Example Question 7

Example corresponding to question 7 in college algebra 2.

Example 1


If $(-x^2+3) \cdot f(x) = -x^3+2x^2+3x-6$, then $f(x)=$

  1. $x+1$
  2. $x-2$
  3. $x+3$
  4. $x-4$
  5. $x$

Solution


  1. One way to answer this question is to divide both sides of the equation by $(-x^2+3)$ to obtain

    $f(x) = \dfrac{-x^3+2x^2+3x-6}{-x^2+3}$

  2. According to the 5 possible answers, we need to simplify the rational expression on the right side. In order to simplify the right term, we need to factor its numerator and one of its factors must be $-x^2+3$ otherwise simplification in not possible. Let us factor the numerator KNOWING that one of the factors must be $-x^2+3$.

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

  3. Substitute $-x^3+2x^2+3x-6$ by $(x-2)(-x^2+3)$ and simplify

    $f(x) = \dfrac{-x^3+2x^2+3x-6}{-x^2+3}=\dfrac{(x-2)\colorcancel{red}{(-x^2+3)}}{\colorcancel{red}{-x^2+3}}=x-2$

    Answer B