Results 1 to 4 of 4

Thread: Remainder theorem

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    66

    Exclamation Remainder theorem

    The expression $\displaystyle f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $\displaystyle x^2 - 2x$. When $\displaystyle f(x)$ is divided by $\displaystyle x - 1$ , the remainder is 8 more than when it is divided by $\displaystyle x + 1$. Factorise $\displaystyle f(x)$ completely.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,119
    Thanks
    718
    Awards
    1
    Quote Originally Posted by Ilsa View Post
    The expression $\displaystyle f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $\displaystyle x^2 - 2x$. When $\displaystyle f(x)$ is divided by $\displaystyle x - 1$ , the remainder is 8 more than when it is divided by $\displaystyle x + 1$. Factorise $\displaystyle f(x)$ completely.
    Well, f(x) is divisible by $\displaystyle x^2 - 2x = x(x - 2)$ so we know f(x) is divisible by x. Thus c = 0 immediately. Consider then
    $\displaystyle f(x) = ax^3 - (a + 3b)x^2 + 2bx$

    From the original statement (divisible by x(x - 2)), we get
    $\displaystyle f(2) = a(2)^3 - (a + 3b)(2)^2 + 2b(2) = 0$

    Call
    $\displaystyle f(1) = a(1)^3 - (a + 3b)(1)^2 + 2b(1) = d$

    Then we know that
    $\displaystyle f(-1) = a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) = d - 8$

    Three equations, three unknowns.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    66
    Quote Originally Posted by topsquark View Post
    Well, f(x) is divisible by $\displaystyle x^2 - 2x = x(x - 2)$ so we know f(x) is divisible by x. Thus c = 0 immediately. Consider then
    $\displaystyle f(x) = ax^3 - (a + 3b)x^2 + 2bx$

    From the original statement (divisible by x(x - 2)), we get
    $\displaystyle f(2) = a(2)^3 - (a + 3b)(2)^2 + 2b(2) = 0$

    Call
    $\displaystyle f(1) = a(1)^3 - (a + 3b)(1)^2 + 2b(1) = d$

    Then we know that
    $\displaystyle f(-1) = a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) = d - 8$

    Three equations, three unknowns.

    -Dan
    If f(x) is divisible by x, how is c=0?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,119
    Thanks
    718
    Awards
    1
    $\displaystyle f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $\displaystyle x^2 - 2x$.
    If x divides f(x) then
    $\displaystyle \displaystyle \frac{f(x)}{x}$ leaves no remainder. Thus

    $\displaystyle \displaystyle \frac{ax^3 - (a + 3b)x^2 + 2bx + c}{x} = ax^2 - (a + 3b)x + 2b + \frac{c}{x}$

    This gives a remainder of c, but f(x) is divisible by x, thus c = 0.

    You can do this quicker by using the fact that if f(x) is divisible by x - r then f(r) = 0. In this case r = 0, so f(0) = 0.

    -Dan
    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