using the reminder theorm, find the remainder when x³-6x³ +2x-4 is divided by (1-2x)
Answers
Solution :-
1 - 2x = 0
- 2x = - 1
2x = 1
x = 1/2
p(x) = x³ - 6x² + 2x - 4
p(1/2) = (1/2)³ - 6*(1/2)² + 2*(1/2) - 4
= 1/8 - 6*(1/4) + 1 - 4
= 1/8 - 6/4 + 1 - 4
= 1/8 - 3/2 + 1 - 4
Taking L.C.M. of the denominators and then solving it.
= (1 - 12 + 8 - 32)/8
= - 35/8
Remainder is - 35/8
Now, actual division.
- 6x² - 1
_________________
1 - 2x ) x³ - 6x² + 2x - 4 (
12x³ - 6x²
- +
______________
- 11x³ + 2x - 4
+ 2x - 1
- +
_______________
- 11x³ - 3
_______________
Remainder is - 11x³ - 3
Verification :
⇒ - 11x³ - 3
⇒ - 11*(1/2)³ - 3
⇒ - 11*1/8 - 3
⇒ - 11/8 - 3/1
⇒ Taking L.C.M. of the denominator and then solving it.
⇒ (- 11 - 24)/8
⇒ - 35/8
Verified.
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Answer:
By remainder theorem, find the remainder when p(x) is divided by g(x)
(i) p(x) = x3-2x2-4x-1, g(x)=x + 1
(ii) p(x) = x3 -3x2 + 4x + 50, g(x)= x - 3
(iii) p(x) = x3 – 12x2 + 14x -3, g(x)= 2x - 1 - 1
(iv) ‘p(x) = x3-6x2+2x-4, g(x) = 1 -(3/2)x
step by step explanation
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