# To simplify x 2 − 2 x − 15 x 2 − 6 x + 5 ÷ x 2 − x − 12 x 2 − 1 by using appropriate operations.

### Precalculus: Mathematics for Calcu...

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

### Precalculus: Mathematics for Calcu...

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

#### Solutions

Chapter 1, Problem 57RE
To determine

## To simplify x2−2x−15x2−6x+5÷x2−x−12x2−1 by using appropriate operations.

Expert Solution

### Explanation of Solution

Given information:

x22x15x26x+5÷x2x12x21

Formula used:

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 .

Property of Fractions: AB÷CD=AB.DC

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

Calculations:

By factoring both numerator and denominator we get,

First Term x22x15x26x+5

Numerator = x22x15

where r=5 and s=3 .

r+s=5+3=2rs=(5)(3)=15(x+r)(x+s)=(x5)(x+3)  (i)

Denominator = x26x+5

where r=5 and s=1 .

r+s=51=6rs=(5)(1)=5(x+r)(x+s)=(x5)(x1)  (ii)

Second Term x2x12x21

Numerator = x2x12

where r=4 and s=3 .

r+s=4+3=1rs=(4)(3)=12(x+r)(x+s)=(x4)(x+3)  (iii)

Denominator = x21

x212=(x1)(x+1)  (iv)

Now,

=(x5)(x+3)(x5)(x1)÷(x4)(x+3)(x1)(x+1)

=(x5)(x+3)(x5)(x1).(x1)(x+1)(x4)(x+3)

=(x+1)(x4) ( by cancelling (x5)(x1)(x+3) )

After simplifying we get (x+1)(x4) .

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