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

To determine

To calculate: The solution of the equation (x21)2(x+2)3(x21)3(x+2)2=0.

Explanation

Given information:

The equation, (x21)2(x+2)3(x21)3(x+2)2=0.

Formula used:

Zero product property,

If ab=0 then either a=0 or b=0.

Formula for difference of two squares,

(a+b)(ab)=a2b2

Where, a and b are real numbers.

Quadratic formula,

The solution of the quadratic equation ax2+bx+c=0 is,

x=b±b24ac2a

Calculation:

Consider the expression,

(x21)2(x+2)3(x21)3(x+2)2=0

The left side expression is (x21)2(x+2)3(x21)3(x+2)2.

As both the term contains the term (x21)2(x+2)2. Thus, the common factor is (x21)2(x+2)2.

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

(x21)2(x+2)3(x21)3(x+2)2=(x21)2(x+2)2(x+2(x21))=(x21)2(x+2)2(x+2x2+1)=(x21)2(x+2)2(x+3x2)=(x1)(x+1)(x+2)2(x+3x2)

Therefore, the factors of the expression (x21)2(x+2)3(x21)3(x+2)2 are (x1)(x+1)(x+2)2(x+3x2)

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