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# In the polynomial function f(x) = x^3 − px^2 + qx − r, where p, q and

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In the polynomial function f(x) = x^3 − px^2 + qx − r, where p, q and  [#permalink]

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16 Jul 2017, 23:17
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55% (hard)

Question Stats:

69% (02:12) correct 31% (01:52) wrong based on 48 sessions

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In the polynomial function $$f(x) = x^3 − px^2 + qx − r$$, where p, q and r are constants and x is a variable, what is the value of 3p − q?

(1) Two of the roots of f(x) = 0 are 1 and 2.
(2) One root of f(x) = 0 is 3.

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Re: In the polynomial function f(x) = x^3 − px2 + qx − r, what is t  [#permalink]

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16 Jul 2017, 23:28
f(x) = x^3 - px^2 + qx - r

(1) Given that x=1 and x=2 satisfy f(x) = 0. Lets put and find out

f(1) = 0 or 1^3 - p*1^2 + q*1 - r = 0 or
1-p+q-r = 0 or p-q+r = 1

f(2) = 0 or 2^3 - p*2^2 + q*2 - r = 0 or
8-4p+2q-r = 0 or 4p-2q+r = 8

We have two equations, if we subtract first from second we get:
4p-2q+r - (p-q+r) = 8-1 or 3p - q = 7. Got the answer - Sufficient.

(2) x=3 satisfies f(x) = 0 or f(3) = 0
or 3^3 - p*3^2 + q*3 - r = 0 or 27 - 9p + 3q - r = 0

or 9p-3q = 27-r.. if we divide both sides by 3 we get
3p-q = 9 - r/3.. Since we don't know the value of r we cannot answer.

Re: In the polynomial function f(x) = x^3 − px2 + qx − r, what is t   [#permalink] 16 Jul 2017, 23:28
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