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Thread: Polynomial Division + Other question help

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    Polynomial Division + Other question help

    1. When the polynomial P(x) is divided by (x+1)(x-2) its remainder is 18x+17. What is the remainder when P(x) is divided by (x-2)?
    Ans = 51

    2. If a>0 and the function f(x) = ax^3 + bx^2 + cx + d is always increasing, what is the condition on a, b and c?
    a)(b^2)-ac<0
    b)(b^2)-2ac<0
    c)(b^2)-3ac<0
    d)(b^2)-4ac<0

    For question 2, how do you go about solving it

    3. What is the number of solutions of the equation ln|(x^2)-1| (Absolute value)

    EDIT:
    3. What is the number of solutions of the equation ln|(x^2)-1| = 0 (Absolute value) I done goofed lol
    Last edited by Danneo; 6 Oct 2018 at 12:26 AM.

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    Senior Member integral95's Avatar
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    Re: Polynomial Division + Other question help

    1. You have to find P(2), assuming you know your remainder theorem.



    Sub x = 2 to get your answer

    2. You have to see that f'(x) >0 and therefore you use the discriminate on f'(x) and make that negative.

    3. That's not even an equation
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    Re: Polynomial Division + Other question help

    Yikes, my bad, fixed it, Q3 should of been equal to 0
    For question 2, i derived it but not sure how to get an answer out of it
    Thx

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    Re: Polynomial Division + Other question help

    Quote Originally Posted by Danneo View Post
    Yikes, my bad, fixed it, Q3 should of been equal to 0
    For question 2, i derived it but not sure how to get an answer out of it
    Thx
    It should be b^2 -3ac<0 just factor out 4

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    Senior Member integral95's Avatar
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    Re: Polynomial Division + Other question help

    Quote Originally Posted by Danneo View Post
    Yikes, my bad, fixed it, Q3 should of been equal to 0
    For question 2, i derived it but not sure how to get an answer out of it
    Thx
    Well the answer is 3

    since



    First equation has 2 solutions, second equation has one.
    “Smart people learn from their mistakes. But the real sharp ones learn from the mistakes of others.”
    ― Brandon Mull

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