Multiply the fothowing and state the restrictions. \( \frac{t^{2}+19 t+84}{4 t-4} \cdot \frac{2 t-2}{t^{2}+9 t+14} \)
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Beyond the Answer
To multiply the given expressions, we first need to factor them where possible. The first fraction can be factored as: - \( t^2 + 19t + 84 = (t + 12)(t + 7) \) - \( 4t - 4 = 4(t - 1) \) The second fraction can be factored as: - \( 2t - 2 = 2(t - 1) \) - \( t^2 + 9t + 14 = (t + 7)(t + 2) \) Now we can rewrite the multiplication: \[ \frac{(t + 12)(t + 7)}{4(t - 1)} \cdot \frac{2(t - 1)}{(t + 7)(t + 2)} \] Next, we simplify: \[ = \frac{(t + 12)\cancel{(t + 7)}2\cancel{(t - 1)}}{4(t - 1)\cancel{(t + 7)}(t + 2)} \] This simplifies to: \[ = \frac{2(t + 12)}{4(t + 2)} = \frac{t + 12}{2(t + 2)} \] Now for the restrictions. The restrictions on \( t \) come from the values which make the denominators zero in the original expressions: 1. From \( 4t - 4 = 0 \), we get \( t = 1 \). 2. From \( t^2 + 9t + 14 = 0 \), factored as \( (t + 7)(t + 2) = 0 \), we get \( t = -7 \) and \( t = -2 \). Thus, the restrictions are: \( t \neq 1, t \neq -7, t \neq -2 \). The final result is: \[ \frac{t + 12}{2(t + 2)} \quad \text{with restrictions } t \neq 1, t \neq -7, t \neq -2. \]
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