Results 1 to 5 of 5

Thread: Remainder Theorem

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    66

    Exclamation Remainder Theorem

    Given that $\displaystyle x^5 + ax^3 + bx^2 - 3 = (x^2 - 1)Q(x) - x - 2$, where $\displaystyle Q(x)$ is a polynomial.
    State the degree of $\displaystyle Q(x)$ and find the value of $\displaystyle a$ and of $\displaystyle b$.

    (I am able to deduce the degree i.e. 3
    I am also able to find the value of a ,i.e -2 and the value of b, i.e. 1.)


    However, the following question linked to the above is what I am not able to solve:

    Find also the remainder when $\displaystyle Q(x)$ is divided by $\displaystyle x + 2$.

    (The answer given is -5.)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by Ilsa View Post
    Given that $\displaystyle x^5 + ax^3 + bx^2 - 3 = (x^2 - 1)Q(x) - x - 2$, where $\displaystyle Q(x)$ is a polynomial.
    State the degree of $\displaystyle Q(x)$ and find the value of $\displaystyle a$ and of $\displaystyle b$.

    (I am able to deduce the degree i.e. 3
    I am also able to find the value of a ,i.e -2 and the value of b, i.e. 1.)


    However, the following question linked to the above is what I am not able to solve:

    Find also the remainder when $\displaystyle Q(x)$ is divided by $\displaystyle x + 2$.

    (The answer given is -5.)
    Hint: Given a polynomial P(x), when you divide it by x - r, the remainder is equal to P(r).

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    66
    How do we get the answer -5 then?
    If the remainder is Q(-2), how do I further calculate the remainder?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Sambit's Avatar
    Joined
    Oct 2010
    Posts
    355
    Let's proceed step by step.

    (i) Since $\displaystyle Q(x)$ is of degree 3, assume $\displaystyle Q(x)=Ax^3+Bx^2+Cx+D$.

    (ii) Put this in RHS. Then by comparing the coefficients of RHS to the coefficients of LHS find the values of $\displaystyle A,B,C$ and $\displaystyle D$.

    (iii) After getting the values of $\displaystyle A,B,C,D$ you know the exact expression of $\displaystyle Q(x)$. So you can easily put $\displaystyle -2$ in place of $\displaystyle x$ and calculate $\displaystyle Q(-2)$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,775
    Thanks
    3028
    That, actually finding Q, seems to me a very difficult way to do it. Topsquark's suggestion is best. You are told that
    $\displaystyle x^5 + ax^3 + bx^2 - 3 = (x^2 - 1)Q(x) - x - 2$ and you say you have determined that a= -2 and b= 1 so you have $\displaystyle x^5 - 2x^3 + x^2 - 3 = (x^2 - 1)Q(x) - x - 2$.

    Setting x= -2, $\displaystyle (-2)^5- 2(-2)^3+ (-2)^2- 3= ((-2)^2- 1)Q(-2)- (-2)- 2$

    $\displaystyle -32+ 16+ 4- 3= -15= 3Q(-2)$
    so it is easy to find Q(-2) which is, as Topsquark said, the remainder when Q(x) is divided by x+ 2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. remainder theorem
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Nov 14th 2011, 07:16 AM
  2. remainder theorem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Sep 17th 2009, 05:26 AM
  3. remainder theorem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Mar 22nd 2008, 08:32 AM
  4. Remainder theorem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Sep 18th 2007, 01:57 PM
  5. Factor Theorem and Remainder Theorem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Sep 8th 2007, 10:50 AM

Search Tags


/mathhelpforum @mathhelpforum