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Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248

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Chapter
Section
BuyFindarrow_forward

Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248
Textbook Problem

Solve the equations in Exercises 1–26.

( x + 1 ) 2 ( x + 2 ) 3 ( x + 1 ) 3 ( x + 2 ) 2 ( x + 2 ) 6 = 0

To determine

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

Explanation

Given information:

The equation, (x+1)2(x+2)3(x+1)3(x+2)2(x+2)6=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,

(x+1)2(x+2)3(x+1)3(x+2)2(x+2)6=0

As the above equation is in PQ=0 form.

Therefore, the equation will be (x+1)2(x+2)3(x+1)3(x+2)2=0.

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

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