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Question:
What is the value of \(\frac{\mathrm{ (x^2+x+3) } }{\mathrm{(x+2)}}\) when x = -1?
A
3
explanation
\(\frac{\mathrm{ (x^2+x+3) } }{\mathrm{(x+2)}} =((-1)^2+(-1)+3) \div ((-1)+2)\) when x=-1
\(\frac{\mathrm{(1-1+3)} }{\mathrm{(-1+2)}} = \frac{\mathrm{3} }{\mathrm{1}}=3 \)
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