AMaths

• Only-Way-4-Destiny!

  25 Apr 09, 16:56

  Hi all, can anyone help me with this please?

  Qn:

  Given that kx³ + 2x² + 2x + 3 and kx³ - 2x + 9 have a common factor, what are the possible values of k?

   

  Thanks!

• skythewood

  25 Apr 09, 17:29

  this means that if you divide 1st equation by 2nd equation, the remainder will be zero.

  Try doing a long division.

• TenSaru

  25 Apr 09, 17:34

  I'm not really sure about this but I think you have to make

  kx³ + 2x² + 2x + 3 = 0 ----------(1)

  kx³ - 2x + 9 = 0           ----------(2)

  then solve for x by (1)-(2)

  then sub the values of x into any of the 2 equations to get k

   

• skythewood

  25 Apr 09, 17:41

  Originally posted by TenSaru:

  I'm not really sure about this but I think you have to make

  kx³ + 2x² + 2x + 3 = 0 ----------(1)

  kx³ - 2x + 9 = 0           ----------(2)

  then solve for x by (1)-(2)

  then sub the values of x into any of the 2 equations to get k

  i think... no?

• Only-Way-4-Destiny!

  25 Apr 09, 17:50

  i oso think no. the value of x can be of any real value. something like polynomials and identity

   

  btw, sky, i tried dividing. got remainder.

• TenSaru

  25 Apr 09, 17:59

  Ok.. Let (x-a) be the common factor

  f(x) = kx³ + 2x² + 2x + 3

  f(a) = ka³ + 2a² + 2a + 3 =0 -------(1)

  g(x) = kx³ - 2x + 9

  g(a) = ka³ - 2a + 9 = 0 --------(2)

  does this sound more correct? XD

• skythewood

  25 Apr 09, 18:00

  Originally posted by Only-Way-4-Destiny!:

  Hi all, can anyone help me with this please?

  Qn:

  Given that kx³ + 2x² + 2x + 3 and kx³ - 2x + 9 have a common factor, what are the possible values of k?

   

  Thanks!

  hmmm, just noticed that both equations is cube, not square... guess have to revise the method...

  let's imagine (x-Q) is the common factor.

  sub x= Q into both equations, and both equations should be equal to zero, because Q is a factor of both equations...

  kQ³ + 2Q² + 2Q + 3 = 0 ----------(1)

  kQ³ - 2Q + 9 = 0           ----------(2)

  equate equation 1 to be equation 2, and find what is Q....

  Than sub in Q into either equation to solve for k. So TenSaru is right!

   

• skythewood

  25 Apr 09, 18:02

  Originally posted by TenSaru:

  Ok.. Let (x-a) be the common factor

  f(x) = kx³ + 2x² + 2x + 3

  f(a) = ka³ + 2a² + 2a + 3 =0 -------(1)

  g(x) = kx³ - 2x + 9

  g(a) = ka³ - 2a + 9 = 0 --------(2)

  does this sound more correct? XD

  yup, that should be correct.

• Only-Way-4-Destiny!

  25 Apr 09, 18:02

  OHH YA! i see it now.

• Only-Way-4-Destiny!

  25 Apr 09, 18:11

  got the answer. same as answer on textbook. thanks!