Find the remainder when p(x)= x^3-6x^2+2x-4 and g(x)= 1-3/2 x
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1
Answer:
Here, g(x)=3x−1. To apply Remainder theorem, (3x−1) should be converted to (x−a) form.
∴3x−1=
3
1
(3x−1)=x−
3
1
∴g(x)=(x−
3
1
)
By remainder theorem, r(x)=p(a)=p(
3
1
)
p(x)=x
3
−6x
2
+2x−4⇒p(
3
1
)=(
3
1
)
3
−6(
3
1
)
2
+2(
3
1
)−4
=
27
1
−
9
6
+
3
2
−4=
27
1−18+18−108
=
27
−107
∴ the remainder p(
3
1
)=−
27
107
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