Problem 27 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

The result is \(\frac{1}{3}x + 1\).

Step 1: Rewrite the division as multiplication

We can write the division as a multiplication by the reciprocal of the divisor. This gives us \(\frac{4 x^{2}+10}{x-3} \times \frac{x^{2}-9}{6 x^{2}+15}\)

Step 2: Simplify the expressions

Factor out common factors from numerators and denominators. We can factor \(2\) from \((4x^2 + 10)\), \(3\) from \((6x^2 + 15)\), and recognize \(x^2 - 9\) as a difference of squares to give \((x-3)(x+3)\). So our expression becomes \(\frac{2(2 x^{2}+5)}{x-3} \times \frac{(x-3)(x+3)}{3(2 x^{2}+5)}\)

Step 3: Cancel out common factors

We can cancel out common factors in the numerator and the denominator. Both \((2x^2 + 5)\) and \((x - 3)\) are common. This gives us \(\frac{1}{3} \times (x+3)\)

Step 4: Simplify the remaining expression

After multiplying out, we get \(\frac{1}{3}x + 1\) as our final solution