Question

Help please. the question attached below

Answer 1

The rational expressions [(x² + 16xz + 64z²)/(x²z + 8xz²)] ÷ [(x² + 5xz + 24z²)/(x⁴z - 9x²z³)] can be simplified as; x(x - 3z)

How to divide algebraic expressions?

We want to divide the rational expressions;

[(x² + 16xz + 64z²)/(x²z + 8xz²)] ÷ [(x² + 5xz + 24z²)/(x⁴z - 9x²z³)]

Let us break down the expressions one after the other;

(x² + 16xz + 64z²) can be expressed as;

(x + 8z)²

Likewise, (x²z + 8xz²) can be expressed as;

xz(x + 8z)

Thus;

[(x² + 16xz + 64z²)/(x²z + 8xz²)] is;

(x + 8z)²/(xz(x + 8z)) = (x + 8z)/xz

(x² + 5xz + 24z²) can be expressed as;

(x + 3z)(x + 8z)

Similarly;

(x⁴z - 9x²z³) can be expressed as;

x²z(x² - 9z²)

Thus;

[(x² + 5xz + 24z²)/(x⁴z - 9x²z³)] can be expressed as;

(x + 3z)(x + 8z)/(x²z(x² - 9z²))

Then our main expression can now be expressed as;

[(x + 8z)/xz] ÷ (x + 3z)(x + 8z)/(x²z(x² - 9z²))

This can be rewritten as;

[(x + 8z)/xz] * [(x²z(x² - 9z²))]/((x + 3z)(x + 8z))

x(x² - 9z²)/(x + 3z)

This can further be simplified as;

x(x + 3z)(x - 3z)/(x + 3z) = x(x - 3z)

Read more about Algebraic expressions at; brainly.com/question/723406

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