Prealgebra & Introductory Algebra (4th Edition)
Prealgebra & Introductory Algebra (4th Edition)
4th Edition
ISBN: 9780321955791
Author: Elayn Martin-Gay
Publisher: PEARSON
Question
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Chapter B, Problem 1E
To determine

To find:

The factorisation of the given binomial.

Expert Solution & Answer
Check Mark

Answer to Problem 1E

Solution:

The given binomial a3+27 can be factorized as (a+3)(a23a+9).

Explanation of Solution

Formula Used:

1) Factor out common factor from both the terms.

2) By the formula of sum of two cubes,

u3+v3=(u+v)(u2uv+v2)

3) By the formula of difference of two cubes,

u3v3=(uv)(u2+uv+v2)

Calculation:

The given binomial is a3+27.

Use the formula of sum of two cubes,

u3+v3=(u+v)(u2uv+v2)

Substitute u = a and v = 3 in u3+v3=(u+v)(u2uv+v2)

a3+33=(a+3)(a2a3+32)a3+27=(a+3)(a23a+9)

Therefore, the given binomial a3+27 can be factorized as (a+3)(a23a+9).

Final statement:

The given binomial a3+27 can be factorized as (a+3)(a23a+9).

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