Solutions to Problem Set 16

Math 101-06

10-16-2017

[Multiplying and dividing fractions]

1. Multiply the fractions, cancelling any common factors, and write your answer using positive powers:

$$\dfrac{8 x^5 y^8}{12 x^3 y^{14}} \cdot \dfrac{25 x^9 y}{10 x^8 y^2}.$$

$$\dfrac{8 x^5 y^8}{12 x^3 y^{14}} \cdot \dfrac{25 x^9 y}{10 x^8 y^2} = \dfrac{200 x^{14} y^9}{120 x^{11} y^{16}} = \dfrac{5}{3} x^3 y^{-7} = \dfrac{5 x^3}{3 y^7}.\quad\halmos$$


2. Multiply the fractions, cancelling any common factors:

$$\dfrac{(x + 1)^2(x - 2)}{(x + 1)^3(x - 3)} \cdot \dfrac{(x - 3)^2}{(x - 2)^3}.$$

$$\dfrac{(x + 1)^2(x - 2)}{(x + 1)^3(x - 3)} \cdot \dfrac{(x - 3)^2}{(x - 2)^3} = \dfrac{x - 3}{(x + 1)(x - 2)^2}.\quad\halmos$$


3. Multiply the fractions, cancelling any common factors:

$$\dfrac{x^2 - x - 2}{x^2 - 4 x + 4} \cdot \dfrac{x^3 + 3 x^2}{x^3 - x}.$$

$$\dfrac{x^2 - x - 2}{x^2 - 4 x + 4} \cdot \dfrac{x^3 + 3 x^2}{x^3 - x} = \dfrac{(x - 2)(x + 1)}{(x - 2)^2} \cdot \dfrac{x^2(x + 3)}{x(x - 1)(x + 1)} = \dfrac{x(x + 3)}{(x - 1)(x - 2)}. \quad\halmos$$


4. Multiply the fractions, cancelling any common factors:

$$\dfrac{x^2 - 7 x + 12}{5 x^2 - 20} \cdot \dfrac{x^2 + 2 x}{10 x^2}.$$

$$\dfrac{x^2 - 7 x + 12}{5 x^2 - 20} \cdot \dfrac{x^2 + 2 x}{10 x^2} = \dfrac{(x - 3)(x - 4)}{5(x - 2)(x + 2)} \cdot \dfrac{x(x + 2)}{10 x^2} = \dfrac{(x - 3)(x - 4)}{50 x(x - 2)}.\quad\halmos$$


5. Divide the fractions and simplify so your answer contains only positive powers:

$$\dfrac{\dfrac{16 x^5 y^9}{24 x^8 y^{-2}}}{\dfrac{4 x^7 y^{10}}{9 x y^8}}.$$

$$\dfrac{\dfrac{16 x^5 y^9}{24 x^8 y^{-2}}}{\dfrac{4 x^7 y^{10}}{9 x y^8}} = \dfrac{16 x^5 y^9}{24 x^8 y^{-2}} \cdot \dfrac{9 x y^8}{4 x^7 y^{10}} = \dfrac{144 x^6 y^{17}}{96 x^{15} y^8} = \dfrac{3}{2} x^{-9} y^9 = \dfrac{3 y^9}{2 x^9}.\quad\halmos$$


6. Divide the fractions, cancelling any common factors:

$$\dfrac{\left(\dfrac{x^2 - 4}{x^4 - 16}\right)} {\left(\dfrac{x^4 + x^2}{5 x^3 + 20 x}\right)}.$$ Note that

$$x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4).$$

So

$$\dfrac{\left(\dfrac{x^2 - 4}{x^4 - 16}\right)} {\left(\dfrac{x^4 + x^2}{5 x^3 + 20 x}\right)} = \dfrac{x^2 - 4}{x^4 - 16} \cdot \dfrac{5 x^3 + 20 x}{x^4 + x^2} = \dfrac{(x - 2)(x + 2)}{(x - 2)(x + 2)(x^2 + 4)} \cdot \dfrac{5 x(x^2 + 4)}{x^2(x^2 + 1)} = \dfrac{5}{x(x^2 + 1)}.\quad\halmos$$


7. Divide the fractions, cancelling any common factors:

$$\dfrac{\left(\dfrac{x^2 - 5 x + 6}{x^2 - 9 x + 14}\right)} {\left(\dfrac{x^4 + 7 x^3}{x^2 - 49}\right)}.$$

$$\dfrac{\left(\dfrac{x^2 - 5 x + 6}{x^2 - 9 x + 14}\right)} {\left(\dfrac{x^4 + 7 x^3}{x^2 - 49}\right)} = \dfrac{x^2 - 5 x + 6}{x^2 - 9 x + 14} \cdot \dfrac{x^2 - 49}{x^4 + 7 x^3} = \dfrac{(x - 2)(x - 3)}{(x - 2)(x - 7)} \cdot \dfrac{(x - 7)(x + 7)}{x^3(x + 7)} = \dfrac{x - 3}{x^3}.\quad\halmos$$


8. Divide the fractions, cancelling any common factors:

$$\dfrac{\left(\dfrac{8 x^2 + 32 x}{3 x^3 + 6 x^2 + 12 x}\right)} {\left(\dfrac{x^2 - 3 x - 28}{x^3 - 8}\right)}.$$ For $x^3 - 8$, use the formula for a difference of cubes ($A^3 - B^3 = (A - B)(A^2 + A B + B^2)$:

$$x^3 - 8 = (x - 2)(x^2 + 2 x + 4).$$

Now flip the bottom fraction over and multiply, then factor and cancel:

$$\dfrac{\left(\dfrac{8 x^2 + 32 x}{3 x^3 + 6 x^2 + 12 x}\right)} {\left(\dfrac{x^2 - 3 x - 28}{x^3 - 8}\right)} = \dfrac{8 x^2 + 32 x}{3 x^3 + 6 x^2 + 12 x} \cdot \dfrac{x^3 - 8}{x^2 - 3 x - 28} =$$

$$\dfrac{8 x(x + 4)}{3 x(x^2 + 2 x + 4)} \cdot \dfrac{(x - 2)(x^2 + 2 x + 4)}{(x - 7)(x + 4)} = \dfrac{8(x - 2)}{3(x - 7)}.\quad\halmos$$


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