   Chapter 0.5, Problem 31E

Chapter
Section
Textbook Problem

Find all possible real solutions of each equation in Exercises 31–44. x 3 + 6 x 2 + 11 x + 6 = 0

To determine

To calculate: The all real solutions in the polynomial equation x3+6x2+11x+6=0.

Explanation

Given Information:

The provided polynomial equation is x3+6x2+11x+6=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+6x2+11x+6=0.

First determine the single solution in this cubic equation x3+6x2+11x+6=0

Compare the equation x3+6x2+11x+6=0 with the standard cubic equation ax3+bx2+cx+d=0

a=1 and d=6

Therefore, possible factor of 1 is x=±1 and 6 is x=±1, x=±2, x=±3 and x=±6.

First try x=1 substitute in the provided polynomial x3+6x2+11x+6=0, (1)3+6(1)2+11(1)+6=?0                   1+611+6=?0                       0=0

Which is true.

Therefore, x+1 is a factor of the polynomial x3+6x2+11x+6=0.

In order to find other two values of x, divide x3+6x2+11x+6=0 by x+1, x+1x2+5x+6x3+6x2+11x+6         x3+x2_               5x2+11x               5x2+5x_                        6x+6                        6x+6_                               0

Thus,

x3

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