To solve the problem, we first multiply the two rational expressions and then simplify the result. Here are the steps:
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Identify the expressions to multiply: \[ \frac{3x^3 + 10x + 3}{9x^2 - 1} \times \frac{3x^2 + 2x - 1}{x^2 + 2 - 3} \]
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Simplify the denominator \(x^2 + 2 - 3\): \[ x^2 + 2 - 3 = x^2 - 1 \] Therefore, the expression becomes: \[ \frac{3x^3 + 10x + 3}{9x^2 - 1} \times \frac{3x^2 + 2x - 1}{x^2 - 1} \]
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Factor the expressions wherever possible:
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For
9x^2 - 1: This is a difference of squares: \[ 9x^2 - 1 = (3x - 1)(3x + 1) \] -
For
x^2 - 1: This is also a difference of squares: \[ x^2 - 1 = (x - 1)(x + 1) \] -
For
3x^2 + 2x - 1: We can factor this quadratic expression. The factors are: \[ 3x^2 + 2x - 1 = (3x - 1)(x + 1) \]
So, we rewrite the multiplication as: \[ \frac{3x^3 + 10x + 3}{(3x - 1)(3x + 1)} \times \frac{(3x - 1)(x + 1)}{(x - 1)(x + 1)} \]
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Now, multiplly the fractions: \[ \frac{(3x^3 + 10x + 3)(3x - 1)(x + 1)}{(3x - 1)(3x + 1)(x - 1)(x + 1)} \]
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Cancel common factors: The \(3x - 1\) and \(x + 1\) factors cancel out: \[ \frac{(3x^3 + 10x + 3)}{(3x + 1)(x - 1)} \]
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Now simplify \(3x^3 + 10x + 3\): We can attempt polynomial division, but for the sake of simplicity, we will leave it as-is for now.
The simplified expression is: \[ \frac{3x^3 + 10x + 3}{(3x + 1)(x - 1)} \]
Now, we find the restrictions on \(x\):
- Set each denominator to zero:
- \(9x^2 - 1 = 0\) gives \(9x^2 = 1 \Rightarrow x = \pm \frac{1}{3}\)
- \(x^2 - 1 = 0\) gives \(x = \pm 1\)
Thus, the restrictions are: \[ x \neq \frac{1}{3}, , x \neq -\frac{1}{3}, , x \neq 1, , x \neq -1 \]
Final Answer:
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The restrictions on \(x\) are: \[ x \neq \frac{1}{3}, , -\frac{1}{3}, , 1, , -1 \]
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There are no additional restrictions on \(x\).
So, in summary:
- \( \frac{1}{3}, -\frac{1}{3}, 1, -1 \)
- Therefore, the answer is:
- \(x \neq \frac{1}{3}, -\frac{1}{3}, 1, -1\)