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The polynomial ax^4+bx^3+1=0 is divisible by x^2-2x+1. What are the numbers a and b?

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oliviao | Student | (Level 1) eNoter

Posted June 6, 2011 at 3:05 PM via web

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The polynomial ax^4+bx^3+1=0 is divisible by x^2-2x+1. What are the numbers a and b?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted June 6, 2011 at 3:13 PM (Answer #1)

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Since the polynomial P(x)=ax^4 + bx^3 + 1 is divisible by x^2 - 2x + 1, then P(x1)=0 and P(x2)=0, where x1 and x2 are the roots of x^2 - 2x + 1=0.

Since x^2 - 2x + 1 is a perfect square, then x1 = x2 = 1.

We notice that the root x = 1 has the order of multiplicity of 2.

Therefore, the derivative of the polynomial P'(x), at the value x = 1, is cancelling.

P'(1) = 0

We'll re-write the conditions:

P(1) = 0 <=> a + b + 1 = 0 => a = -b - 1 (1)

P'(1) = 0 <=> 4a + 3b = 0 (2)

We'll replace a by (1):

4(-b - 1) + 3b = 0

We'll remove the brackets:

-4b - 4 + 3b = 0

We'll combine like terms:

-b - 4 = 0

b = -4

a = -b - 1 => a = 3

The values of the coefficients a and b are: a =3 and b = -4.

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