(a) f(x)=(x+1)(x−2)(x+3)
→ (x2−2x+x−2)(x+3)
→(x2−x−2)(x+3)
→x3+3x2−x2−3x−2x−6
→x3+2x2−5x−6
For f(-2), put x= -2
f(−2)=(−2)3+2(−2)2−5(−2)−6
→−8+8+10−6 = 4
So f(-2) = 4
(b) f(x)=(2x2−3)(x3+1)
=(2x5+2x2−3x3−3)
=2x5−3x3+2x2−3
For f(-2), put x= -2
f(−2)=2(−2)5−3(−2)3+2(−2)2−3
=−64+24+8−3
=−35
So f(-2) = -35
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