Results 1 to 4 of 4
Like Tree5Thanks
  • 2 Post By topsquark
  • 3 Post By HallsofIvy

Math Help - Factoring linear factors.

  1. #1
    Junior Member
    Joined
    Apr 2012
    From
    Seattle, Washington
    Posts
    72

    Factoring linear factors.

    Hi, I'm working through an exam pretest and I came across this problem. Looking through my notes, I can't find examples that show me how to do it. I'm not totally sure what it's asking.

    a) Use the Remainder Theorem to find the remainder when x^3-2x^2+8x-16 is divided by x-2.

    For this I did P(2) = (2)^3 -2(2)^2+8(2)-16

    I got P(2) = 0 but was I supposed to plug in -2?

    This is the part I'm stuck on:

    b) Use your answer in part a to factor x^3-2x^2+8x-16 into linear factors allowing complex numbers. Clearly explain your reasoning.

    Thanks in advance for any help I can get on this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1

    Re: Factoring linear factors.

    Quote Originally Posted by FatimaA View Post
    Hi, I'm working through an exam pretest and I came across this problem. Looking through my notes, I can't find examples that show me how to do it. I'm not totally sure what it's asking.

    a) Use the Remainder Theorem to find the remainder when x^3-2x^2+8x-16 is divided by x-2.

    For this I did P(2) = (2)^3 -2(2)^2+8(2)-16

    I got P(2) = 0 but was I supposed to plug in -2?
    Nope. You got it right. So now you know that x - 2 divides x^3 - 2x^2 + 8x - 16

    Quote Originally Posted by FatimaA View Post
    b) Use your answer in part a to factor x^3-2x^2+8x-16 into linear factors allowing complex numbers. Clearly explain your reasoning.

    Thanks in advance for any help I can get on this.
    Divide your polynomial by x - 2. Then note that the remaining factor is quadratic. How do you split a quadratic into its factors?

    -Dan
    Thanks from FatimaA and MarkFL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,417
    Thanks
    1330

    Re: Factoring linear factors.

    Quote Originally Posted by FatimaA View Post
    Hi, I'm working through an exam pretest and I came across this problem. Looking through my notes, I can't find examples that show me how to do it. I'm not totally sure what it's asking.

    a) Use the Remainder Theorem to find the remainder when x^3-2x^2+8x-16 is divided by x-2.

    For this I did P(2) = (2)^3 -2(2)^2+8(2)-16

    I got P(2) = 0 but was I supposed to plug in -2?
    No, the remainder when a polynomial, p(x), is divided by x- a is p(a) so 2 is correct.
    And, in fact, if you checked "the hard way" by actually dividing, you would see that x- 2 divides into x^3- 2x^2= x^2(x- 2) exactly " x^2" times, leaving a remainder of 8x- 16= 8(x- 2) and x- 2 obviously divides evenly into that. I don't know why you are questioning you answer.

    This is the part I'm stuck on:

    b) Use your answer in part a to factor x^3-2x^2+8x-16 into linear factors allowing complex numbers. Clearly explain your reasoning.
    You know that x- 2 is one factor. What is the quotient when you divide x^3- 2x^2+ 8x- 16 by x- 2? (Of course, I have given you the answer above.)
    The quotient will be quadratic which you can factor once you know the zeros, if necessary by using the quadratic formula.

    Thanks in advance for any help I can get on this.
    Last edited by HallsofIvy; April 28th 2013 at 02:29 PM.
    Thanks from FatimaA, MarkFL and topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2012
    From
    Seattle, Washington
    Posts
    72

    Re: Factoring linear factors.

    Thanks. I got:

    (x^3-2x^2+8x-16)/(x-2) = x^2+8

    So the answer is (x-2)(x+2i(square root 2))(x-2i(square root 2))
    Last edited by FatimaA; April 28th 2013 at 02:37 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 28th 2010, 10:33 AM
  2. Nonrepeated Linear Factors
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 24th 2010, 02:24 PM
  3. factoring radicands using perfect cube factors
    Posted in the Calculators Forum
    Replies: 0
    Last Post: November 19th 2008, 06:07 PM
  4. Finding Linear Factors Over C
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 5th 2008, 11:32 PM
  5. Linear factors of polynomial
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 17th 2008, 11:03 PM

Search Tags


/mathhelpforum @mathhelpforum