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Chapter 0: Problem 27

Multiply or divide as indicated. $$\frac{4 x^{2}+10}{x-3} \div \frac{6 x^{2}+15}{x^{2}-9}$$

Short Answer

Expert verified
The result after simplifying the expression becomes \(\frac{2(x+3)}{3}\).

Step by step solution

01

Reciprocal of Second Fraction

This step involves flipping the second fraction or changing the places of its numerator and its denominator. This will convert division problem into multiplication. The reciprocal of \(\frac{6 x^{2}+15}{x^{2}-9}\) is \(\frac{x^{2}-9}{6 x^{2}+15}\).
02

Multiply the given fractions

Now, multiply the first fraction \(\frac{4 x^{2}+10}{x-3}\) by the reciprocal of the second fraction \(\frac{x^{2}-9}{6 x^{2}+15}\). The result is \(\frac{(4 x^{2}+10)(x^{2}-9)}{(x-3)(6 x^{2}+15)}\).
03

Factor the expressions

The expressions can be factored further to simplify the fraction. Factor the expressions: \(4x^{2}+10 = 2(2x^2+5)\), \(x^2-9 = (x-3)(x+3)\), \(x^2-9 = (x-3)(x+3)\), \(6x^2+15 = 3(2x^2+5)\). So the expression becomes \(\frac {2(2x^2+5)(x-3)(x+3)}{(x-3)3(2x^2+5)}\).
04

Simplify the expression

Cancel out the common terms in the numerator and the denominator which are \((2x^2+5)\) and \((x-3)\). The final expression becomes \(\frac{2(x+3)}{3}\).

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