To simplify the expression \(\frac{-10x^{2}+20x+80}{x+2}\), we can start by factoring the numerator \(-10x^{2}+20x+80\).
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First, factor out \(-10\) from the numerator: \[ -10(x^{2} - 2x - 8) \]
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Next, we need to factor the quadratic \(x^{2} - 2x - 8\). We are looking for two numbers that multiply to \(-8\) and add to \(-2\). These numbers are \(-4\) and \(2\). Therefore, we can factor it as: \[ (x-4)(x+2) \]
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Now we can rewrite the numerator: \[ -10(x-4)(x+2) \]
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Now substitute this back into the fraction: \[ \frac{-10(x-4)(x+2)}{x+2} \]
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The \(x+2\) terms can be cancelled out (assuming \(x \neq -2\)): \[ -10(x-4) \]
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Finally, distribute the \(-10\): \[ -10x + 40 \]
Thus, the simplified form of \(\frac{-10x^{2}+20x+80}{x+2}\) is: \[ -10x + 40 \quad \text{(for } x \neq -2\text{)} \]