   Chapter 0.5, Problem 44E

Chapter
Section
Textbook Problem

Find all possible real solutions of each equation in Exercises 31–44. ( x 2 − 4 x + 4 ) 2 ( x 2 + 6 x + 5 ) 3 = 0

To determine

To calculate: The all real solutions in the polynomial equation (x24x+4)2(x2+6x+5)3=0.

Explanation

Given Information:

The provided polynomial equation is (x24x+4)2(x2+6x+5)3=0.

Formula used:

Principle of 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±b4ac2a where a,b and c are real numbers.

Calculation:

Consider the polynomial equation (x24x+4)2(x2+6x+5)3=0.

Apply the zero-product rule,

If (x24x+4)2=0

Take square root from both sides,

x24x+4=0

Compare the above equation with the equation ax2+bx+c=0,

a=1

, b=4 and c=4

Put the values of a,b,c in the quadratic formula x=b±b4ac2a

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