   Chapter 0.6, Problem 16E

Chapter
Section
Textbook Problem

Solve the equations in Exercises 1–26. 6 x ( x 2 + 1 ) 2 ( x 2 + 2 ) 4 − 8 x ( x 2 + 1 ) 3 ( x 2 + 2 ) 3 ( x 2 + 2 ) 8 = 0

To determine

To calculate: The solution of the equation 6x(x2+1)2(x2+2)48x(x2+1)3(x2+2)3(x2+2)8=0.

Explanation

Given information:

The equation, 6x(x2+1)2(x2+2)48x(x2+1)3(x2+2)3(x2+2)8=0.

Formula used:

Zero product property,

If a product of two number is zero, one of the two must be zero.

Condition to solve an equation of the form PQ=0, If PQ=0, then P=0.

Calculation:

Consider the expression,

6x(x2+1)2(x2+2)48x(x2+1)3(x2+2)3(x2+2)8=0

As the above equation is in PQ=0 form. Therefore, the equation will be 6x(x2+1)2(x2+2)48x(x2+1)3(x2+2)3=0.

As both the term contains the term 2x(x2+1)2(x2+2)3. Thus, the common factor is 2x(x2+1)2(x2+2)3.

Now, factor out the common factor and simplify as below,

6x(x2+1)2(x2+2)48x(x2+1)3(x2+2)3=02x(x2+1)2(x2+2)3(3(x2+2)4(x2+1))=02x(x2+1)2(<

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