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Example 1
If $\dfrac{(x-5)(3x+k)}{x+3}=3(x-5)$, then $k=$?
- $-9$
- $0$
- $7$
- $8$
- $9$
Solution
- 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 $A=B \cdot C$
- Hence if $\dfrac{(x-5)(3x+k)}{x+3}=3(x-5)$ then
$(x-5)(3x+k) = 3(x-5)(x+3)$
- We now simplify the above expression by cancelling the term $x-5$ from bothe sides
$\colorcancel{red}{(x-5)}(3x+k) = 3\colorcancel{red}{(x-5)}(x+3)$
- Which gives
$3x+k = 3(x+3)$
- Expand term on the right
$3x+k = 3x+9$
- For two polynomials to be equal, all their corresponding coeficients must be equal. Hence
$k=9$
Answer E
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