x+3$

Concept Involved:

Synthetic Division (for a Cubic Polynomial):To divide $a{x}^3+b{x}^2+cx+d$ by $x-k$ , use this pattern.

$\begin{array}{l}\begin{array}{l}k\\ \end{array}|\stackrel{Coefficientofdividend}{\overbrace{\underset{\_}{\begin{array}{cccc}\boxed{a}& b& c& d\\ \downarrow & \underset{\downarrow }{\boxed{ka}}& \underset{\downarrow }{\boxed{}}& \underset{\downarrow }{\boxed{}}\end{array}}}}\\ \begin{array}{ccccc}& \nearrow & \nearrow & \nearrow & \end{array}\\ \underset{Coefficientofquotient}{\underbrace{\begin{array}{ccc}\boxed{a}& \boxed{}& \boxed{}\end{array}}}\begin{array}{cc}\boxed{r}& \leftarrow \text{Remainder}\end{array}\end{array}$

In case when we have a polynomial with a missing term, insert placeholders with zero coefficients for missing powers of the variable. Vertical pattern: Add terms in columns Diagonal pattern: Multiply results by k. This algorithm for synthetic division works only for divisors of the form x - k. Remember that $x+k=x-\left(-k\right)$ .

The Division Algorithm: If $f\left(x\right)$  and   $d\left(x\right)$ are polynomials such that $d\left(x\right)\ne 0$ , and the degree of $d\left(x\right)$ isless than or equal to the degree of $f\left(x\right)$ , then there exist unique polynomials $q\left(x\right)$  and   $r\left(x\right)$ such that

  $\underset{\stackrel{\uparrow }{{\displaystyle Dividend}}}{{\displaystyle f\left(x\right)}}\text{ = }\underset{\stackrel{\uparrow }{{\displaystyle Divisor}}}{{\displaystyle d\left(x\right)}}\cdot \underset{\stackrel{\uparrow }{{\displaystyle Quotient}}}{{\displaystyle q\left(x\right)}}+\underset{\stackrel{\uparrow }{{\displaystyle Remainder}}}{{\displaystyle r\left(x\right)}}$

…where $r\left(x\right)=0$ or the degree of $r\left(x\right)$ is less than the degree of $d\left(x\right)$ .

If theremainder $r\left(x\right)$ is zero, then $d\left(x\right)$ divides evenly into $f\left(x\right)$ and k can be called as zero of the polynomial.

Graph:

  

Interpretation:

The point where the graph of the function cuts the x-axis are called zeros of the polynomial. From the above graph we see at $x=-0.5,x=0.5\&x=0.75$ the graph of the function cuts the x -axis. So those are the zeros of the polynomial

Calculation:

Description	Steps
Use synthetic division to check whether $x=-0.5$ is zero of the polynomial	$\begin{array}{l}\begin{array}{l}-0.5\\ \end{array}|\underset{\_}{\begin{array}{c}16\\ \end{array}\begin{array}{c}-12\\ -8\end{array}\begin{array}{c}-4\\ 10\end{array}\begin{array}{c}3\\ -3\end{array}}\\ 16-206\overline{)0}\leftarrow Remainder\end{array}$
Use synthetic division to check whether $x=0.5$ is zero of the polynomial	$\begin{array}{l}\begin{array}{l}0.5\\ \end{array}|\underset{\_}{\begin{array}{c}16\\ \end{array}\begin{array}{c}-12\\ 8\end{array}\begin{array}{c}-4\\ -2\end{array}\begin{array}{c}3\\ -3\end{array}}\\ 16-4-6\overline{)0}\leftarrow Remainder\end{array}$
Use synthetic division to check whether $x=0.75$ is zero of the polynomial	$\begin{array}{l}\begin{array}{l}0.75\\ \end{array}|\underset{\_}{\begin{array}{c}16\\ \end{array}\begin{array}{c}-12\\ 12\end{array}\begin{array}{c}-4\\ 0\end{array}\begin{array}{c}3\\ -3\end{array}}\\ 160-4\overline{)0}\leftarrow Remainder\end{array}$

Conclusion:

The zeros of the given function $f\left(x\right)=16x^3-12x^2-4x+3$ are: $x=-0.5;0.5;0.75$