Polynomials question (1 Viewer)

[byakuya kuchiki]

Let P(x) = x^(n+1) - (n+1)x + n, where n is a positive integer.

show that P(x) has a double zero at x = 1.

(2008 4U HSC, Q5 b) )

 

[buchanan]

This is a 2 mark question.

P(1)=1-(n+1)+n=0 (This gets you the first mark)

P'(x)=(n+1)(xⁿ-1) ∴ P'(1)=0 (This gets you the second mark)

∴ x=1 is a double zero.

For 1 < k < n+2, P^(k)(x)=(n+1)n(n-1)...(n-k+2)x^n-k+1 &there4; P^(k)(1)=(n+1)n(n-1)...(n-k+2)&ne;0 (Note: this step is not required to get full marks, but nevertheless does show it isn't a triple, quadruple, quintuple zero, etc.)

 

[Drongoski]

Let P(x) = x^(n+1) - (n+1)x + n, where n is a positive integer.

show that P(x) has a double zero at x = 1.

(2008 4U HSC, Q5 b) )

P(x) = xⁿ⁺¹ - (n+1)x + n

P'(x) = (n+1) xⁿ - (n+1)

.: P(1) = 1 - (n+1) + n = 0

& P'(1) = (n+1) -(n+1) = 0

i.e. 1 is a double (at least) root

To show not a triple root, show P''(1) =/= 0

 

[byakuya kuchiki]

thanks alot to the both of you.