Results 1 to 13 of 13

Math Help - Dividing Polynomials....

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    13

    Dividing Polynomials....

    I'm having problems with two questions from my hw. I have no idea where to even start to solve the problems. Please help as much as you can!! Thanks!

    1. When x^3 - 7x + 4 is divided by the polynomial D(x), the quotient is x^2 - 3x + 2 and the remainder is -2. Find D(x).


    2. Find k so that when x^3 + kx^2 + k^2x + 14 is divided by x+2, the remainder is 0.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by EyesThatSparkle02 View Post

    2. Find k so that when x^3 + kx^2 + k^2x + 14 is divided by x+2, the remainder is 0.

    x^3+kx^2+k^2x+14=Q(x)(x+2)+R

    (-2)^3+k(-2)^2+k^2(-2)+14=Q(x)(-2+2)+0

    Continue doing and solve for k .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    13
    Quote Originally Posted by mathaddict View Post
    x^3+kx^2+k^2x+14=Q(x)(x+2)+R

    (-2)^3+k(-2)^2+k^2(-2)+14=Q(x)(-2+2)+0

    Continue doing and solve for k .
    Thank you so much!!!

    I'm still kind of confused with the simplifying though...

    I get: -8 +4k - 2k^2 + 14 = Q(x)

    Q(x) = -2 (k^2-2k-3)

    Is that right? Or did I make a mistake somewhere?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by EyesThatSparkle02 View Post

    1. When x^3 - 7x + 4 is divided by the polynomial D(x), the quotient is x^2 - 3x + 2 and the remainder is -2. Find D(x).

    I can see that D(x) is a polynomial of degree 1 , (x-a)

    Expressing the whole thing in terms of quotient , divisor and remainder :

    x^3-7x+4=(x^2-3x+2)(x-a)-2

     x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2

    By comparing ,

    3xa+2x=-7x

    a=-3

    -2a-2=4

    a=-3

    Thus , D(x) = (x+3)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by EyesThatSparkle02 View Post
    Thank you so much!!!

    I'm still kind of confused with the simplifying though...

    I get: -8 +4k - 2k^2 + 14 = Q(x)

    Q(x) = -2 (k^2-2k-3)

    Is that right? Or did I make a mistake somewhere?

    Q(x)(-2+2)+0=0

    So -8 +4k - 2k^2 + 14 = 0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2009
    Posts
    13
    Quote Originally Posted by mathaddict View Post
    I can see that D(x) is a polynomial of degree 1 , (x-a)

    Expressing the whole thing in terms of quotient , divisor and remainder :

    x^3-7x+4=(x^2-3x+2)(x-a)-2

     x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2

    By comparing ,

    3xa+2x=-7x

    a=-3

    -2a-2=4

    a=-3

    Thus , D(x) = (x+3)
    Thank you!! One question though, how did you compare?? I mean, how did you go from that to 3xa+2x = -7x ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by EyesThatSparkle02 View Post
    Thank you!! One question though, how did you compare?? I mean, how did you go from that to 3xa+2x = -7x ?

    Firstly , you must make sure that the degree of 'x' on both sides are the same .

    In this case , it is 1 so we can compare .

    (3a+2)x=(-7)x , we can cancel the x and then solve for a .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Feb 2009
    Posts
    13
    Quote Originally Posted by mathaddict View Post
    Firstly , you must make sure that the degree of 'x' on both sides are the same .

    In this case , it is 1 so we can compare .

    (3a+2)x=(-7)x , we can cancel the x and then solve for a .

    How did you get 3ax+2x = -7x in the first place??
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Feb 2009
    Posts
    13
    Quote Originally Posted by mathaddict View Post
    Q(x)(-2+2)+0=0

    So -8 +4k - 2k^2 + 14 = 0

    I'm so sorry for having so many questions, but you're really helping!!

    The answer is supposed to be - 9/2 .

    I keep getting k= 3 or -1 ??
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by EyesThatSparkle02 View Post
    How did you get 3ax+2x = -7x in the first place??
    Alright , lets come back to this equation .

     x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2

    DO some regrouping .

     x^3+0x^2-7x+4=x^3-(a+3)x^2+(3a+2)x-21-2

    So now we compare ..

    x^3=x^3 ( there is nothing to compare)

    0x^2=-(a+3)x^2 , cancel the x^2 and solve for a .

    -7x=(3a+2)x , cancel the x and solve for a

    the last one , the constant , do the same comparing .

    All of these comparisons give a so you will only have to compare one of them .

    Hope this helps. Any more doubts ?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by EyesThatSparkle02 View Post
    How did you get 3ax+2x = -7x in the first place??
    Alright , lets come back to this equation .

     x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2

    DO some regrouping .

     x^3+0x^2-7x+4=x^3-(a+3)x^2+(3a+2)x-2a-2

    So now we compare ..

    x^3=x^3 ( there is nothing to compare)

    0x^2=-(a+3)x^2 , cancel the x^2 and solve for a .

    -7x=(3a+2)x , cancel the x and solve for a

    the last one , the constant , do the same comparing .

    All of these comparisons give a so you will only have to compare one of them .

    Hope this helps. Any more doubts ?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Feb 2009
    Posts
    13
    Quote Originally Posted by mathaddict View Post
    Alright , lets come back to this equation .

     x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2

    DO some regrouping .

     x^3+0x^2-7x+4=x^3-(a+3)x^2+(3a+2)x-2a-2

    So now we compare ..

    x^3=x^3 ( there is nothing to compare)

    0x^2=-(a+3)x^2 , cancel the x^2 and solve for a .

    -7x=(3a+2)x , cancel the x and solve for a

    the last one , the constant , do the same comparing .

    All of these comparisons give a so you will only have to compare one of them .

    Hope this helps. Any more doubts ?
    Thank you so much for all your help!!! It's really helped!!!
    I had a question about the 2nd problem. The answer is supposed to be -9/2 yet I keep getting 3 or -1.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by EyesThatSparkle02 View Post
    Thank you so much for all your help!!! It's really helped!!!
    I had a question about the 2nd problem. The answer is supposed to be -9/2 yet I keep getting 3 or -1.

    I think 3 or -1 should be correct because when you try doing the long division , their remainders are 0 . But it gives a non-zero remainder when k=-9/2 .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. help dividing polynomials?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 6th 2010, 04:24 PM
  2. Dividing Polynomials
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 17th 2009, 10:41 AM
  3. [SOLVED] dividing polynomials
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 10th 2007, 07:32 PM
  4. dividing polynomials
    Posted in the Algebra Forum
    Replies: 6
    Last Post: November 8th 2006, 05:20 AM
  5. Replies: 5
    Last Post: November 29th 2005, 04:22 PM

Search Tags


/mathhelpforum @mathhelpforum