Chapter 2.5, Problem 97E

To determine

To find : the rational zeros of the given function

Answer to Problem 97E

The rational zeros of the function are: $\boxed{\boxed{-1,1/4,\&1}}$

Explanation of Solution

Given information: $f\left(x\right)=x^3-\frac{1}{4}{x}^2-x+\frac{1}{4}$

Concept Involved:

The zeros of the polynomial are the values of x where the graph cuts the x-axis and makes $P\left(x\right)=0$

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 $\underset{\stackrel{­}{{\displaystyle Dividend}}}{{\displaystyle f(x)}}=\underset{\stackrel{­}{{\displaystyle Divisor}}}{{\displaystyle d(x)}}\times \underset{\stackrel{­}{{\displaystyle Quotient}}}{{\displaystyle q(x)}}+\underset{\stackrel{­}{{\displaystyle Remainder}}}{{\displaystyle r(x)}}$ 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.

Calculation:

To find the zeros of the function set the polynomial equal to zero $f\left(x\right)=0$

  $x^3-\frac{1}{4}{x}^2-x+\frac{1}{4}=0$

To get rid of the fraction in the equation multiply 4 throughout the equation

  $4x^3-x^2-4x+1=0$

Factor the GCF of first two terms and GCF of last two terms in left side of the equation

  $x^2\left(4x-1\right)-1\left(4x-1\right)=0$

Factor the common binomial from left side of the equation

  $\left(4x-1\right)\left(x^2-1\right)=0$

Rewrite the binomials inside each parenthesis

  $\left(4x-1\right)\left(x^2-1^2\right)=0$

Use the algebraic identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$ to factor the left side of the equation

  $\left(4x-1\right)\left(x+1\right)\left(x-1\right)=0$

We need to set each factor equal to zero using the zero factor property which states that

“If $ab=0$ , then $a=0$ or $b=0$ .”

  $x+1=0$  ▸1^st equation

  $x-1=0$  ▸2^nd equation

  $4x-1=0$  ▸3^rdequation

Solving the 1st equation $x+1=0$

• By subtracting 1 on both sides
• By simplifying the equation on both sides

  $x+1-1=0-1$

  $x=-1$

Solving the 2nd equation $x-1=0$

• By adding 1on both sides
• By simplifying the equation on both sides

  $x-1+1=0+1$

  $x=1$

Solving the 3rdequation $4x-1=0$

• By adding 1 on both sides, then
• By simplifying on both sides, then
• By dividing 4 on both sides, and then
• By simplifying fraction on both sides

  $4x-1+1=0+1$

  $4x=1$

  $\frac{4x}{4}=\frac{1}{4}$

  $x=\frac{1}{4}$

Graph:

   Interpretation:

From the graph of the function we can pick possible zeros of the function as $x=-1,\frac{1}{4},\&1$

Conclusion:

The rational zeros of the function are: $x=-1,\frac{1}{4},\&1$