BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.3, Problem 132E
To determine

To factor: The algebraic expression 4a2c2(a2b2+c2)2 completely.

Expert Solution

Answer to Problem 132E

The algebraic expression 4a2c2(a2b2+c2)2 can be completely factored as (a+b+c)(a+bc)(ab+c)(a+b+c)_.

Explanation of Solution

The given algebraic expression is, 4a2c2(a2b2+c2)2.

Factorize 4a2c2(a2b2+c2)2 as follows.

4a2c2(a2b2+c2)2=(2ac)2(a2b2+c2)2=([2ac(a2b2+c2)][2ac+(a2b2+c2)])[(A2B2)=(AB)(A+B)]=[(2aca2+b2c2)(2ac+a2b2+c2)]

Rewrite the above equation as follows.

[(2aca2+b2c2)(2ac+a2b2+c2)]=([b2(a2+c22ac)][(a2+c2+2ac)b2])[(A+B)2=(A2+B2+2AB)(AB)2=(A2+B22AB)]=[b2(ac)2][(a+c)2b2]=[(ba+c)(b+ac)(a+cb)(a+c+b)][(A2B2)=(AB)(A+B)]=(a+b+c)(a+bc)(ab+c)(a+b+c)

Thus, the algebraic expression 4a2c2(a2b2+c2)2 can be completely factored as (a+b+c)(a+bc)(ab+c)(a+b+c)_.

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