   Chapter 0.6, Problem 7E

Chapter
Section
Textbook Problem

Solve the equations in Exercises 1–26. ( x 2 + 1 ) 5 ( x + 3 ) 4 + ( x 2 + 1 ) 6 ( x + 3 ) 3 = 0

To determine

To calculate: The solution of the equation (x2+1)5(x+3)4+(x2+1)6(x+3)3=0.

Explanation

Given information:

The equation, (x2+1)5(x+3)4+(x2+1)6(x+3)3=0.

Formula used:

Zero product property,

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

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

x=b±b24ac2a

Calculation:

Consider the expression,

(x2+1)5(x+3)4+(x2+1)6(x+3)3=0

The left side expression is (x2+1)5(x+3)4+(x2+1)6(x+3)3.

Rewrite the above expression as below,

(x2+1)5(x+3)3(x+3)+(x2+1)5(x2+1)(x+3)3

As both the term contains the term (x2+1)5(x+3)3. Thus, the common factor is (x2+1)5(x+3)3.

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

(x2+1)5(x+3)4+(x2+1)6(x+3)3=(x2+1)5(x+3)3{(x+3)+(x2+1)}=(x2+1)5(x+3)3(x2+x+4)

Therefore, the factors of the expression (x2+1)5(x+3)4+(x2+1)6(x+3)3 are (x2+1)5(x+3)3(x2+x+4)

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