   Chapter 0.5, Problem 37E

Chapter
Section
Textbook Problem

Find all possible real solutions of each equation in Exercises 31–44. y 3 + 3 y 2 + 3 y + 2 = 0

To determine

To calculate: The all real solutions in the polynomial equation y3+3y2+3y+2=0.

Explanation

Given Information:

The provided polynomial equation is y3+3y2+3y+2=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 y3+3y2+3y+2=0.

First determine the single solution in this cubic equation y3+3y2+3y+2=0

Compare the equation y3+3y2+3y+2=0 with the standard cubic equation ax3+bx2+cx+d=0.

a=1 and d=2

Therefore, possible factor of 1 is x=±1 and 2 is x=±1, x=±2.

First try y=2 substitute in the provided polynomial equation y3+3y2+3y+2=0, (2)3+3(2)2+3(2)+2=?0                  8+126+2=?0                                0=0

Which is true

Therefore, y+2 is a factor of the polynomial y3+3y2+3y+2=0.

In order to find other two values of x, divide y3+3y2+3y+2=0 by y+2, y+2y2+y+1y3+3y2+3y+2         y3+2y2_               y2+3y               y</

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