Results 1 to 4 of 4

Thread: A cubic question

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Melbourne
    Posts
    275
    Thanks
    4

    A cubic question

    The polynomial $\displaystyle P(x)$ has a remainder of 2 when divided by $\displaystyle x-1$ and a remainder of 3, when divided by $\displaystyle x-2$. The remainder when $\displaystyle P(x)$ is divided by$\displaystyle (x-1)(x-2)$ is $\displaystyle ax + b$, i.e. $\displaystyle P(x)$ can be written as $\displaystyle P(x) = (x-1)(x-2)Q(x)+ax+b$

    Find a and b.

    Well I got this already a= 1 and b=1

    but the second question I don't know what to do...

    Given that$\displaystyle P(x)$ is a cubic polynomial with coefficient of $\displaystyle x^3$ being 1, and −1 is a solution of the equation $\displaystyle P(x) = 0$, find $\displaystyle P(x)$.

    I'm not too sure how to start... If -1 is a solution of the question when $\displaystyle P(x) = 0$ that means $\displaystyle (x+1)$ must be a factor right? I don't know what to do from here...

    All replies are appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Is it the same $\displaystyle \displaystyle P(x)$ as it is in Part 1? If so, $\displaystyle \displaystyle P(1) = 2$ and $\displaystyle \displaystyle P(2) = 3$ by the Remainder Theorem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2009
    From
    Melbourne
    Posts
    275
    Thanks
    4
    Oh wow, I really should think a bit before I start posting.

    Using $\displaystyle P(x) = (x-1)(x-2)Q(x)+ax+b$

    Let $\displaystyle Q(x) = x+1 $ because -1 is a solution of x

    Then multiply out and sub $\displaystyle a=b=1 $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by jgv115 View Post
    The polynomial $\displaystyle P(x)$ has a remainder of 2 when divided by $\displaystyle x-1$ and a remainder of 3, when divided by $\displaystyle x-2$. The remainder when $\displaystyle P(x)$ is divided by$\displaystyle (x-1)(x-2)$ is $\displaystyle ax + b$, i.e. $\displaystyle P(x)$ can be written as $\displaystyle P(x) = (x-1)(x-2)Q(x)+ax+b$

    Find a and b.

    Well I got this already a= 1 and b=1

    but the second question I don't know what to do...

    Given that$\displaystyle P(x)$ is a cubic polynomial with coefficient of $\displaystyle x^3$ being 1, and −1 is a solution of the equation $\displaystyle P(x) = 0$, find $\displaystyle P(x)$.

    I'm not too sure how to start... If -1 is a solution of the question when $\displaystyle P(x) = 0$ that means $\displaystyle (x+1)$ must be a factor right? I don't know what to do from here...

    All replies are appreciated
    As $\displaystyle P(-1)=-6Q(-1)-1+1=-6Q(-1)=0$, this implies that $\displaystyle $$-1$ is a root of $\displaystyle $$Q(x)$..

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cubic Polynomial Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 10th 2011, 04:10 PM
  2. Replies: 3
    Last Post: Mar 24th 2010, 05:41 PM
  3. Replies: 1
    Last Post: Jan 31st 2010, 12:12 PM
  4. Cubic root question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Nov 7th 2009, 07:08 AM
  5. General Question about Cubic Polynomials
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 11th 2009, 05:54 AM

Search Tags


/mathhelpforum @mathhelpforum