Example Question 2

Example corresponding to question 2 in algebra_4.

Example 1


If $\dfrac{(x-5)(3x+k)}{x+3}=3(x-5)$, then $k=$?

  1. $-9$
  2. $0$
  3. $7$
  4. $8$
  5. $9$

Solution


  1. One way to find $k$ is to rewrite the given division of polynomials as the product using the following: if $\dfrac{A}{B}=C$, then $A=B \cdot C$

  2. Hence if $\dfrac{(x-5)(3x+k)}{x+3}=3(x-5)$ then

    $(x-5)(3x+k) = 3(x-5)(x+3)$

  3. We now simplify the above expression by cancelling the term $x-5$ from bothe sides

    $\colorcancel{red}{(x-5)}(3x+k) = 3\colorcancel{red}{(x-5)}(x+3)$

  4. Which gives


    $3x+k = 3(x+3)$

  5. Expand term on the right


    $3x+k = 3x+9$



  6. For two polynomials to be equal, all their corresponding coeficients must be equal. Hence


    $k=9$

    Answer E