Multiplying & Dividing Rational Expressions Practice

Multiplying & Dividing Rational Expressions

Interactive practice with step-by-step solutions

Key Principles for Multiplying Rational Expressions

Factor all numerators and denominators completely.
Divide out (cancel) common factors.
Multiply numerators together and multiply denominators together.

Multiplication Problems (1-28)

Question 1:

$\frac{5y^2}{3} \cdot \frac{9x}{10y}$

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Step 1: Write the problem

$\frac{5y^2}{3} \cdot \frac{9x}{10y}$

Step 2: Factor terms if needed

The terms are already factored.

Step 3: Cancel Common factors in Red

$\frac{5y^2}{3} \cdot \frac{9x}{10y} = \frac{5y \cdot y}{3} \cdot \frac{9x}{10 \cdot y}$

We can cancel one

$y$ from numerator and denominator:

$\frac{5y \cdot \color{red}{y}}{3} \cdot \frac{9x}{10 \cdot \color{red}{y}} = \frac{5y}{3} \cdot \frac{9x}{10}$

Step 4: Multiply remaining terms

$\frac{5y}{3} \cdot \frac{9x}{10} = \frac{5y \cdot 9x}{3 \cdot 10} = \frac{45xy}{30} = \frac{3xy}{2}$

Final Answer:

$\frac{3xy}{2}$

Question 2:

$\frac{2x^3y}{3z^4} \cdot \frac{6xz^5}{10y^5}$

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Step 1: Write the problem

$\frac{2x^3y}{3z^4} \cdot \frac{6xz^5}{10y^5}$

Step 2: Factor terms

$\frac{2x^3y}{3z^4} \cdot \frac{6xz^5}{10y^5} = \frac{2x^3y}{3z^4} \cdot \frac{6xz^5}{2 \cdot 5 \cdot y^5}$

Step 3: Cancel Common factors in Red

We can cancel a factor of 2:

$\frac{\color{red}{2}x^3y}{3z^4} \cdot \frac{6xz^5}{\color{red}{2} \cdot 5 \cdot y^5} = \frac{x^3y}{3z^4} \cdot \frac{6xz^5}{5 \cdot y^5}$

No more common factors to cancel.

Step 4: Multiply remaining terms

$\frac{x^3y}{3z^4} \cdot \frac{6xz^5}{5 \cdot y^5} = \frac{x^3y \cdot 6xz^5}{3z^4 \cdot 5 \cdot y^5} = \frac{6x^4yz^5}{15z^4y^5} = \frac{6x^4z^5}{15z^4y^4} = \frac{6x^4z}{15y^4}$

Simplifying further:

$\frac{6x^4z}{15y^4} = \frac{2x^4z}{5y^4}$

Final Answer:

$\frac{2x^4z}{5y^4}$

Question 3:

$\frac{9y^2}{8} \cdot \frac{32x}{27y}$

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Step 1: Write the problem

$\frac{9y^2}{8} \cdot \frac{32x}{27y}$

Step 2: Factor terms

$\frac{9y^2}{8} \cdot \frac{32x}{27y} = \frac{3^2 \cdot y^2}{2^3} \cdot \frac{2^5 \cdot x}{3^3 \cdot y}$

Step 3: Cancel Common factors in Red

We can cancel

$y$ once from numerator and denominator:

$\frac{3^2 \cdot y \cdot \color{red}{y}}{2^3} \cdot \frac{2^5 \cdot x}{3^3 \cdot \color{red}{y}} = \frac{3^2 \cdot y}{2^3} \cdot \frac{2^5 \cdot x}{3^3}$

Step 4: Multiply remaining terms

$\frac{3^2 \cdot y}{2^3} \cdot \frac{2^5 \cdot x}{3^3} = \frac{3^2 \cdot 2^5 \cdot y \cdot x}{2^3 \cdot 3^3} = \frac{9 \cdot 32 \cdot xy}{8 \cdot 27} = \frac{288xy}{216} = \frac{4xy}{3}$

Final Answer:

$\frac{4xy}{3}$

Question 4:

$\frac{2x^2y}{3z^3} \cdot \frac{12xz^4}{6y^3}$

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Step 1: Write the problem

$\frac{2x^2y}{3z^3} \cdot \frac{12xz^4}{6y^3}$

Step 2: Factor terms

$\frac{2x^2y}{3z^3} \cdot \frac{12xz^4}{6y^3} = \frac{2x^2y}{3z^3} \cdot \frac{2 \cdot 6 \cdot xz^4}{6y^3} = \frac{2x^2y}{3z^3} \cdot \frac{2 \cdot 6 \cdot xz^4}{6y^3}$

Step 3: Cancel Common factors in Red

We can cancel 6 in the second fraction:

$\frac{2x^2y}{3z^3} \cdot \frac{2 \cdot \color{red}{6} \cdot xz^4}{\color{red}{6}y^3} = \frac{2x^2y}{3z^3} \cdot \frac{2xz^4}{y^3}$

Step 4: Multiply remaining terms

$\frac{2x^2y}{3z^3} \cdot \frac{2xz^4}{y^3} = \frac{2x^2y \cdot 2xz^4}{3z^3 \cdot y^3} = \frac{4x^3yz^4}{3z^3y^3} = \frac{4x^3z^4}{3z^3y^2} = \frac{4x^3z}{3y^2}$

Final Answer:

$\frac{4x^3z}{3y^2}$

Question 5:

$\frac{3y^2}{5} \cdot \frac{10x}{15y}$

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Step 1: Write the problem

$\frac{3y^2}{5} \cdot \frac{10x}{15y}$

Step 2: Factor terms

$\frac{3y^2}{5} \cdot \frac{10x}{15y} = \frac{3y^2}{5} \cdot \frac{2 \cdot 5 \cdot x}{3 \cdot 5 \cdot y}$

Step 3: Cancel Common factors in Red

We can cancel 5 and 3 and one y:

$\frac{\color{red}{3}y \cdot \color{red}{y}}{\color{red}{5}} \cdot \frac{2 \cdot \color{red}{5} \cdot x}{\color{red}{3} \cdot 5 \cdot \color{red}{y}} = \frac{y}{1} \cdot \frac{2x}{5}$

Step 4: Multiply remaining terms

$\frac{y}{1} \cdot \frac{2x}{5} = \frac{y \cdot 2x}{1 \cdot 5} = \frac{2xy}{5}$

Final Answer:

$\frac{2xy}{5}$

Question 6:

$\frac{4x^2y}{2z^2} \cdot \frac{6xz^3}{20y^4}$

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Step 1: Write the problem

$\frac{4x^2y}{2z^2} \cdot \frac{6xz^3}{20y^4}$

Step 2: Factor terms

$\frac{4x^2y}{2z^2} \cdot \frac{6xz^3}{20y^4} = \frac{2^2 \cdot x^2y}{2 \cdot z^2} \cdot \frac{6xz^3}{2^2 \cdot 5 \cdot y^4}$

Step 3: Cancel Common factors in Red

We can cancel 2 from the first fraction and 2² from the second:

$\frac{\color{red}{2^2} \cdot x^2y}{\color{red}{2} \cdot z^2} \cdot \frac{6xz^3}{\color{red}{2^2} \cdot 5 \cdot y^4} = \frac{2 \cdot x^2y}{z^2} \cdot \frac{6xz^3}{5 \cdot y^4}$

Step 4: Multiply remaining terms

$\frac{2 \cdot x^2y}{z^2} \cdot \frac{6xz^3}{5 \cdot y^4} = \frac{2 \cdot 6 \cdot x^2y \cdot xz^3}{z^2 \cdot 5 \cdot y^4} = \frac{12x^3yz^3}{5z^2y^4} = \frac{12x^3z}{5y^3}$

Final Answer:

$\frac{12x^3z}{5y^3}$

Question 7:

$\frac{x+4}{3x+4y} \cdot \frac{9x^2-16y^2}{2x^2+3x-20}$

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Step 1: Write the problem

$\frac{x+4}{3x+4y} \cdot \frac{9x^2-16y^2}{2x^2+3x-20}$

Step 2: Factor all expressions

$x+4$ is already factored

$3x+4y = 3x+4y$ is already factored

$9x^2-16y^2 = (3x)^2-(4y)^2 = (3x+4y)(3x-4y)$

$2x^2+3x-20 = 2x^2+8x-5x-20 = 2x(x+4)-(5)(x+4) = (x+4)(2x-5)$

Step 3: Cancel Common factors in Red

$\frac{\color{red}{x+4}}{3x+4y} \cdot \frac{(3x+4y)(3x-4y)}{(2x-5)\color{red}{(x+4)}} = \frac{1}{3x+4y} \cdot \frac{(3x+4y)(3x-4y)}{2x-5}$

We can also cancel

$3x+4y$ :

$\frac{1}{\color{red}{3x+4y}} \cdot \frac{\color{red}{(3x+4y)}(3x-4y)}{2x-5} = \frac{3x-4y}{2x-5}$

Final Answer:

$\frac{3x-4y}{2x-5}$

Question 8:

$\frac{a^2-10a+21}{a-7} \cdot \frac{a^2+a-12}{(a-3)^2}$

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Step 1: Write the problem

$\frac{a^2-10a+21}{a-7} \cdot \frac{a^2+a-12}{(a-3)^2}$

Step 2: Factor all expressions

$a^2-10a+21 = (a-7)(a-3)$

$a-7$ is already factored

$a^2+a-12 = (a+4)(a-3)$

$(a-3)^2 = (a-3)(a-3)$

Step 3: Cancel Common factors in Red

$\frac{(a-7)\color{red}{(a-3)}}{\color{red}{a-7}} \cdot \frac{(a+4)\color{red}{(a-3)}}{\color{red}{(a-3)}(a-3)} = \frac{a-3}{1} \cdot \frac{a+4}{a-3} = \frac{\color{red}{a-3} \cdot (a+4)}{\color{red}{a-3}} = a+4$

Final Answer:

$a+4$

Question 9:

$\frac{x+1}{3x+y} \cdot \frac{9x^2-y^2}{2x^2+3x+1}$

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Step 1: Write the problem

$\frac{x+1}{3x+y} \cdot \frac{9x^2-y^2}{2x^2+3x+1}$

Step 2: Factor all expressions

$x+1$ is already factored

$3x+y$ is already factored

$9x^2-y^2 = (3x)^2-y^2 = (3x+y)(3x-y)$

$2x^2+3x+1 = 2x^2+2x+x+1 = 2x(x+1)+(x+1) = (x+1)(2x+1)$

Step 3: Cancel Common factors in Red

$\frac{\color{red}{x+1}}{3x+y} \cdot \frac{(3x+y)(3x-y)}{(2x+1)\color{red}{(x+1)}} = \frac{1}{3x+y} \cdot \frac{(3x+y)(3x-y)}{2x+1}$

We can also cancel

$3x+y$ :

$\frac{1}{\color{red}{3x+y}} \cdot \frac{\color{red}{(3x+y)}(3x-y)}{2x+1} = \frac{3x-y}{2x+1}$

Final Answer:

$\frac{3x-y}{2x+1}$

Question 10:

$\frac{a^2-6a+9}{a-3} \cdot \frac{a^2+3a-18}{(a-3)^2}$

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Step 1: Write the problem

$\frac{a^2-6a+9}{a-3} \cdot \frac{a^2+3a-18}{(a-3)^2}$

Step 2: Factor all expressions

$a^2-6a+9 = (a-3)^2$

$a-3$ is already factored

$a^2+3a-18 = (a+6)(a-3)$

$(a-3)^2 = (a-3)(a-3)$

Step 3: Cancel Common factors in Red

$\frac{(a-3)\color{red}{(a-3)}}{\color{red}{a-3}} \cdot \frac{(a+6)\color{red}{(a-3)}}{\color{red}{(a-3)}(a-3)} = \frac{a-3}{1} \cdot \frac{a+6}{a-3} = \frac{\color{red}{a-3} \cdot (a+6)}{\color{red}{a-3}} = a+6$

Final Answer:

$a+6$

Question 11:

$\frac{5x^2}{3t^2} \cdot \frac{9t^8}{25x}$

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Step 1: Write the problem

$\frac{5x^2}{3t^2} \cdot \frac{9t^8}{25x}$

Step 2: Factor terms

$\frac{5x^2}{3t^2} \cdot \frac{9t^8}{25x} = \frac{5x^2}{3t^2} \cdot \frac{3^2 \cdot t^8}{5^2 \cdot x}$

Step 3: Cancel Common factors in Red

$\frac{\color{red}{5}x^2}{3t^2} \cdot \frac{3^2 \cdot t^8}{\color{red}{5^2} \cdot x} = \frac{x^2}{3t^2} \cdot \frac{9 \cdot t^8}{5 \cdot x}$

We can also cancel one power of

$x$ :

$\frac{\color{red}{x} \cdot x}{3t^2} \cdot \frac{9 \cdot t^8}{5 \cdot \color{red}{x}} = \frac{x}{3t^2} \cdot \frac{9t^8}{5}$

Step 4: Multiply remaining terms

$\frac{x}{3t^2} \cdot \frac{9t^8}{5} = \frac{x \cdot 9t^8}{3t^2 \cdot 5} = \frac{9xt^8}{15t^2} = \frac{9xt^6}{15}$

Simplifying further:

$\frac{9xt^6}{15} = \frac{3xt^6}{5}$

Final Answer:

$\frac{3xt^6}{5}$

Question 12:

$\frac{7a^3}{10b^7} \cdot \frac{5b^3}{3a}$

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Step 1: Write the problem

$\frac{7a^3}{10b^7} \cdot \frac{5b^3}{3a}$

Step 2: Factor terms

$\frac{7a^3}{10b^7} \cdot \frac{5b^3}{3a} = \frac{7a^3}{2 \cdot 5 \cdot b^7} \cdot \frac{5b^3}{3a}$

Step 3: Cancel Common factors in Red

$\frac{7a^3}{2 \cdot \color{red}{5} \cdot b^7} \cdot \frac{\color{red}{5}b^3}{3a} = \frac{7a^3}{2 \cdot b^7} \cdot \frac{b^3}{3a}$

We can also cancel one power of

$a$ :

$\frac{7a^2 \cdot \color{red}{a}}{2 \cdot b^7} \cdot \frac{b^3}{3 \cdot \color{red}{a}} = \frac{7a^2}{2 \cdot b^7} \cdot \frac{b^3}{3}$

Step 4: Multiply remaining terms

$\frac{7a^2}{2 \cdot b^7} \cdot \frac{b^3}{3} = \frac{7a^2 \cdot b^3}{2 \cdot b^7 \cdot 3} = \frac{7a^2b^3}{6b^7} = \frac{7a^2}{6b^4}$

Final Answer:

$\frac{7a^2}{6b^4}$

Question 13:

$\frac{3x-6}{5x} \cdot \frac{x^3}{5x-10}$

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Step 1: Write the problem

$\frac{3x-6}{5x} \cdot \frac{x^3}{5x-10}$

Step 2: Factor all expressions

$3x-6 = 3(x-2)$

$5x = 5x$ is already factored

$x^3 = x^3$ is already factored

$5x-10 = 5(x-2)$

Step 3: Cancel Common factors in Red

$\frac{3\color{red}{(x-2)}}{5x} \cdot \frac{x^3}{5\color{red}{(x-2)}} = \frac{3}{5x} \cdot \frac{x^3}{5} = \frac{3 \cdot x^3}{5x \cdot 5} = \frac{3x^3}{25x} = \frac{3x^2}{25}$

Final Answer:

$\frac{3x^2}{25}$

Question 14:

$\frac{5t^3}{4t-8} \cdot \frac{6t-12}{10t}$

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Step 1: Write the problem

$\frac{5t^3}{4t-8} \cdot \frac{6t-12}{10t}$

Step 2: Factor all expressions

$5t^3 = 5t^3$ is already factored

$4t-8 = 4(t-2)$

$6t-12 = 6(t-2)$

$10t = 10t$ is already factored

Step 3: Cancel Common factors in Red

$\frac{5t^3}{4\color{red}{(t-2)}} \cdot \frac{6\color{red}{(t-2)}}{10t} = \frac{5t^3}{4} \cdot \frac{6}{10t} = \frac{5t^3}{4} \cdot \frac{6}{10t}$

We can also simplify

$\frac{6}{10} = \frac{3}{5}$ :

$\frac{5t^3}{4} \cdot \frac{3}{5t} = \frac{5t^3 \cdot 3}{4 \cdot 5t} = \frac{15t^3}{20t} = \frac{15t^2}{20} = \frac{3t^2}{4}$

Final Answer:

$\frac{3t^2}{4}$

Question 15:

$\frac{y^2-16}{2y+6} \cdot \frac{y+3}{y-4}$

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Step 1: Write the problem

$\frac{y^2-16}{2y+6} \cdot \frac{y+3}{y-4}$

Step 2: Factor all expressions

$y^2-16 = (y+4)(y-4)$

$2y+6 = 2(y+3)$

$y+3$ is already factored

$y-4$ is already factored

Step 3: Cancel Common factors in Red

$\frac{(y+4)\color{red}{(y-4)}}{2\color{red}{(y+3)}} \cdot \frac{\color{red}{(y+3)}}{\color{red}{(y-4)}} = \frac{(y+4)}{2}$

Final Answer:

$\frac{y+4}{2}$

Question 16:

$\frac{m^2-n^2}{4m+4n} \cdot \frac{m+n}{m-n}$

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Step 1: Write the problem

$\frac{m^2-n^2}{4m+4n} \cdot \frac{m+n}{m-n}$

Step 2: Factor all expressions

$m^2-n^2 = (m+n)(m-n)$

$4m+4n = 4(m+n)$

$m+n$ is already factored

$m-n$ is already factored

Step 3: Cancel Common factors in Red

$\frac{\color{red}{(m+n)}(m-n)}{4\color{red}{(m+n)}} \cdot \frac{\color{red}{(m+n)}}{\color{red}{(m-n)}} = \frac{m-n}{4} \cdot \frac{1}{1} = \frac{m-n}{4}$

Final Answer:

$\frac{m-n}{4}$

Question 17:

$\frac{x^2-16}{x^2} \cdot \frac{x^2-4x}{x^2-x-12}$

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Step 1: Write the problem

$\frac{x^2-16}{x^2} \cdot \frac{x^2-4x}{x^2-x-12}$

Step 2: Factor all expressions

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

$x^2 = x^2$ is already factored

$x^2-4x = x(x-4)$

$x^2-x-12 = (x+3)(x-4)$

Step 3: Cancel Common factors in Red

$\frac{(x+4)\color{red}{(x-4)}}{x^2} \cdot \frac{x\color{red}{(x-4)}}{(x+3)\color{red}{(x-4)}} = \frac{(x+4)}{x^2} \cdot \frac{x}{(x+3)}$

$\frac{(x+4)}{x^2} \cdot \frac{x}{(x+3)} = \frac{(x+4) \cdot x}{x^2 \cdot (x+3)} = \frac{x(x+4)}{x^2(x+3)} = \frac{x+4}{x(x+3)}$

Final Answer:

$\frac{x+4}{x(x+3)}$

Question 18:

$\frac{y^2+10y+25}{y^2-9} \cdot \frac{y^2+3y}{y+5}$

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Step 1: Write the problem

$\frac{y^2+10y+25}{y^2-9} \cdot \frac{y^2+3y}{y+5}$

Step 2: Factor all expressions

$y^2+10y+25 = (y+5)^2$

$y^2-9 = (y+3)(y-3)$

$y^2+3y = y(y+3)$

$y+5$ is already factored

Step 3: Cancel Common factors in Red

$\frac{\color{red}{(y+5)}(y+5)}{(y+3)(y-3)} \cdot \frac{y\color{red}{(y+3)}}{\color{red}{(y+5)}} = \frac{(y+5)}{(y-3)} \cdot \frac{y}{1} = \frac{y(y+5)}{y-3}$

Final Answer:

$\frac{y(y+5)}{y-3}$

Question 19:

$\frac{6-2t}{t^2+4t+4} \cdot \frac{t^3+2t^2}{t^8-9t^6}$

▼

Step 1: Write the problem

$\frac{6-2t}{t^2+4t+4} \cdot \frac{t^3+2t^2}{t^8-9t^6}$

Step 2: Factor all expressions

$6-2t = 2(3-t)$

$t^2+4t+4 = (t+2)^2$

$t^3+2t^2 = t^2(t+2)$

$t^8-9t^6 = t^6(t^2-9) = t^6(t-3)(t+3)$

Step 3: Cancel Common factors in Red

$\frac{2(3-t)}{(t+2)^2} \cdot \frac{t^2\color{red}{(t+2)}}{t^6(t-3)(t+3)}$

To handle (3-t), we can rewrite it as -(t-3):

$\frac{2(-(t-3))}{(t+2)^2} \cdot \frac{t^2\color{red}{(t+2)}}{t^6\color{red}{(t-3)}(t+3)} = \frac{-2(t-3)}{(t+2)(t+2)} \cdot \frac{t^2}{t^6(t+3)}$

Canceling the common factor (t-3):

$\frac{-2\color{red}{(t-3)}}{(t+2)^2} \cdot \frac{t^2}{t^6\color{red}{(t-3)}(t+3)} = \frac{-2}{(t+2)^2} \cdot \frac{t^2}{t^6(t+3)}$

Step 4: Multiply remaining terms

$\frac{-2}{(t+2)^2} \cdot \frac{t^2}{t^6(t+3)} = \frac{-2 \cdot t^2}{(t+2)^2 \cdot t^6(t+3)} = \frac{-2t^2}{t^6(t+2)^2(t+3)} = \frac{-2}{t^4(t+2)^2(t+3)}$

Final Answer:

$\frac{-2}{t^4(t+2)^2(t+3)}$

Question 20:

$\frac{x^2-6x+9}{12-4x} \cdot \frac{x^6-9x^4}{x^3-3x^2}$

▼

Step 1: Write the problem

$\frac{x^2-6x+9}{12-4x} \cdot \frac{x^6-9x^4}{x^3-3x^2}$

Step 2: Factor all expressions

$x^2-6x+9 = (x-3)^2$

$12-4x = 4(3-x) = -4(x-3)$

$x^6-9x^4 = x^4(x^2-9) = x^4(x+3)(x-3)$

$x^3-3x^2 = x^2(x-3)$

Step 3: Cancel Common factors in Red

$\frac{(x-3)^2}{-4(x-3)} \cdot \frac{x^4(x+3)\color{red}{(x-3)}}{x^2\color{red}{(x-3)}} = \frac{(x-3)}{-4} \cdot \frac{x^4(x+3)}{x^2}$

$\frac{(x-3)}{-4} \cdot \frac{x^4(x+3)}{x^2} = \frac{(x-3) \cdot x^4(x+3)}{-4 \cdot x^2} = \frac{-x^2(x-3)(x+3)}{4}$

Step 4: Simplify further

$\frac{-x^2(x-3)(x+3)}{4} = \frac{-x^2(x^2-9)}{4}$

Final Answer:

$\frac{-x^2(x^2-9)}{4}$

Key Principles for Dividing Rational Expressions

Invert the divisor (the second fraction) and multiply.
Factor all numerators and denominators completely.
Divide out (cancel) common factors.
Multiply numerators together and multiply denominators together.

Question 21:

$\frac{x^2-2x-35}{2x^3-3x^2} \cdot \frac{4x^3-9x}{7x-49}$

▼

Step 1: Write the problem

$\frac{x^2-2x-35}{2x^3-3x^2} \cdot \frac{4x^3-9x}{7x-49}$

Step 2: Factor all expressions

$x^2-2x-35 = (x+5)(x-7)$

$2x^3-3x^2 = x^2(2x-3)$

$4x^3-9x = x(4x^2-9) = x(2x+3)(2x-3)$

$7x-49 = 7(x-7)$

Step 3: Cancel Common factors in Red

$\frac{(x+5)\color{red}{(x-7)}}{x^2(2x-3)} \cdot \frac{x(2x+3)\color{red}{(2x-3)}}{7\color{red}{(x-7)}} = \frac{(x+5)}{x^2\color{red}{(2x-3)}} \cdot \frac{x(2x+3)}{\color{red}{7}} = \frac{(x+5)(2x+3)}{7x}$

Final Answer:

$\frac{(x+5)(2x+3)}{7x}$

Question 22:

$\frac{y^2-10y+9}{y^2-1} \cdot \frac{y+4}{y^2-5y-36}$

▼

Step 1: Write the problem

$\frac{y^2-10y+9}{y^2-1} \cdot \frac{y+4}{y^2-5y-36}$

Step 2: Factor all expressions

$y^2-10y+9 = (y-9)(y-1)$

$y^2-1 = (y+1)(y-1)$

$y+4$ is already factored

$y^2-5y-36 = (y+4)(y-9)$

Step 3: Cancel Common factors in Red

$\frac{(y-9)\color{red}{(y-1)}}{(y+1)\color{red}{(y-1)}} \cdot \frac{\color{red}{(y+4)}}{\color{red}{(y+4)}(y-9)} = \frac{\color{red}{(y-9)}}{(y+1)} \cdot \frac{1}{\color{red}{(y-9)}} = \frac{1}{(y+1)}$

Final Answer:

$\frac{1}{y+1}$

Question 23:

$\frac{c^3+8}{c^5-4c^3} \cdot \frac{c^6-4c^5+4c^4}{c^2-2c+4}$

▼

Step 1: Write the problem

$\frac{c^3+8}{c^5-4c^3} \cdot \frac{c^6-4c^5+4c^4}{c^2-2c+4}$

Step 2: Factor all expressions

$c^3+8 = c^3+2^3 = (c+2)(c^2-2c+4)$

$c^5-4c^3 = c^3(c^2-4) = c^3(c+2)(c-2)$

$c^6-4c^5+4c^4 = c^4(c^2-4c+4) = c^4(c-2)^2$

$c^2-2c+4$ is already factored

Step 3: Cancel Common factors in Red

$\frac{(c+2)\color{red}{(c^2-2c+4)}}{c^3(c+2)(c-2)} \cdot \frac{c^4(c-2)^2}{\color{red}{(c^2-2c+4)}} = \frac{\color{red}{(c+2)}}{c^3\color{red}{(c+2)}(c-2)} \cdot \frac{c^4(c-2)^2}{1}$

$\frac{1}{c^3(c-2)} \cdot \frac{c^4(c-2)^2}{1} = \frac{c^4(c-2)^2}{c^3(c-2)} = \frac{c^4(c-2)}{c^3} = \frac{c \cdot (c-2)}{1} = c(c-2)$

Final Answer:

$c(c-2)$

Question 24:

$\frac{x^3-27}{x^4-9x^2} \cdot \frac{x^5-6x^4+9x^3}{x^2+3x+9}$

▼

Step 1: Write the problem

$\frac{x^3-27}{x^4-9x^2} \cdot \frac{x^5-6x^4+9x^3}{x^2+3x+9}$

Step 2: Factor all expressions

$x^3-27 = (x-3)(x^2+3x+9)$

$x^4-9x^2 = x^2(x^2-9) = x^2(x+3)(x-3)$

$x^5-6x^4+9x^3 = x^3(x^2-6x+9) = x^3(x-3)^2$

$x^2+3x+9$ is already factored

Step 3: Cancel Common factors in Red

$\frac{(x-3)\color{red}{(x^2+3x+9)}}{x^2(x+3)(x-3)} \cdot \frac{x^3(x-3)^2}{\color{red}{(x^2+3x+9)}} = \frac{\color{red}{(x-3)}}{x^2(x+3)\color{red}{(x-3)}} \cdot \frac{x^3(x-3)^2}{1}$

$\frac{1}{x^2(x+3)} \cdot \frac{x^3(x-3)^2}{1} = \frac{x^3(x-3)^2}{x^2(x+3)} = \frac{x(x-3)^2}{(x+3)}$

Final Answer:

$\frac{x(x-3)^2}{(x+3)}$

Question 25:

$\frac{a^3-b^3}{3a^2+9ab+6b^2} \cdot \frac{a^2+2ab+b^2}{a^2-b^2}$

▼

Step 1: Write the problem

$\frac{a^3-b^3}{3a^2+9ab+6b^2} \cdot \frac{a^2+2ab+b^2}{a^2-b^2}$

Step 2: Factor all expressions

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

$3a^2+9ab+6b^2 = 3(a^2+3ab+2b^2) = 3(a+2b)(a+b)$

$a^2+2ab+b^2 = (a+b)^2$

$a^2-b^2 = (a+b)(a-b)$

Step 3: Cancel Common factors in Red

$\frac{(a-b)(a^2+ab+b^2)}{3(a+2b)\color{red}{(a+b)}} \cdot \frac{\color{red}{(a+b)}^2}{\color{red}{(a+b)}(a-b)} = \frac{(a^2+ab+b^2)}{3(a+2b)} \cdot \frac{(a+b)}{1}$

$\frac{(a^2+ab+b^2)(a+b)}{3(a+2b)} = \frac{(a^2+ab+b^2)(a+b)}{3(a+2b)}$

Final Answer:

$\frac{(a^2+ab+b^2)(a+b)}{3(a+2b)}$

Question 26:

$\frac{x^3+y^3}{x^2+2xy-3y^2} \cdot \frac{x^2-y^2}{3x^2+6xy+3y^2}$

▼

Step 1: Write the problem

$\frac{x^3+y^3}{x^2+2xy-3y^2} \cdot \frac{x^2-y^2}{3x^2+6xy+3y^2}$

Step 2: Factor all expressions

$x^3+y^3 = (x+y)(x^2-xy+y^2)$

$x^2+2xy-3y^2 = (x+3y)(x-y)$

$x^2-y^2 = (x+y)(x-y)$

$3x^2+6xy+3y^2 = 3(x^2+2xy+y^2) = 3(x+y)^2$

Step 3: Cancel Common factors in Red

$\frac{\color{red}{(x+y)}(x^2-xy+y^2)}{(x+3y)\color{red}{(x-y)}} \cdot \frac{\color{red}{(x+y)}\color{red}{(x-y)}}{3\color{red}{(x+y)}^2} = \frac{(x^2-xy+y^2)}{(x+3y)} \cdot \frac{1}{3(x+y)}$

$\frac{(x^2-xy+y^2)}{(x+3y) \cdot 3(x+y)} = \frac{(x^2-xy+y^2)}{3(x+3y)(x+y)}$

Final Answer:

$\frac{x^2-xy+y^2}{3(x+3y)(x+y)}$

Question 27:

$\frac{4x^2-9y^2}{8x^3-27y^3} \cdot \frac{4x^2+6xy+9y^2}{4x^2+12xy+9y^2}$

▼

Step 1: Write the problem

$\frac{4x^2-9y^2}{8x^3-27y^3} \cdot \frac{4x^2+6xy+9y^2}{4x^2+12xy+9y^2}$

Step 2: Factor all expressions

$4x^2-9y^2 = (2x+3y)(2x-3y)$

$8x^3-27y^3 = (2x-3y)(4x^2+6xy+9y^2)$

$4x^2+6xy+9y^2 = (2x+3y)^2$

$4x^2+12xy+9y^2 = (2x+3y)^2$

Step 3: Cancel Common factors in Red

We notice that

$4x^2+12xy+9y^2$ factors differently from what was given. Let's double-check:

$4x^2+12xy+9y^2 = 4x^2+12xy+9y^2 = (2x)^2+2(2x)(3y)+(3y)^2$

This is

$(2x+3y)^2$ . Let's continue with the correct factorization:

$\frac{\color{red}{(2x+3y)}(2x-3y)}{\color{red}{(2x-3y)}(4x^2+6xy+9y^2)} \cdot \frac{\color{red}{(2x+3y)}^2}{\color{red}{(2x+3y)}^2} = \frac{(2x+3y)}{(4x^2+6xy+9y^2)} = \frac{1}{(2x+3y)}$

Final Answer:

$\frac{1}{2x+3y}$

Question 28:

$\frac{3x^2-3y^2}{27x^3-8y^3} \cdot \frac{6x^2+5xy-6y^2}{6x^2+12xy+6y^2}$

▼

Step 1: Write the problem

$\frac{3x^2-3y^2}{27x^3-8y^3} \cdot \frac{6x^2+5xy-6y^2}{6x^2+12xy+6y^2}$

Step 2: Factor all expressions

$3x^2-3y^2 = 3(x^2-y^2) = 3(x+y)(x-y)$

$27x^3-8y^3 = (3x-2y)(9x^2+6xy+4y^2)$

$6x^2+5xy-6y^2 = (3x+6y)(2x-y)$

$6x^2+12xy+6y^2 = 6(x^2+2xy+y^2) = 6(x+y)^2$

Step 3: Cancel Common factors in Red

$\frac{3\color{red}{(x+y)}(x-y)}{(3x-2y)(9x^2+6xy+4y^2)} \cdot \frac{(3x+6y)(2x-y)}{6\color{red}{(x+y)}^2}$

Let's factor further to find common terms:

$(2x-y)$ and

$(x-y)$ have a common factor of

$(x-y)$ if we write

$(2x-y) = 2(x-\frac{y}{2})$ .

Similarly,

$(3x+6y) = 3(x+2y)$ and

$(3x-2y)$ don't have obvious common factors.

Let's continue with our cancellation:

$\frac{3(x-y)}{(3x-2y)(9x^2+6xy+4y^2)} \cdot \frac{(3x+6y)(2x-y)}{6(x+y)}$

Step 4: Simplify further

Without more common factors to cancel, we have:

$\frac{3(x-y)(3x+6y)(2x-y)}{6(x+y)(3x-2y)(9x^2+6xy+4y^2)}$

$\frac{3(x-y)(3x+6y)(2x-y)}{6(x+y)(3x-2y)(9x^2+6xy+4y^2)} = \frac{(x-y)(3x+6y)(2x-y)}{2(x+y)(3x-2y)(9x^2+6xy+4y^2)}$

$\frac{(x-y)(3(x+2y))(2x-y)}{2(x+y)(3x-2y)(9x^2+6xy+4y^2)}$

Final Answer:

$\frac{(x-y)(3(x+2y))(2x-y)}{2(x+y)(3x-2y)(9x^2+6xy+4y^2)}$

Division Problems (29-56)

Question 29:

$28p^2q^4 \div \frac{4pq^4}{5r}$

▼

Step 1: Write the problem

$28p^2q^4 \div \frac{4pq^4}{5r}$

Step 2: Convert first term to fraction if needed

$28p^2q^4 = \frac{28p^2q^4}{1}$

Step 3: Invert the divisor and multiply

$\frac{28p^2q^4}{1} \div \frac{4pq^4}{5r} = \frac{28p^2q^4}{1} \cdot \frac{5r}{4pq^4}$

Step 4: Factor terms

$\frac{28p^2q^4}{1} \cdot \frac{5r}{4pq^4} = \frac{4 \cdot 7 \cdot p^2q^4}{1} \cdot \frac{5r}{4 \cdot pq^4}$

Step 5: Cancel Common factors in Red

$\frac{\color{red}{4} \cdot 7 \cdot p^2\color{red}{q^4}}{1} \cdot \frac{5r}{\color{red}{4} \cdot p\color{red}{q^4}} = \frac{7 \cdot p^2}{1} \cdot \frac{5r}{p} = \frac{7 \cdot 5 \cdot p^2 \cdot r}{p} = \frac{35pr}{1}$

Final Answer:

$35pr$

Question 30:

$\frac{r^3s}{t} \div \frac{rs^3}{t^3}$

▼

Step 1: Write the problem

$\frac{r^3s}{t} \div \frac{rs^3}{t^3}$

Step 2: Invert the divisor and multiply

$\frac{r^3s}{t} \div \frac{rs^3}{t^3} = \frac{r^3s}{t} \cdot \frac{t^3}{rs^3}$

Step 3: Cancel Common factors in Red

$\frac{r^3\color{red}{s}}{\color{red}{t}} \cdot \frac{\color{red}{t^3}}{\color{red}{r}s^3} = \frac{r^2 \cdot t^2}{s^2}$

Final Answer:

$\frac{r^2t^2}{s^2}$

Question 31:

$24e^2d^4 \div \frac{3cd^4}{5f}$

▼

Step 1: Write the problem

$24e^2d^4 \div \frac{3cd^4}{5f}$

Step 2: Convert first term to fraction if needed

$24e^2d^4 = \frac{24e^2d^4}{1}$

Step 3: Invert the divisor and multiply

$\frac{24e^2d^4}{1} \div \frac{3cd^4}{5f} = \frac{24e^2d^4}{1} \cdot \frac{5f}{3cd^4}$

Step 4: Factor terms

$\frac{24e^2d^4}{1} \cdot \frac{5f}{3cd^4} = \frac{24e^2\color{red}{d^4}}{1} \cdot \frac{5f}{3c\color{red}{d^4}}$

Step 5: Cancel Common factors in Red

$\frac{24e^2\color{red}{d^4}}{1} \cdot \frac{5f}{3c\color{red}{d^4}} = \frac{24e^2 \cdot 5f}{3c} = \frac{120e^2f}{3c} = \frac{40e^2f}{c}$

Final Answer:

$\frac{40e^2f}{c}$

Question 32:

$\frac{u^5x}{y} \div \frac{ux^2}{y^3}$

▼

Step 1: Write the problem

$\frac{u^5x}{y} \div \frac{ux^2}{y^3}$

Step 2: Invert the divisor and multiply

$\frac{u^5x}{y} \div \frac{ux^2}{y^3} = \frac{u^5x}{y} \cdot \frac{y^3}{ux^2}$

Step 3: Cancel Common factors in Red

$\frac{u^5\color{red}{x}}{\color{red}{y}} \cdot \frac{\color{red}{y^3}}{\color{red}{u}\color{red}{x^2}} = \frac{u^4 \cdot y^2}{x}$

Final Answer:

$\frac{u^4y^2}{x}$

Question 33:

$\frac{m^5n}{p} \div \frac{mn^3}{p^4}$

▼

Step 1: Write the problem

$\frac{m^5n}{p} \div \frac{mn^3}{p^4}$

Step 2: Invert the divisor and multiply

$\frac{m^5n}{p} \div \frac{mn^3}{p^4} = \frac{m^5n}{p} \cdot \frac{p^4}{mn^3}$

Step 3: Cancel Common factors in Red

$\frac{m^5\color{red}{n}}{\color{red}{p}} \cdot \frac{\color{red}{p^4}}{\color{red}{m}\color{red}{n^3}} = \frac{m^4 \cdot p^3}{n^2}$

Final Answer:

$\frac{m^4p^3}{n^2}$

Question 34:

$\frac{3x^2+4x+1}{3x^2-5x-2} \div \frac{x^2-2x-3}{-5x^2+25x-30}$

▼

Step 1: Write the problem

$\frac{3x^2+4x+1}{3x^2-5x-2} \div \frac{x^2-2x-3}{-5x^2+25x-30}$

Step 2: Factor all expressions

$3x^2+4x+1 = (3x+1)(x+1)$

$3x^2-5x-2 = (3x+1)(x-2)$

$x^2-2x-3 = (x+1)(x-3)$

$-5x^2+25x-30 = -5(x^2-5x+6) = -5(x-2)(x-3)$

Step 3: Invert the divisor and multiply

$\frac{(3x+1)(x+1)}{(3x+1)(x-2)} \div \frac{(x+1)(x-3)}{-5(x-2)(x-3)} = \frac{(3x+1)(x+1)}{(3x+1)(x-2)} \cdot \frac{-5(x-2)(x-3)}{(x+1)(x-3)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(3x+1)}\color{red}{(x+1)}}{\color{red}{(3x+1)}\color{red}{(x-2)}} \cdot \frac{-5\color{red}{(x-2)}\color{red}{(x-3)}}{\color{red}{(x+1)}\color{red}{(x-3)}} = 1 \cdot (-5) = -5$

Final Answer:

$-5$

Question 35:

$\frac{2x^2+5x+3}{2x^2+7x+6} \div \frac{x^2+6x+5}{-5x^2-35x-50}$

▼

Step 1: Write the problem

$\frac{2x^2+5x+3}{2x^2+7x+6} \div \frac{x^2+6x+5}{-5x^2-35x-50}$

Step 2: Factor all expressions

$2x^2+5x+3 = (2x+3)(x+1)$

$2x^2+7x+6 = (2x+3)(x+2)$

$x^2+6x+5 = (x+5)(x+1)$

$-5x^2-35x-50 = -5(x^2+7x+10) = -5(x+5)(x+2)$

Step 3: Invert the divisor and multiply

$\frac{(2x+3)(x+1)}{(2x+3)(x+2)} \div \frac{(x+5)(x+1)}{-5(x+5)(x+2)} = \frac{(2x+3)(x+1)}{(2x+3)(x+2)} \cdot \frac{-5(x+5)(x+2)}{(x+5)(x+1)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(2x+3)}\color{red}{(x+1)}}{\color{red}{(2x+3)}\color{red}{(x+2)}} \cdot \frac{-5\color{red}{(x+5)}\color{red}{(x+2)}}{\color{red}{(x+5)}\color{red}{(x+1)}} = \frac{1}{1} \cdot (-5) = -5$

Final Answer:

$-5$

Question 36:

$\frac{30}{y^2+4y-12} \div \frac{6y}{y-2}$

▼

Step 1: Write the problem

$\frac{30}{y^2+4y-12} \div \frac{6y}{y-2}$

Step 2: Factor all expressions

$30 = 30$ is already factored

$y^2+4y-12 = (y+6)(y-2)$

$6y = 6y$ is already factored

$y-2 = y-2$ is already factored

Step 3: Invert the divisor and multiply

$\frac{30}{(y+6)(y-2)} \div \frac{6y}{y-2} = \frac{30}{(y+6)(y-2)} \cdot \frac{y-2}{6y}$

Step 4: Cancel Common factors in Red

$\frac{30}{(y+6)\color{red}{(y-2)}} \cdot \frac{\color{red}{(y-2)}}{6y} = \frac{30}{(y+6)} \cdot \frac{1}{6y} = \frac{30}{6(y+6)y} = \frac{5}{(y+6)y}$

Final Answer:

$\frac{5}{y(y+6)}$

Question 37:

$\frac{15}{y^2+2y-8} \div \frac{5y}{y-2}$

▼

Step 1: Write the problem

$\frac{15}{y^2+2y-8} \div \frac{5y}{y-2}$

Step 2: Factor all expressions

$15 = 15$ is already factored

$y^2+2y-8 = (y+4)(y-2)$

$5y = 5y$ is already factored

$y-2 = y-2$ is already factored

Step 3: Invert the divisor and multiply

$\frac{15}{(y+4)(y-2)} \div \frac{5y}{y-2} = \frac{15}{(y+4)(y-2)} \cdot \frac{y-2}{5y}$

Step 4: Cancel Common factors in Red

$\frac{15}{(y+4)\color{red}{(y-2)}} \cdot \frac{\color{red}{(y-2)}}{5y} = \frac{15}{(y+4)} \cdot \frac{1}{5y} = \frac{15}{5(y+4)y} = \frac{3}{(y+4)y}$

Final Answer:

$\frac{3}{y(y+4)}$

Question 38:

$\frac{x^2+3x-28}{x^2+4x+4} \div \frac{x^2-49}{x^2-5x-14}$

▼

Step 1: Write the problem

$\frac{x^2+3x-28}{x^2+4x+4} \div \frac{x^2-49}{x^2-5x-14}$

Step 2: Factor all expressions

$x^2+3x-28 = (x+7)(x-4)$

$x^2+4x+4 = (x+2)^2$

$x^2-49 = (x+7)(x-7)$

$x^2-5x-14 = (x+2)(x-7)$

Step 3: Invert the divisor and multiply

$\frac{(x+7)(x-4)}{(x+2)^2} \div \frac{(x+7)(x-7)}{(x+2)(x-7)} = \frac{(x+7)(x-4)}{(x+2)^2} \cdot \frac{(x+2)(x-7)}{(x+7)(x-7)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(x+7)}(x-4)}{(x+2)\color{red}{(x+2)}} \cdot \frac{\color{red}{(x+2)}\color{red}{(x-7)}}{\color{red}{(x+7)}\color{red}{(x-7)}} = \frac{(x-4)}{(x+2)} \cdot 1 = \frac{x-4}{x+2}$

Final Answer:

$\frac{x-4}{x+2}$

Question 39:

$\frac{16a^7}{3b^5} \div \frac{8a^3}{6b}$

▼

Step 1: Write the problem

$\frac{16a^7}{3b^5} \div \frac{8a^3}{6b}$

Step 2: Invert the divisor and multiply

$\frac{16a^7}{3b^5} \div \frac{8a^3}{6b} = \frac{16a^7}{3b^5} \cdot \frac{6b}{8a^3}$

Step 3: Factor terms

$\frac{16a^7}{3b^5} \cdot \frac{6b}{8a^3} = \frac{2^4 \cdot a^7}{3b^5} \cdot \frac{6b}{2^3 \cdot a^3}$

Step 4: Cancel Common factors in Red

$\frac{2^4 \cdot a^7}{3b^5} \cdot \frac{6b}{2^3 \cdot a^3} = \frac{2 \cdot 2^3 \cdot a^7}{3b^5} \cdot \frac{6b}{2^3 \cdot a^3} = \frac{2 \cdot \color{red}{2^3} \cdot a^4 \cdot a^3}{3b^5} \cdot \frac{6b}{\color{red}{2^3} \cdot a^3}$

$\frac{2 \cdot a^4 \cdot \color{red}{a^3}}{3b^5} \cdot \frac{6b}{\color{red}{a^3}} = \frac{2 \cdot a^4 \cdot 6b}{3b^5} = \frac{12a^4b}{3b^5} = \frac{4a^4}{b^4}$

Final Answer:

$\frac{4a^4}{b^4}$

Question 40:

$\frac{9x^5}{8y^2} \div \frac{3x}{16y^9}$

▼

Step 1: Write the problem

$\frac{9x^5}{8y^2} \div \frac{3x}{16y^9}$

Step 2: Invert the divisor and multiply

$\frac{9x^5}{8y^2} \div \frac{3x}{16y^9} = \frac{9x^5}{8y^2} \cdot \frac{16y^9}{3x}$

Step 3: Factor terms

$\frac{9x^5}{8y^2} \cdot \frac{16y^9}{3x} = \frac{3^2 \cdot x^5}{2^3 \cdot y^2} \cdot \frac{2^4 \cdot y^9}{3 \cdot x}$

Step 4: Cancel Common factors in Red

$\frac{3^2 \cdot x^5}{2^3 \cdot y^2} \cdot \frac{2^4 \cdot y^9}{3 \cdot x} = \frac{3 \cdot 3 \cdot x^4 \cdot \color{red}{x}}{2^3 \cdot y^2} \cdot \frac{2^4 \cdot y^9}{\color{red}{3} \cdot \color{red}{x}} = \frac{3 \cdot x^4 \cdot 2^4 \cdot y^9}{2^3 \cdot y^2} = \frac{3 \cdot x^4 \cdot 2 \cdot y^9}{y^2} = \frac{6x^4y^9}{y^2} = 6x^4y^7$

Final Answer:

$6x^4y^7$

Question 41:

$\frac{3y+15}{y^5} \div \frac{y+5}{y^2}$

▼

Step 1: Write the problem

$\frac{3y+15}{y^5} \div \frac{y+5}{y^2}$

Step 2: Factor all expressions

$3y+15 = 3(y+5)$

$y^5 = y^5$ is already factored

$y+5 = y+5$ is already factored

$y^2 = y^2$ is already factored

Step 3: Invert the divisor and multiply

$\frac{3(y+5)}{y^5} \div \frac{y+5}{y^2} = \frac{3(y+5)}{y^5} \cdot \frac{y^2}{y+5}$

Step 4: Cancel Common factors in Red

$\frac{3\color{red}{(y+5)}}{y^5} \cdot \frac{y^2}{\color{red}{y+5}} = \frac{3 \cdot y^2}{y^5} = \frac{3y^2}{y^5} = \frac{3}{y^3}$

Final Answer:

$\frac{3}{y^3}$

Question 42:

$\frac{6x+12}{x^9} \div \frac{x+2}{x^3}$

▼

Step 1: Write the problem

$\frac{6x+12}{x^9} \div \frac{x+2}{x^3}$

Step 2: Factor all expressions

$6x+12 = 6(x+2)$

$x^9 = x^9$ is already factored

$x+2 = x+2$ is already factored

$x^3 = x^3$ is already factored

Step 3: Invert the divisor and multiply

$\frac{6(x+2)}{x^9} \div \frac{x+2}{x^3} = \frac{6(x+2)}{x^9} \cdot \frac{x^3}{x+2}$

Step 4: Cancel Common factors in Red

$\frac{6\color{red}{(x+2)}}{x^9} \cdot \frac{x^3}{\color{red}{x+2}} = \frac{6 \cdot x^3}{x^9} = \frac{6x^3}{x^9} = \frac{6}{x^6}$

Final Answer:

$\frac{6}{x^6}$

Question 43:

$\frac{y^2-9}{y^2} \div \frac{y^5+3y^4}{y+2}$

▼

Step 1: Write the problem

$\frac{y^2-9}{y^2} \div \frac{y^5+3y^4}{y+2}$

Step 2: Factor all expressions

$y^2-9 = (y+3)(y-3)$

$y^2 = y^2$ is already factored

$y^5+3y^4 = y^4(y+3)$

$y+2 = y+2$ is already factored

Step 3: Invert the divisor and multiply

$\frac{(y+3)(y-3)}{y^2} \div \frac{y^4(y+3)}{y+2} = \frac{(y+3)(y-3)}{y^2} \cdot \frac{y+2}{y^4(y+3)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(y+3)}(y-3)}{y^2} \cdot \frac{y+2}{y^4\color{red}{(y+3)}} = \frac{(y-3)}{y^2} \cdot \frac{y+2}{y^4} = \frac{(y-3)(y+2)}{y^6}$

Final Answer:

$\frac{(y-3)(y+2)}{y^6}$

Question 44:

$\frac{x^2-4}{x^3} \div \frac{x^5-2x^4}{x+3}$

▼

Step 1: Write the problem

$\frac{x^2-4}{x^3} \div \frac{x^5-2x^4}{x+3}$

Step 2: Factor all expressions

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

$x^3 = x^3$ is already factored

$x^5-2x^4 = x^4(x-2)$

$x+3 = x+3$ is already factored

Step 3: Invert the divisor and multiply

$\frac{(x+2)(x-2)}{x^3} \div \frac{x^4(x-2)}{x+3} = \frac{(x+2)(x-2)}{x^3} \cdot \frac{x+3}{x^4(x-2)}$

Step 4: Cancel Common factors in Red

$\frac{(x+2)\color{red}{(x-2)}}{x^3} \cdot \frac{x+3}{x^4\color{red}{(x-2)}} = \frac{(x+2)}{x^3} \cdot \frac{x+3}{x^4} = \frac{(x+2)(x+3)}{x^7}$

Final Answer:

$\frac{(x+2)(x+3)}{x^7}$

Question 45:

$\frac{4a^2-1}{a^2-4} \div \frac{2a-1}{a-2}$

▼

Step 1: Write the problem

$\frac{4a^2-1}{a^2-4} \div \frac{2a-1}{a-2}$

Step 2: Factor all expressions

$4a^2-1 = (2a+1)(2a-1)$

$a^2-4 = (a+2)(a-2)$

$2a-1 = 2a-1$ is already factored

$a-2 = a-2$ is already factored

Step 3: Invert the divisor and multiply

$\frac{(2a+1)(2a-1)}{(a+2)(a-2)} \div \frac{2a-1}{a-2} = \frac{(2a+1)(2a-1)}{(a+2)(a-2)} \cdot \frac{a-2}{2a-1}$

Step 4: Cancel Common factors in Red

$\frac{(2a+1)\color{red}{(2a-1)}}{(a+2)\color{red}{(a-2)}} \cdot \frac{\color{red}{(a-2)}}{\color{red}{(2a-1)}} = \frac{(2a+1)}{(a+2)}$

Final Answer:

$\frac{2a+1}{a+2}$

Question 46:

$\frac{25x^2-4}{x^2-9} \div \frac{5x-2}{x-3}$

▼

Step 1: Write the problem

$\frac{25x^2-4}{x^2-9} \div \frac{5x-2}{x-3}$

Step 2: Factor all expressions

$25x^2-4 = (5x+2)(5x-2)$

$x^2-9 = (x+3)(x-3)$

$5x-2 = 5x-2$ is already factored

$x-3 = x-3$ is already factored

Step 3: Invert the divisor and multiply

$\frac{(5x+2)(5x-2)}{(x+3)(x-3)} \div \frac{5x-2}{x-3} = \frac{(5x+2)(5x-2)}{(x+3)(x-3)} \cdot \frac{x-3}{5x-2}$

Step 4: Cancel Common factors in Red

$\frac{(5x+2)\color{red}{(5x-2)}}{(x+3)\color{red}{(x-3)}} \cdot \frac{\color{red}{(x-3)}}{\color{red}{(5x-2)}} = \frac{(5x+2)}{(x+3)}$

Final Answer:

$\frac{5x+2}{x+3}$

Question 47:

$\frac{x^2-y^2}{4x+4y} \div \frac{2y-3x}{x^2-2xy+y^2}$

▼

Step 1: Write the problem

$\frac{x^2-y^2}{4x+4y} \div \frac{2y-3x}{x^2-2xy+y^2}$

Step 2: Factor all expressions

$x^2-y^2 = (x+y)(x-y)$

$4x+4y = 4(x+y)$

$2y-3x = -1(3x-2y)$

$x^2-2xy+y^2 = (x-y)^2$

Step 3: Invert the divisor and multiply

$\frac{(x+y)(x-y)}{4(x+y)} \div \frac{-1(3x-2y)}{(x-y)^2} = \frac{(x+y)(x-y)}{4(x+y)} \cdot \frac{(x-y)^2}{-1(3x-2y)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(x+y)}(x-y)}{4\color{red}{(x+y)}} \cdot \frac{(x-y)(x-y)}{-1(3x-2y)} = \frac{(x-y)}{4} \cdot \frac{(x-y)^2}{-1(3x-2y)} = \frac{(x-y)^3}{-4(3x-2y)}$

Final Answer:

$\frac{-(x-y)^3}{4(3x-2y)}$

Question 48:

$\frac{x+y}{z^2-z} \div \frac{x^2-y^2}{z^3-z^2}$

▼

Step 1: Write the problem

$\frac{x+y}{z^2-z} \div \frac{x^2-y^2}{z^3-z^2}$

Step 2: Factor all expressions

$x+y = x+y$ is already factored

$z^2-z = z(z-1)$

$x^2-y^2 = (x+y)(x-y)$

$z^3-z^2 = z^2(z-1)$

Step 3: Invert the divisor and multiply

$\frac{x+y}{z(z-1)} \div \frac{(x+y)(x-y)}{z^2(z-1)} = \frac{x+y}{z(z-1)} \cdot \frac{z^2(z-1)}{(x+y)(x-y)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(x+y)}}{z\color{red}{(z-1)}} \cdot \frac{z^2\color{red}{(z-1)}}{\color{red}{(x+y)}(x-y)} = \frac{z^2}{z(x-y)} = \frac{z}{x-y}$

Final Answer:

$\frac{z}{x-y}$

Question 49:

$\frac{x^2-16}{x^2-10x+25} \div \frac{3x-12}{x-5}$

▼

Step 1: Write the problem

$\frac{x^2-16}{x^2-10x+25} \div \frac{3x-12}{x-5}$

Step 2: Factor all expressions

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

$x^2-10x+25 = (x-5)^2$

$3x-12 = 3(x-4)$

$x-5 = x-5$ is already factored

Step 3: Invert the divisor and multiply

$\frac{(x+4)(x-4)}{(x-5)^2} \div \frac{3(x-4)}{x-5} = \frac{(x+4)(x-4)}{(x-5)^2} \cdot \frac{x-5}{3(x-4)}$

Step 4: Cancel Common factors in Red

$\frac{(x+4)\color{red}{(x-4)}}{(x-5)\color{red}{(x-5)}} \cdot \frac{\color{red}{(x-5)}}{3\color{red}{(x-4)}} = \frac{(x+4)}{(x-5)} \cdot \frac{1}{3} = \frac{x+4}{3(x-5)}$

Final Answer:

$\frac{x+4}{3(x-5)}$

Question 50:

$\frac{y^2-36}{y^2-8y+16} \div \frac{3y-18}{y^2-y-12}$

▼

Step 1: Write the problem

$\frac{y^2-36}{y^2-8y+16} \div \frac{3y-18}{y^2-y-12}$

Step 2: Factor all expressions

$y^2-36 = (y+6)(y-6)$

$y^2-8y+16 = (y-4)^2$

$3y-18 = 3(y-6)$

$y^2-y-12 = (y+3)(y-4)$

Step 3: Invert the divisor and multiply

$\frac{(y+6)(y-6)}{(y-4)^2} \div \frac{3(y-6)}{(y+3)(y-4)} = \frac{(y+6)(y-6)}{(y-4)^2} \cdot \frac{(y+3)(y-4)}{3(y-6)}$

Step 4: Cancel Common factors in Red

$\frac{(y+6)\color{red}{(y-6)}}{(y-4)\color{red}{(y-4)}} \cdot \frac{(y+3)\color{red}{(y-4)}}{3\color{red}{(y-6)}} = \frac{(y+6)}{(y-4)} \cdot \frac{(y+3)}{3} = \frac{(y+6)(y+3)}{3(y-4)}$

Final Answer:

$\frac{(y+6)(y+3)}{3(y-4)}$

Question 51:

$\frac{y^3+3y}{y^2-9} \div \frac{y^2+5y-14}{y^2+4y-21}$

▼

Step 1: Write the problem

$\frac{y^3+3y}{y^2-9} \div \frac{y^2+5y-14}{y^2+4y-21}$

Step 2: Factor all expressions

$y^3+3y = y(y^2+3) = y(y^2+3)$

$y^2-9 = (y+3)(y-3)$

$y^2+5y-14 = (y+7)(y-2)$

$y^2+4y-21 = (y+7)(y-3)$

Step 3: Invert the divisor and multiply

$\frac{y(y^2+3)}{(y+3)(y-3)} \div \frac{(y+7)(y-2)}{(y+7)(y-3)} = \frac{y(y^2+3)}{(y+3)(y-3)} \cdot \frac{(y+7)(y-3)}{(y+7)(y-2)}$

Step 4: Cancel Common factors in Red

$\frac{y(y^2+3)}{(y+3)\color{red}{(y-3)}} \cdot \frac{\color{red}{(y+7)}\color{red}{(y-3)}}{\color{red}{(y+7)}(y-2)} = \frac{y(y^2+3)}{(y+3)(y-2)}$

Final Answer:

$\frac{y(y^2+3)}{(y+3)(y-2)}$

Question 52:

$\frac{a^3+4a}{a^2-16} \div \frac{a^2+8a+15}{a^2+a-20}$

▼

Step 1: Write the problem

$\frac{a^3+4a}{a^2-16} \div \frac{a^2+8a+15}{a^2+a-20}$

Step 2: Factor all expressions

$a^3+4a = a(a^2+4) = a(a^2+4)$

$a^2-16 = (a+4)(a-4)$

$a^2+8a+15 = (a+5)(a+3)$

$a^2+a-20 = (a+5)(a-4)$

Step 3: Invert the divisor and multiply

$\frac{a(a^2+4)}{(a+4)(a-4)} \div \frac{(a+5)(a+3)}{(a+5)(a-4)} = \frac{a(a^2+4)}{(a+4)(a-4)} \cdot \frac{(a+5)(a-4)}{(a+5)(a+3)}$

Step 4: Cancel Common factors in Red

$\frac{a(a^2+4)}{(a+4)\color{red}{(a-4)}} \cdot \frac{\color{red}{(a+5)}\color{red}{(a-4)}}{\color{red}{(a+5)}(a+3)} = \frac{a(a^2+4)}{(a+4)(a+3)}$

Final Answer:

$\frac{a(a^2+4)}{(a+4)(a+3)}$

Question 53:

$\frac{x^3-64}{x^3+64} \div \frac{x^2-16}{x^2-4x+16}$

▼

Step 1: Write the problem

$\frac{x^3-64}{x^3+64} \div \frac{x^2-16}{x^2-4x+16}$

Step 2: Factor all expressions

$x^3-64 = (x-4)(x^2+4x+16)$

$x^3+64$ cannot be factored further

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

$x^2-4x+16 = (x-2)^2+12 = (x-2)^2+12$ cannot be factored further in real numbers

Step 3: Invert the divisor and multiply

$\frac{(x-4)(x^2+4x+16)}{x^3+64} \div \frac{(x+4)(x-4)}{x^2-4x+16} = \frac{(x-4)(x^2+4x+16)}{x^3+64} \cdot \frac{x^2-4x+16}{(x+4)(x-4)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(x-4)}(x^2+4x+16)}{x^3+64} \cdot \frac{x^2-4x+16}{(x+4)\color{red}{(x-4)}} = \frac{(x^2+4x+16)(x^2-4x+16)}{(x^3+64)(x+4)}$

Final Answer:

$\frac{(x^2+4x+16)(x^2-4x+16)}{(x^3+64)(x+4)}$

Question 54:

$\frac{8y^3-27}{64y^3-1} \div \frac{4y^2-9}{16y^2-1}$

▼

Step 1: Write the problem

$\frac{8y^3-27}{64y^3-1} \div \frac{4y^2-9}{16y^2-1}$

Step 2: Factor all expressions

$8y^3-27 = (2y-3)(4y^2+6y+9)$

$64y^3-1 = (4y-1)(16y^2+4y+1)$

$4y^2-9 = (2y+3)(2y-3)$

$16y^2-1 = (4y+1)(4y-1)$

Step 3: Invert the divisor and multiply

$\frac{(2y-3)(4y^2+6y+9)}{(4y-1)(16y^2+4y+1)} \div \frac{(2y+3)(2y-3)}{(4y+1)(4y-1)} = \frac{(2y-3)(4y^2+6y+9)}{(4y-1)(16y^2+4y+1)} \cdot \frac{(4y+1)(4y-1)}{(2y+3)(2y-3)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(2y-3)}(4y^2+6y+9)}{\color{red}{(4y-1)}(16y^2+4y+1)} \cdot \frac{(4y+1)\color{red}{(4y-1)}}{(2y+3)\color{red}{(2y-3)}} = \frac{(4y^2+6y+9)(4y+1)}{(16y^2+4y+1)(2y+3)}$

Final Answer:

$\frac{(4y^2+6y+9)(4y+1)}{(16y^2+4y+1)(2y+3)}$

Question 55:

$\frac{8a^3+b^3}{2a^2+3ab+b^2} \div \frac{8a^2-4ab+2b^2}{2a+b}$

▼

Step 1: Write the problem

$\frac{8a^3+b^3}{2a^2+3ab+b^2} \div \frac{8a^2-4ab+2b^2}{2a+b}$

Step 2: Factor all expressions

$8a^3+b^3 = (2a+b)(4a^2-2ab+b^2)$

$2a^2+3ab+b^2 = (2a+b)(a+b)$

$8a^2-4ab+2b^2 = 2(4a^2-2ab+b^2)$

$2a+b = 2a+b$ is already factored

Step 3: Invert the divisor and multiply

$\frac{(2a+b)(4a^2-2ab+b^2)}{(2a+b)(a+b)} \div \frac{2(4a^2-2ab+b^2)}{2a+b} = \frac{(2a+b)(4a^2-2ab+b^2)}{(2a+b)(a+b)} \cdot \frac{2a+b}{2(4a^2-2ab+b^2)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(2a+b)}\color{red}{(4a^2-2ab+b^2)}}{\color{red}{(2a+b)}(a+b)} \cdot \frac{\color{red}{(2a+b)}}{2\color{red}{(4a^2-2ab+b^2)}} = \frac{1}{(a+b)} \cdot \frac{1}{2} = \frac{1}{2(a+b)}$

Final Answer:

$\frac{1}{2(a+b)}$

Question 56:

$\frac{x^3+8y^3}{x^2+2xy+4y^2} \div \frac{x^3-2x^2y+4xy^2-8y^3}{x-2y}$

▼

Step 1: Write the problem

$\frac{x^3+8y^3}{x^2+2xy+4y^2} \div \frac{x^3-2x^2y+4xy^2-8y^3}{x-2y}$

Step 2: Factor all expressions

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

$x^2+2xy+4y^2 = (x+2y)^2$

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

$x-2y = x-2y$ is already factored

Step 3: Invert the divisor and multiply

$\frac{(x+2y)(x^2-2xy+4y^2)}{(x+2y)^2} \div \frac{(x-2y)(x^2+2xy+4y^2)}{x-2y} = \frac{(x+2y)(x^2-2xy+4y^2)}{(x+2y)^2} \cdot \frac{x-2y}{(x-2y)(x^2+2xy+4y^2)}$

Step 4: Cancel Common factors in Red

$\frac{\color{red}{(x+2y)}(x^2-2xy+4y^2)}{(x+2y)\color{red}{(x+2y)}} \cdot \frac{\color{red}{(x-2y)}}{\color{red}{(x-2y)}(x^2+2xy+4y^2)} = \frac{(x^2-2xy+4y^2)}{(x+2y)} \cdot \frac{1}{(x^2+2xy+4y^2)}$

$\frac{(x^2-2xy+4y^2)}{(x+2y)(x^2+2xy+4y^2)}$

Final Answer:

$\frac{(x^2-2xy+4y^2)}{(x+2y)(x^2+2xy+4y^2)}$

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