deskochan
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The marking scheme said from part a to show P(x)-x is a quadratic factor of P(P(x))-x =0 in part (b) (i). I don't figure out the reason.
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mqb2766
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(Original post by deskochan)
The marking scheme said from part a to show P(x)-x is a quadratic factor of P(P(x))-x =0 in part (b) (i). I don't figure out the reason.
That gives you one of the quadratic factors of the quartic if you do that?
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deskochan
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(Original post by mqb2766)
That gives you one of the quadratic factors of the quartic if you do that?
Yes, P(x)-x = x^2+(a-1)x+b and I divide P(P(x))-x (that is x^4+..+b^2+ab+b) by it and get x^2+(a+1)x+(a+b+1). But I don't know why P(x)-x is a quadratic factor from part a and the marking scheme does not explain how to deduce this. Moreover, the marking scheme also mentions by inspection to get another quadratic factor (x^2+(a+1)x+(a+b+1)) from P(x)-x and P(P(x))-x but I only can do it by long division. I cannot do it by inspection.
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mqb2766
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What did you do for part a?
Bi) simply notes that each factor will be a factor of the quartic, so one quadratic factor is the product of the two linear factors which is P(x)-x.
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deskochan
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For part a, P(a)-a =0, P(a) = a, P(P(a)) -a = P(a)-a =0, thus a is also a root of P(P(x)) - x =0. I don't know why P(x)-x is a quadratic factor of P((x))-x from part a.
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mqb2766
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(Original post by deskochan)
For part a, P(a)-a =0, P(a) = a, P(P(a)) -a = P(a)-a =0, thus a is also a root of P(P(x)) - x =0. I don't know why P(x)-x is a quadratic factor of P((x))-x from part a.
Both alpha and beta are roots (factors) of both the quadratic and the quartic (from part a).So P(x)-x must be one of the quadratic factors of P(P(x))-x.

P(x)-x = (x-alpha)(x-beta)
and similarly for
P(P(x))-x = (x-alpha)(x-beta)Q(x) = (P(x)-x)Q(x)
for some quadratic Q().
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deskochan
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(Original post by mqb2766)
Both alpha and beta are roots (factors) of both the quadratic and the quartic (from part a).So P(x)-x must be one of the quadratic factors of P(P(x))-x.
P(x)-x = (x-alpha)(x-beta)
and similarly for
P(P(x))-x = (x-alpha)(x-beta)Q(x) = (P(x)-x)Q(x)
for some quadratic Q().
Ah. I understand now. By the way, how by inspection to find out another quadratic factor from x^2+(a-1)x+b and x^4+2ax^3+(a^2+2b+a)x^2+(2ab+a^2-1)x +(b^2+ab+b)
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mqb2766
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a

(Original post by deskochan)
Ah. I understand now. By the way, how by inspection to find out another quadratic factor from x^2+(a-1)x+b and x^4+2ax^3+(a^2+2b+a)x^2+(2ab+a^2-1)x +(b^2+ab+b)
Just match the x^4, constant and one other coefficient. For Q(x) = dx^2 + ex + f
x^4 gives d = 1
1 gives f = b+a+1
x^3 (probably the easiest) gives 2a = (a-1) + e so e = a+1.

You're just matching coefficients given that the polynomial is divisible / can be factorized.
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deskochan
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(Original post by mqb2766)
a


Just match the x^4, constant and one other coefficient. For Q(x) = dx^2 + ex + f
x^4 gives d = 1
1 gives f = b+a+1
x^3 (probably the easiest) gives 2a = (a-1) + e so e = a+1.

You're just matching coefficients given that the polynomial is divisible / can be factorized.
Thank you and I understand now.
Last edited by deskochan; 1 month ago
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