Rational Expression. How to multiply and divide rational expressions

Problem 1

Evaluate $$ \displaystyle \frac{x-4}{x^2 + 6x -7} \cdot \frac{x^2 + 4x - 5}{x^2 + 7x + 10} $$

Step 1

Factor each numerator and denominator.

$$ \begin{align*} \frac{x-4}{x^2 + 6x -7} \cdot \frac{x^2 + 4x - 5}{x^2 + 7x + 10} = \frac{x-4}{(x + 7)(x - 1)} \cdot \frac{(x + 5)(x - 1)}{(x+5)(x+2)} \end{align*} $$

Step 2

Write the product as a single fraction.

$$ \begin{align*} \frac{x-4}{(x + 7)(x - 1)} \cdot \frac{(x + 5)(x - 1)}{(x+5)(x+2)} = \frac{(x-4)(x + 5)(x - 1)}{(x + 7)(x - 1)(x+5)(x+2)} \end{align*} $$

Step 3

Divide out the common factors.

$$ \begin{align*} \frac{(x-4)(x + 5)(x - 1)}{(x + 7)(x - 1)(x+5)(x+2)} & = \frac{(x-4)\cancelred{(x + 5)}\cancelred{(x - 1)}}{(x + 7)\cancelred{(x - 1)}\cancelred{(x+5)}(x+2)}\\[6pt] & = \frac{x-4}{(x + 7)(x+2)};\quad x \neq -5, 1 \end{align*} $$

Note that we keep track of the $$x$$-values that would cause the calculation to be undefined at any step.

Step 4

(Optional) Some instructors will ask you to expand the numerator and denominators when possible.

$$ \begin{align*} \frac{x-4}{(x + 7)(x+2)} = \frac{x-4}{x^2 + 9x + 14} \end{align*} $$

Answer

Note: The answer should include the restrictions we found in Step 3.

$$ \begin{align*} \frac{x-4}{x^2 + 6x -7} \cdot \frac{x^2 + 4x - 5}{x^2 + 7x + 10} = \frac{x-4}{(x + 7)(x+2)};\quad x \neq -5, 1 \end{align*} $$

Problem 2

Evaluate $$ \displaystyle \frac{4x^2+12x}{x^2 - 3x - 18}\cdot \frac{x-6}{x^2 - 8x} $$

Step 1

Factor the numerators and denominators.

$$ \begin{align*} \frac{4x^2+12x}{x^2 - 3x - 18}\cdot \frac{x-6}{x^2 - 8x} = \frac{4x(x + 3)}{(x - 6)(x + 3)}\cdot \frac{x-6}{x(x - 8)} \end{align*} $$

Step 2

Write the product as a single fraction.

$$ \begin{align*} \frac{4x(x + 3)}{(x - 6)(x + 3)}\cdot \frac{x-6}{x(x - 8)} = \frac{4x(x + 3)(x-6)}{x(x - 8)(x - 6)(x + 3)} \end{align*} $$

Step 3

Divide out the common factors.

$$ \begin{align*} \frac{4x(x + 3)(x-6)}{x(x - 8)(x - 6)(x + 3)} & = \frac{4\cancelred{x}\cancelred{(x + 3)}\cancelred{(x-6)}}{\cancelred{x}(x - 8)\cancelred{(x - 6)}\cancelred{(x + 3)}}\\[6pt] & = \frac{4}{x - 8}; \quad x \neq -3, 0, 6 \end{align*} $$

Answer

$$ \begin{align*} \frac{4x^2+12x}{x^2 - 3x - 18}\cdot \frac{x-6}{x^2 - 8x} = \frac{4}{x - 8}; \quad x \neq -3, 0, 6 \end{align*} $$

Problem 3

Evaluate $$ \displaystyle \frac{2x^2 + 5x + 2}{4x^2 +21x + 27}\cdot \frac{16x^2 - 81}{8x^2 - 14x - 9} $$

Step 1

Factor the numerators and the denominators.

$$ \begin{align*} \frac{2x^2 + 5x + 2}{4x^2 + 21x + 27}\cdot \frac{16x^2 - 81}{8x^2 - 14x - 9} = \frac{(2x+ 1)(x + 2)}{(x + 3)(4x + 9)}\cdot \frac{(4x - 9)(4x + 9)}{(2x+1)(4x-9)} \end{align*} $$

Step 2

Rewrite the product as a single fraction.

$$ \begin{align*}{(2x+ 1)(x + 2)}{(x + 3)(4x + 9)}\cdot \frac{(4x - 9)(4x + 9)}{(2x+1)(4x-9)} = \frac{(2x+ 1)(x + 2)(4x - 9)(4x + 9)}{(x + 3)(4x + 9)(2x+1)(4x-9)} \end{align*} $$

Step 3

Divide out the common factors.

$$ \begin{align*} \frac{(2x+ 1)(x + 2)(4x - 9)(4x + 9)}{(x + 3)(4x + 9)(2x+1)(4x-9)} & = \frac{\cancelred{(2x+ 1)}(x + 2)\cancelred{(4x - 9)}\cancelred{(4x + 9)}}{(x + 3)\cancelred{(4x + 9)}\cancelred{(2x+ 1)}\cancelred{(4x - 9)}}\\[6pt] & = \frac{x + 2}{x + 3}; \quad x \neq -\frac 9 4, -\frac 1 2, \frac 9 4 \end{align*} $$

Answer

$$ \begin{align*} \frac{2x^2 + 5x + 2}{4x^2 +21x + 27}\cdot \frac{16x^2 - 81}{8x^2 - 14x - 9} = \frac{x + 2}{x + 3}; \quad x \neq -\frac 9 4, -\frac 1 2, \frac 9 4 \end{align*} $$

Problem 4

Evaluate $$ \displaystyle \frac{7x^2 - 12x - 4}{5x+1}\cdot \frac{5x^2 - 19x - 4}{3x^2 - 18x + 24} $$

Step 1

Factor the numerators and denominators.

$$ \begin{align*} \frac{7x^2 - 12x - 4}{5x+1}\cdot \frac{5x^2 - 19x - 4}{3x^2 - 18x + 24} = \frac{(x - 2)(7x + 2)}{5x+1}\cdot \frac{(x - 4)(5x + 1)}{3(x - 4)(x - 2)} \end{align*} $$

Step 2

Rewrite the product as a single fraction.

$$ \begin{align*} \frac{(x - 2)(7x + 2)}{5x+1}\cdot \frac{(x - 4)(5x + 1)}{3(x - 4)(x - 2)} = \frac{(x - 2)(7x + 2)(x - 4)(5x + 1)}{3(5x+1)(x - 4)(x - 2)} \end{align*} $$

Step 3

Divide out the common factors.

$$ \begin{align*} \frac{(x - 2)(7x + 2)(x - 4)(5x + 1)}{3(5x+1)(x - 4)(x - 2)} & = \frac{\cancelred{(x - 2)}(7x + 2)\cancelred{(x - 4)}\cancelred{(5x + 1)}}{3\cancelred{(5x + 1)}\cancelred{(x - 4)}\cancelred{(x - 2)}}\\[6pt] & = \frac{7x + 2} 3;\quad x \neq -\frac 1 5, 2, 4 \end{align*} $$

Answer

$$ \begin{align*} \frac{7x^2 - 12x - 4}{5x+1}\cdot \frac{5x^2 - 19x - 4}{3x^2 - 18x + 24} = \frac{7x + 2} 3;\quad x \neq -\frac 1 5, 2, 4 \end{align*} $$

Problem 5

Evaluate $$ \displaystyle \frac{x^2 - 5x + 6}{x^2+5x} \div \frac{x^2 - 4}{x+5} $$

Step 1

Factor the numerators and denominators.

$$ \begin{align*} \frac{x^2 - 5x + 6}{x^2+5x} \div \frac{x^2 - 4}{x+5} = \frac{(x - 3)(x - 2)}{x(x+5)} \div \frac{(x + 2)(x - 2)}{x+5} \end{align*} $$

Note that our final answer will need to restrict the values of $$x$$ so that $$x \neq -5, 0$$.

Step 2

Rewrite the division as a product with the reciprocal of the second expression.

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

Note that we now also need to make sure our answer restricts $$x$$ so $$x \neq -2, 2$$.

Step 3

Rewrite the product as a single fraction.

$$ \begin{align*} \frac{(x-3)(x-2)}{x(x+5)} \cdot \frac{x+5}{(x-2)(x+2)} ={(x-3)(x-2)(x+5)}{x(x+5)(x-2)(x+2)} \end{align*} $$

Step 4

Divide out the common factors.

$$ \begin{align*} \frac{(x-3)(x-2)(x+5)}{x(x+5)(x-2)(x+2)} & = \frac{(x-3)\cancelred{(x-2)}\cancelred{(x+5)}}{x\cancelred{(x+5)}\cancelred{(x-2)}(x+2)}\\[6pt] & = \frac{x-3}{x(x+2)};\quad x \neq -5, 2 \end{align*} $$

Note that the other restrictions we identified ($$x \neq -2$$ and $$x\neq 0$$) are still implied in the expression itself.

Step 5

(Optional) Some instructors will require you to expand the denominator.

$$ \begin{align*} \frac{x-3}{x(x+2)} = \frac{x-3}{x^2+2x};\quad x \neq -5, 2 \end{align*} $$

Answer

$$ \begin{align*} \frac{x^2 - 5x + 6}{x^2+5x} \div \frac{x^2 - 4}{x+5} = \frac{x-3}{x(x+2)};\quad x \neq -5, 2 \end{align*} $$

Problem 6

Evaluate $$ \displaystyle \frac{x^2 + 9x + 18}{x^2 -6x + 5}\div \frac{x^2 + 7x +12}{x^2 - 3x - 10} $$

Step 1

Factor the numerators and denominators.

$$ \begin{align*} \frac{x^2 + 9x + 18}{x^2 -6x + 5}\div \frac{x^2 + 7x +12}{x^2 - 3x - 10} = \frac{(x+3)(x+6)}{(x-5)(x-1)}\div \frac{(x+3)(x+4)}{(x-5)(x+2)} \end{align*} $$

Note that our answer will need to restrict the $$x$$-values so that $$x \neq -2, 5, 1$$.

Step 2

Rewrite the division as a product with the reciprocal of the second expression.

$$ \begin{align*} \frac{(x+3)(x+6)}{(x-5)(x-1)}\div \frac{(x+3)(x+4)}{(x-5)(x+2)} = \frac{(x+3)(x+6)}{(x-5)(x-1)}\cdot \frac{(x-5)(x+2)}{(x+3)(x+4)} \end{align*} $$

Note that our answer will also need to restrict $$x$$ so that $$x \neq -4, -3$$.

Step 3

Rewrite the product as a single fraction.

$$ \begin{align*} \frac{(x+3)(x+6)}{(x-5)(x-1)}\cdot \frac{(x-5)(x+2)}{(x+3)(x+4)} = \frac{(x+3)(x+6)(x-5)(x+2)}{(x-5)(x-1)(x+3)(x+4)} \end{align*} $$

Step 4

Divide out the common factors.

$$ \begin{align*} \frac{(x+3)(x+6)(x-5)(x+2)}{(x-5)(x-1)(x+3)(x+4)} & = \frac{\cancelred{(x+3)}(x+6)\cancelred{(x-5)}(x+2)}{\cancelred{(x-5)}(x-1)\cancelred{(x+3)}(x+4)}\\[6pt] & = \frac{(x+6)(x+2)}{(x-1)(x+4)}; \quad x\neq -3,-2, 5 \end{align*} $$

The other restrictions are still implied in the expression itself.

Step 5

(Optional) Some instructors will require you to expand the numerator and denominator.

$$ \begin{align*} \frac{(x+6)(x+2)}{(x-1)(x+4)} = \frac{x^2 + 8x + 12}{x^2 + 3x - 4}; \quad x\neq -3,-2, 5 \end{align*} $$

Answer

$$ \begin{align*} \frac{x^2 + 9x + 18}{x^2 -6x + 5}\div \frac{x^2 + 7x +12}{x^2 - 3x - 10} = \frac{(x+6)(x+2)}{(x-1)(x+4)}; \quad x\neq -3,-2, 5 \end{align*} $$

Problem 7

Evaluate $$ \displaystyle \frac{2x - 5}{x^2 - 4x + 3}\div \frac{2x^2 + 9x + 9}{2x^2 + x - 3} $$

Step 1

Factor the numerators and denominators.

$$ \begin{align*} \frac{2x - 5}{x^2 - 4x + 3}\div \frac{2x^2 + 9x + 9}{2x^2 + x - 3} = \frac{2x - 5}{(x - 3)(x - 1)}\div \frac{(x+3)(2x+3)}{(2x+3)(x-1)} \end{align*} $$

Note that we will need to restrict the $$x$$-values of our answer so $$x \neq -\frac 3 2, 1, 3$$.

Step 2

Rewrite the division as a product with the reciprocal of the second expression.

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

Note that we will also need to restrict our answer so that $$x\neq -3$$.

Step 3

Rewrite the product as a single fraction.

$$ \begin{align*} \frac{2x - 5}{(x - 3)(x - 1)}\cdot \frac{(2x+3)(x-1)}{(x+3)(2x+3)} = \frac{(2x - 5)(2x+3)(x-1)}{(x - 3)(x - 1)(x+3)(2x+3)} \end{align*} $$

Step 4

Divide out the common factors.

$$ \begin{align*} \frac{(2x - 5)(2x+3)(x-1)}{(x - 3)(x - 1)(x+3)(2x+3)} & = \frac{(2x - 5)\cancelred{(2x+3)}\cancelred{(x-1)}}{(x - 3)\cancelred{(x-1)}(x+3)\cancelred{(2x+3)}}\\[6pt] & = \frac{2x - 5}{(x - 3)(x+3)};\quad x\neq -\frac 3 2, 1 \end{align*} $$

The other restrictions we found are still implied in the expression itself.

Step 5

(Optional) Some instructors will require you to expand the denominator.

$$ \begin{align*} \frac{2x - 5}{(x - 3)(x+3)} = \frac{2x - 5}{x^2 - 9};\quad x\neq -\frac 3 2, 1 \end{align*} $$

Answer

$$ \begin{align*} \frac{2x - 5}{x^2 - 4x + 3}\div \frac{2x^2 + 9x + 9}{2x^2 + x - 3} = \frac{2x - 5}{(x - 3)(x+3)};\quad x\neq -\frac 3 2, 1 \end{align*} $$

Problem 8

Evaluate $$ \displaystyle \frac{16x^2 + 16x + 3}{8x^2 +18x - 5}\div\frac{12x^2 + 11x + 2}{12x^2 + 5x - 2} $$

Step 1

Factor the numerators and denominators.

$$ \begin{align*} \frac{16x^2 + 16x + 3}{8x^2 +18x - 5}\div\frac{12x^2 + 11x + 2}{12x^2 + 5x - 2} = \frac{(4x + 3)(4x + 1)}{(4x-1)(2x+5)}\div\frac{(4x+1)(3x+2)}{(3x + 2)(4x - 1)} \end{align*} $$

We will need to restrict our answer so that $$x \neq -\frac 5 2, -\frac 2 3, \frac 1 4$$

Step 2

Rewrite the division as a product with the reciprocal of the second expression.

$$ \begin{align*} \frac{(4x + 3)(4x + 1)}{(4x-1)(2x+5)}\div\frac{(4x+1)(3x+2)}{(3x + 2)(4x - 1)} = \frac{(4x + 3)(4x + 1)}{(4x-1)(2x+5)}\cdot\frac{(3x + 2)(4x - 1)}{(4x+1)(3x+2)} \end{align*} $$

We will also need to restrict our answer so that $$x \neq -\frac 1 4$$.

Step 3

Rewrite the product as a single fraction.

$$ \begin{align*} \frac{(4x + 3)(4x + 1)}{(4x-1)(2x+5)}\cdot\frac{(3x + 2)(4x - 1)}{(4x+1)(3x+2)} = \frac{(4x + 3)(4x + 1)(3x + 2)(4x - 1)}{(4x-1)(2x+5)(4x+1)(3x+2)} \end{align*} $$

Step 4

Divide out the common factors.

$$ \begin{align*} \frac{(4x + 3)(4x + 1)(3x + 2)(4x - 1)}{(4x-1)(2x+5)(4x+1)(3x+2)} & = \frac{(4x + 3)\cancelred{(4x + 1)}\cancelred{(3x + 2)}\cancelred{(4x - 1)}}{\cancelred{(4x - 1)}(2x+5)\cancelred{(4x + 1)}\cancelred{(3x + 2)}}\\[6pt] & = \frac{4x + 3}{2x + 5}; \quad x \neq -\frac 2 3, -\frac 1 4, \frac 1 4 \end{align*} $$

The other restriction is still implied in the expression itself.

Answer

$$ \begin{align*} \frac{16x^2 + 16x + 3}{8x^2 +18x - 5}\div\frac{12x^2 + 11x + 2}{12x^2 + 5x - 2} = \frac{4x + 3}{2x + 5}; \quad x \neq -\frac 2 3, -\frac 1 4, \frac 1 4 \end{align*} $$