Chapter 2.6, Problem 5QR

Solution

To Calculate:

The real zeroes of the function $f\left(x\right)=x^3-1$ using factoring.

  $x=1$

Given:

The function $f\left(x\right)=x^3-1$ .

Concepts Used:

The algebraic identity $a^3-b^3=\left(a-b\right)\left(a^2+b^2+ab\right)$ .

Completing the square of a quadratic polynomial.

The square of a real number cannot be negative.

Calculations:

Find the zeroes of $f\left(x\right)=x^3-1$ by solving the equation $x^3-1=0$ :

  $\begin{array}{cc}& x^3-1=0\\ \Rightarrow & x^3-1^3=0\\ \Rightarrow & \left(x-1\right)\left(x^2+1^2+\left(x\right)\left(1\right)\right)=0\\ \Rightarrow & \left(x-1\right)\left(x^2+x+1\right)=0\\ \Rightarrow & x=1\end{array}$

Note that the factor $x^2+x+1$ has no real zeroes demonstrate it by completing the square:

  $\begin{array}{c}{x}^2+x+1=x^2+x+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2+1\\ ={\left(x+\frac{1}{2}\right)}^2-\frac{1}{4}+1\\ ={\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}\end{array}$

The expression ${\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}$ has a squared real number added to a positive quantity. So this expression is not zero for any real value of $x$ . Its minimum value is $\frac{3}{4}$

Conclusion:

  $x=1$is the only real zero of the function $f\left(x\right)=x^3-1$ .