How do you evaluate #\frac { x ^ { 2} - 4} { x + 2} + \frac { x ^ { 3} - 3x + 6} { x - 1} # when x = - 3?

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Answer 1 · 2017-07-14T19:34:20

#-2#

We could substitute #-3# for each #x#, but let's simplify the expression first.

Factorise the first term and cancel like factors.

#(cancel((x+2))(x-2))/cancel((x+2)) + (x^3 -3x+6)/(x-1)#

Now substitute #-3 " for "x#

#x-2 + (x^3 -3x+6)/(x-1)#

#=-3-2 + ((-3)^3 -3(-3)+6)/((-3)-1)#

#=-5 +(-27+9+6)/(-4)#

#=-5 +(-12)/-4#

#=-5 +3#

#=-2#