Chapter 4.4, Problem 7PS

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

Chapter
Section

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

# For Problems 1-40, perform the indicated operations, and express your answers in simplest form. (Objective 1) 6 a + 4 a 2 − 1 − 5 a − 1

To determine

To Find:

The simplest expression of 6a+4a215a1.

Explanation

Approach:

The expression of a rational is defined as the quotient obtained by a division of two polynomials in the form of p(x)q(x) where p(x) and q(x) are polynomials such that the variable x does not assume values such that q(x)=0.

For values of x where q(x) and k(x) are both nonzero expressions, then by the principle of fractions, for all polynomials p(x), the following holds.

p(x)â‹…k(x)q(x)â‹…k(x)=p(x)q(x).

Calculation:

The given expression is 6a+4a2âˆ’1âˆ’5aâˆ’1.

Factorise the terms,

6a+4a2âˆ’1âˆ’5aâˆ’1=6a+4(aâˆ’1)(a+1)âˆ’5(aâˆ’1)

The LCD of the denominator is (aâˆ’1)(a+1) where aâ‰ 1â€‰orâ€‰aâ‰ âˆ’1

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