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, Problem 60RE
To determine

To simplify 1x+2+1x24+2x2x2 by using appropriate operations.

Expert Solution

Answer to Problem 60RE

The answer is 2x2+2x2x(x2)2 .

Explanation of Solution

Given information:

  1x+2+1x24+2x2x2

Formula used:

Properties of fractions:- AC+BC=A+BC

Difference of squares : A2B2=(AB)(A+B)

Factoring Trinomials:- To factor the form x2+bx+c

  (x+r)(x+s)=x2(r+s)x+rs where r+s=b and rs=c .

Calculations:

First Term 1x+2

Second Term

  =1x24=1(x+2)(x2)

Now,

  =1(x+2)+1(x+2)(x2)=1(x+2){1+1(x2)} ( Taking 1(x2) common)

writing the fraction in LCD (least common denominator) form we have,

  =1(x2){x2+1x2}=1(x2){x1x2}

  =2x4+2x(x2)=2x2x(x2)

New term 2x2x(x2) (i)

Third term

  =2x2x2

Denominator = x2x2

where r=2 and s=1 .

  r+s=2+1=1rs=(2)(1)=2(x+r)(x+s)=(x2)(x+1)  (ii)

Solving equation (i) and (ii)

  =2x2x(x2)+2(x2)(x+1)

Taking 1(x2) common

  =1(x2){2x2x+2(x+1)}

writing the fraction in LCD (least common denominator) form we have,

  =1(x2){(2x2)(x+1)x(x+1)+2xx(x+1)}=1(x2){2x2+2x2x2x(x+1)+2xx(x+1)}=1(x2){2x22+2xx(x+1)}=1(x2){2x2+2x2x(x2)}

  =2x2+2x2x(x2)2

After simplifying we get 2x2+2x2x(x2)2 .

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