if x≠-1 then find the quetient of x⁵+x⁴+x³+x²/x³+x²+x+1
Answers
Step-by-step explanation:
10th
Maths
Polynomials
Division Algorithm for Polynomials
Let P(x) = 1 + x + x^2 + x^...
MATHS
Let P(x)=1+x+x
2
+x
3
+x
4
+x
5
. What is the remainder when P(x
12
) is divided by P(x)?
MEDIUM
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ANSWER
As P(x) is the sum of GP. =
1−x
1−x
6
It has 5 roots , let a
1
,a
2
,a
3
,a
4
,a
5
, and they are the 6th roots of unity except unity.
NowP(x
12
)=1+x
12
+x
24
+x
36
+x
48
+x
60
=P(x).Q(x)+R(x).
Here R(x) is a remainder and a polynomial of maximum degree 4.
Put x=a
1
,a
2
...............,a
5
We get,
R(a
1
)=6, R(a
2
)=6 ,R(a
3
)=6, R(a
4
)=6, R(a
5
)=6
i.e, R(x)−6=0 has 6 roots.
Which contradict that R(x) is maximum of degree 4.
So, it is an identity
Therefore, R(x)=6.
Answer:
As P(x) is the sum of GP. =
1−x
1−x
6
It has 5 roots , let a
1
,a
2
,a
3
,a
4
,a
5
, and they are the 6th roots of unity except unity.
NowP(x
12
)=1+x
12
+x
24
+x
36
+x
48
+x
60
=P(x).Q(x)+R(x).
Here R(x) is a remainder and a polynomial of maximum degree 4.
Put x=a
1
,a
2
...............,a
5
We get,
R(a
1
)=6, R(a
2
)=6 ,R(a
3
)=6, R(a
4
)=6, R(a
5
)=6
i.e, R(x)−6=0 has 6 roots.