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Perform the indicated operation. Simplify if possible. \frac{ x^2-6x }{x-2}+\frac{ 8 }{ x-2 }
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\[\frac{ x ^{2}-6x }{ x-2 }+\frac{ 8 }{ x-2 }=\frac{ x ^{2}-6x+8 }{ x-2 }\] -6=-4-2 -4*-2=8 1*8=8 factorise the numerators and denominators and cancel the common factor.
?
\[\frac{ x ^{2}-4x-2x+8 }{ x-2}=\frac{ x \left( x-4 \right)-2\left( x-4 \right) }{ x-2 }\] \[=\frac{ \left( x-2 \right)\left( x-4 \right) }{x-2 }\] =x-4
thanks I was confused
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