   Chapter 0.5, Problem 33E

Chapter
Section
Textbook Problem

Find all possible real solutions of each equation in Exercises 31–44. x 3 + 4 x 2 + 4 x + 3 = 0

To determine

To calculate: The all real solutions in the polynomial equation x3+4x2+4x+3=0.

Explanation

Given Information:

The provided polynomial equation is x3+4x2+4x+3=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 x3+4x2+4x+3=0.

First determine the single solution in this cubic equation x3+4x2+4x+3=0

Compare the equation x3+4x2+4x+3=0 with the standard cubic equation ax3+bx2+cx+d=0

a=1 and d=3

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

First try x=3 substitute in the provided polynomial x3+4x2+4x+3=0, (3)3+4(3)2+4(3)+3=?0             27+3612+3=?0    0=0

This is true.

Therefore, x+3 is a factor of the polynomial x3+4x2+4x+3=0.

In order to find other two values of x, divide x3+4x2+4x+3=0 by x+3, x+3x2+x+1x3+4x2+4x+3         x3+3x2_                  x2+4x                 x2+3x_                        �

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