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Multiply and Divide Rational Expressions - College Algebra

  1. Which expression represents x2+5x+6x2−x−2÷x2+4x+3x2−4\frac{x^2+5x+6}{x^2-x-2} \div \frac{x^2+4x+3}{x^2-4} in lowest terms?
    Answer: x−2x+1\frac{x-2}{x+1} Convert to multiplication: x2+5x+6x2−x−2⋅x2−4x2+4x+3\frac{x^2+5x+6}{x^2-x-2} \cdot \frac{x^2-4}{x^2+4x+3}. Factor each polynomial: x2+5x+6=(x+2)(x+3)x^2+5x+6 = (x+2)(x+3), x2−x−2=(x−2)(x+1)x^2-x-2 = (x-2)(x+1), x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2), and x2+4x+3=(x+1)(x+3)x^2+4x+3 = (x+1)(x+3). The expression becomes (x+2)(x+3)(x−2)(x+1)⋅(x−2)(x+2)(x+1)(x+3)\frac{(x+2)(x+3)}{(x-2)(x+1)} \cdot \frac{(x-2)(x+2)}{(x+1)(x+3)}. Cancel (x+2)(x+2), (x+3)(x+3), and (x+1)(x+1) to get x−2x+1\frac{x-2}{x+1}.
  2. Which expression is equivalent to x2+3x−10x2−25⋅x2−10x+25x2−4x+4\frac{x^2+3x-10}{x^2-25} \cdot \frac{x^2-10x+25}{x^2-4x+4}?
    Answer: x−5x−2\frac{x-5}{x-2} Factor each polynomial: x2+3x−10=(x+5)(x−2)x^2+3x-10 = (x+5)(x-2), x2−25=(x−5)(x+5)x^2-25 = (x-5)(x+5), x2−10x+25=(x−5)2x^2-10x+25 = (x-5)^2, and x2−4x+4=(x−2)2x^2-4x+4 = (x-2)^2. The expression becomes (x+5)(x−2)(x−5)(x+5)⋅(x−5)2(x−2)2\frac{(x+5)(x-2)}{(x-5)(x+5)} \cdot \frac{(x-5)^2}{(x-2)^2}. Cancel (x+5)(x+5) and one factor of (x−2)(x-2) and one factor of (x−5)(x-5) to get x−5x−2\frac{x-5}{x-2}.
  3. Simplify x3−8x2+2x+4÷x2−4x2+4x+4\frac{x^3-8}{x^2+2x+4} \div \frac{x^2-4}{x^2+4x+4}.
    Answer: x+2x+2 Convert division to multiplication: x3−8x2+2x+4⋅x2+4x+4x2−4\frac{x^3-8}{x^2+2x+4} \cdot \frac{x^2+4x+4}{x^2-4}. Factor using difference of cubes and other patterns: x3−8=(x−2)(x2+2x+4)x^3-8 = (x-2)(x^2+2x+4), x2+4x+4=(x+2)2x^2+4x+4 = (x+2)^2, and x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2). The expression becomes (x−2)(x2+2x+4)x2+2x+4⋅(x+2)2(x−2)(x+2)\frac{(x-2)(x^2+2x+4)}{x^2+2x+4} \cdot \frac{(x+2)^2}{(x-2)(x+2)}. Cancel (x2+2x+4)(x^2+2x+4), (x−2)(x-2), and one factor of (x+2)(x+2) to get x+2x+2.
  4. If f(x)=x2−4x+3x2−9f(x) = \frac{x^2-4x+3}{x^2-9} and g(x)=x2+x−12x2−2x−3g(x) = \frac{x^2+x-12}{x^2-2x-3}, what is f(x)⋅g(x)f(x) \cdot g(x) in simplest form?
    Answer: (x−1)(x+4)(x+3)(x+1)\frac{(x-1)(x+4)}{(x+3)(x+1)} First factor each polynomial: x2−4x+3=(x−1)(x−3)x^2-4x+3 = (x-1)(x-3), x2−9=(x−3)(x+3)x^2-9 = (x-3)(x+3), x2+x−12=(x+4)(x−3)x^2+x-12 = (x+4)(x-3), and x2−2x−3=(x−3)(x+1)x^2-2x-3 = (x-3)(x+1). So f(x)=(x−1)(x−3)(x−3)(x+3)=x−1x+3f(x) = \frac{(x-1)(x-3)}{(x-3)(x+3)} = \frac{x-1}{x+3} and g(x)=(x+4)(x−3)(x−3)(x+1)=x+4x+1g(x) = \frac{(x+4)(x-3)}{(x-3)(x+1)} = \frac{x+4}{x+1}. Therefore, f(x)⋅g(x)=x−1x+3⋅x+4x+1=(x−1)(x+4)(x+3)(x+1)f(x) \cdot g(x) = \frac{x-1}{x+3} \cdot \frac{x+4}{x+1} = \frac{(x-1)(x+4)}{(x+3)(x+1)}.

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