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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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#2

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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.

**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.

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That gives you one of the quadratic factors of the quartic if you do that?

**mqb2766**)That gives you one of the quadratic factors of the quartic if you do that?

**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.

Last edited by deskochan; 1 month ago

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#4

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.

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.

Last edited by mqb2766; 1 month ago

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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.

Last edited by deskochan; 1 month ago

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#6

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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.

**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.

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().

Last edited by mqb2766; 1 month ago

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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().

**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().

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#8

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.

(Original post by

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)

**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)

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.

Last edited by mqb2766; 1 month ago

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(Original post by

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.

**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.

Last edited by deskochan; 1 month ago

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