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Simplify: x + 7/x^2 + 4x - 21 A. 1/x - 3; where x doesn't equal 3, 7 B. x - 3: where x doesn't equal 3 C. 1/x - 7; where x doesn't equal 7 D. x - 7 Thanks.
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\[\frac{x + 7}{x^2 + 4x - 21}\] Factor the denominator such that \[x^2+4x-21=(x+a)(x+b)\]where (a + b) = 4 and ab = -21. I'm guessing one of the factors will be (x + 7) so it can cancel with the numerator.
I got C. Is that right? Thanks for your answer by the way.
No, I didn't get C. How did you get that?
\[\frac{x+7}{x^2 +4x-21}\\ \frac{x+7}{(x+7 )( x-3)}\\ \frac{ \cancel{x+7} }{(\cancel{x+7})( x-3)}\\ \ anzwer : \ \frac{1}{x-3} \]
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