Pure Mathematics(P3) qp help needed!

[abudaud001]

please post the solution to this problem with explanation.

 

[iKhaled]

please post the solution to this problem with explanation.

mark scheme for this question pls ??!!

 

[abudaud001]

here:

 

[iKhaled]

here:

i am sorry i can't do it till now..wait i will ask for help cuz i also need to know the answer of the first part..i found the remainder and the quotient but i am not able to find a yet :S

 

[abudaud001]

i just got the solution, but it's really complicated, need a better solution.

 

[CaptainDanger]

Do it by comparing method.

As the maximum power is 4. So the other factor will be a quadratic as well.

x^4 + 3x^3 + ax + 3 = (x^2 - x + 1 ) (bx^2 + cx + d)

Open RHS and compare.

 

[iKhaled]

Do it by comparing method.

As the maximum power is 4. So the other factor will be a quadratic as well.

x^4 + 3x^3 + ax + 3 = (x^2 - x + 1 ) (bx^2 + cx + d)

Open RHS and compare.

isn't it suppose to be x^4 + 3x^3 + ax + 3 = (x^2 - x + 1 ) (bx^2 + cx + d) + ( REMAINDER ) ???

 

[iKhaled]

Do it by comparing method.

As the maximum power is 4. So the other factor will be a quadratic as well.

x^4 + 3x^3 + ax + 3 = (x^2 - x + 1 ) (bx^2 + cx + d)

Open RHS and compare.

tried to solve it but ur way is so wrong..correct me if i am wrong :S

 

[CaptainDanger]

(bx^2 + cx + d) is a factor too. So no remainder is left.

Solution :

x^4 + 3x^3 + ax + 3 = (x^2 - x + 1 ) (bx^2 + cx + d)
---------------------....= bx^4 + cx^3 + dx^2 - bx^3 - cx^2 - xd + bx^2 + cx + d
---------------------....= bx^4 + cx^3 - bx^3 + dx^2 - cx^2 + bx^2 - xd + cx + d
---------------------....= bx^4 + (c - b)x^3 + (d - c + b)x^2 + (c - d)x + d (Taking commons)

When you compare the RHS to LHS equation, you get 5 different equations.

1) b = 1

2) c - b = 3

3) d - c + b = 0

4) c - d = a

5 ) d = 3

From 2nd equation you can get c
c - b = 3 (Put b =1)
c - 1 = 3
c = 4

You have to find a that is in equation 4.

c - d = a (Put c = 4 & d = 3)
4 - 3 = a

a = 1 (Answer)

 

[iKhaled]

(bx^2 + cx + d) is a factor too. So no remainder is left.

Solution :

x^4 + 3x^3 + ax + 3 = (x^2 - x + 1 ) (bx^2 + cx + d)
---------------------....= bx^4 + cx^3 + dx^2 - bx^3 - cx^2 - xd + bx^2 + cx + d
---------------------....= bx^4 + cx^3 - bx^3 + dx^2 - cx^2 + bx^2 - xd + cx + d
---------------------....= bx^4 + (c - b)x^3 + (d - c + b)x^2 + (c - d)x + d (Taking commons)

When you compare the RHS to LHS equation, you get 5 different equations.

1) b = 1

2) c - b = 3

3) d - c + b = 0

4) c - d = a

5 ) d = 3

From 2nd equation you can get c
c - b = 3 (Put b =1)
c - 1 = 3
c = 4

You have to find a that is in equation 4.

c - d = a (Put c = 4 & d = 3)
4 - 3 = a

a = 1 (Answer)

can u explain y it is not (x^2 - x + 1 ) (bx^2 + cx + d) + (remainder) ?

 

[CaptainDanger]

Questions says its divisible by (a factor given) so when your are finding the other factor, no remainder is left.