Ruiz Ray
03/27/2023 · Senior High School
\( \frac{2x+1}{x-3}.\frac{3x}{p(x)}=\frac{6x^{2}+3x}{x^{2}-2x-3} \)
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Step-by-step Solution
To solve this problem, we need to find the polynomial \( p(x) \) such that the given equation holds true for all values of \( x \) except possibly at \( x = 3 \) (since \( x - 3 \) is in the denominator of the left-hand side and cannot be zero).
The equation is:
\[ \frac{2x+1}{x-3} \cdot \frac{3x}{p(x)} = \frac{6x^2 + 3x}{x^2 - 2x - 3} \]
First, let's simplify the right-hand side by factoring the numerator and the denominator:
The numerator \( 6x^2 + 3x \) can be factored as \( 3x(2x + 1) \).
The denominator \( x^2 - 2x - 3 \) can be factored as \( (x - 3)(x + 1) \).
So the right-hand side becomes:
\[ \frac{3x(2x + 1)}{(x - 3)(x + 1)} \]
Now, we can equate the numerators and the denominators of the left-hand side and the right-hand side, respectively:
For the numerators:
\[ (2x + 1) \cdot 3x = 3x(2x + 1) \]
For the denominators:
\[ (x - 3) \cdot p(x) = (x - 3)(x + 1) \]
Since the numerators are already equal, we can focus on the denominators. We have:
\[ (x - 3) \cdot p(x) = (x - 3)(x + 1) \]
To find \( p(x) \), we can divide both sides by \( (x - 3) \), assuming \( x \neq 3 \) to avoid division by zero:
\[ p(x) = x + 1 \]
Therefore, the polynomial \( p(x) \) that satisfies the given equation is \( p(x) = x + 1 \).
Quick Answer
The polynomial \( p(x) \) is \( x + 1 \).
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