   Chapter 0.5, Problem 36E

Chapter
Section
Textbook Problem

Find all possible real solutions of each equation in Exercises 31–44. x 3 − 27 = 0

To determine

To calculate: The all real solutions in the polynomial equation x327=0.

Explanation

Given Information:

The provided polynomial equation is x327=0.

Formula used:

Zero product rule:

An equation ab=0 is true if and only if a=0 or b=0, both

A product is zero if and only if at least one factor is zero

The value of x in the equation ax2+bx+c=0 can be given by the quadratic formula,

x=b±b24ac2a where a,b and c are real numbers with a0.

Calculation:

Consider the polynomial equation x327=0.

First determine the single solution in this cubic equation x327=0

Compare the equation x327=0 with the standard cubic equation ax3+bx2+cx+d=0.

a=1 and d=27

Therefore, possible factor of 1 is x=±1 and 27 is x=±1, x=±3, x=±9 and x=±27.

First try x=3 substitute in the provided polynomial equation x327=0, (3)327=?0  2727=?0           0=0

Which is true

Therefore, x3 is a factor of the polynomial x327=0.

In order to find other two values of x, divide x327=0 by x3, x3x2+3x+9x3+0x2+0x27x33x2_       3x2+0x       3x29x_<

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