polynomials question (1 Viewer)

[thoth1]

http://4unitmaths.com/2001moriah.pdf

question 5 (d)

 

[xV1P3R]

For simplicity in typing it out, I'll just replace phi with P.

P(x) = x³Q(x)-----(1a)
P(x) - 1 = (x-1)³q(x) -----(1b)
P(1) - 1 = 0
P(1) = 1 -----------(2)
P(1) = 1³Q(1) = 1 ----subbing (2) into (1a)
Q(1) = 1 ----------(2b)

P(x) - 1 = (x-1)³q(x) = x³Q(x) - 1
sub x = 0
-q(0) = -1
q(0) = 1 -----------(3)

q and Q are both of degree 2 as P is of degree 5
Let q = ax² + bx + c
c = 1 from (3)
Q = Ax² + Bx + C
A + B + C = 1 from (2b)

(x³-3x²+3x-1)q(x) = x³Q(x) - 1
(x³-3x²+3x-1)(ax²+bx+1) = x³(Ax²+Bx+C) - 1
Equating coefficients of x^5
a = A
-of x^4
b -3a = B
-of x³
1 - 3b + 3a = C
3a = 3b
a = b
-of x²
-3 + 3b - a = 0
-of x
3 + b = 0

Solving all these equations should give it to you.

[I haven't checked over them, so they could be wrong.]

 

[thoth1]

For simplicity in typing it out, I'll just replace phi with P.

P(x) = x³Q(x)
P(x) - 1 = (x-1)³q(x) -----(1)
P(1) - 1 = 0
P(1) = 1 -----------(2)
P(1) = 1³Q(1) = 1
Q(1) = 1 ----------(2b)

P(x) - 1 = (x-1)³q(x) = x³Q(x) - 1
sub x = 0
-q(0) = -1
q(0) = 1 -----------(3)

q and Q are both of degree 2 as P is of degree 5
Let q = ax² + bx + c
c = 1 from (3)
Q = Ax² + Bx + C
C = 1 from (2b)

(x³-3x²+3x-1)q(x) = x³Q(x) - 1
(x³-3x²+3x-1)(ax²+bx+1) = x³(Ax²+Bx+1) - 1
Equating coefficients of x^5
a = A
-of x^4
b -3a = B
-of x³
1 - 3b + 3a = 1
3a = 3b
-of x²
-3 + 3b - a = 0
-of x
3 + b = 0

Solving all these equations should give it to you.

[I haven't checked over them, so they could be wrong.]

tnx m8.

 

[xV1P3R]

made a small change. C =/= 1

 

[Drongoski]