ii) f(x) = 2x3+ 4x +6, g(x) = x +1
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Answer:
(i) Apply factor theorem
x+1=0
So x=−1
2x
3
+x
2
−2x−1
Replace x by −1, we get
2(−1)
3
+(−1)
2
−2(−1)−1=−2+1+2−1=0
Reminder is 0 so that x+1 is a factor of 2x
3
+x
2
−2x−1
(ii) Apply factor theorem
x+2=0
So x=−2
x
3
+3x
2
+3x+1
Replace x by −2, we get
(−2)
3
+3(−2)
2
+3(−2)+1=−8+12−6+1=1
Reminder is 1 so that x+2 is not a factor of x
3
+3x
2
+3x+1
(iii) Apply factor theorem
x−3=0
So x=3
x
3
−4x
2
+x+6
Replace x by 3, we get
(3)
3
−4(3)
2
+(3)−1=27−36+3+6=0
Reminder is 0 so that x−3 is a factor of x
3
−4x
2
+x+6
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