Chapter 4.4, Problem 16PS

### 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 2 − 3 a − 54 − 10 a 2 + 5 a − 6

To determine

To Find:

The simplest expression of 6a23a5410a2+5a6.

Explanation

Approach:

The rational expression 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, while for any values of x the denominator q(x)â‰ 0.

For values of x, where q(x) and k(x) are both nonzero expressions, then 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: 6a2âˆ’3aâˆ’54âˆ’10a2+5aâˆ’6.

Factorise the terms.

6a2âˆ’3aâˆ’54âˆ’10a2+5aâˆ’6=6a2âˆ’9a+6aâˆ’54âˆ’10a2+6aâˆ’aâˆ’6=6a(aâˆ’9)+6(aâˆ’9)âˆ’10a(a+6)âˆ’1(a+6)=6(a+6)(aâˆ’9)âˆ’10(aâˆ’1)(a+6)

The LCD of the denominator is (a+6)(aâˆ’9)(aâˆ’1) where aâ‰ âˆ’6â€‰,â€‰aâ‰ 9â€‰orâ€‰aâ‰ 1

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