Amath Question

• PUrePLs

  11 May 09, 17:58

  Find the value of a and b for which x^2 + 4x -1 is a factor of x^3 + ax^2 - 5x + b.

  Anyone can help ? Thanks.

• nightzip

  11 May 09, 18:25

  Dont have paper with me, just some ideas. If x^2+4x-1 is a factor of the x^3+ax^2-5x+b. so we can factorize x^2+4x-1 term to obtain the two factors (x+/-..)(x+/-..) Then since this two are factors of the x^3 expression, we use f(..)= sub in the values of x for X^3 expressions = 0. I think as you will have two factors, and two unknowns, should be able to solve.
• Lee012lee

  12 May 09, 13:06

  If the question is changed slightly, it will be a bit more fun.

  Find the value of a and b for which x^2 + x + 4 is a factor of x^3 + ax^2 - 5x + b.

   

• Mikethm

  12 May 09, 14:54

  Originally posted by PUrePLs:

  Find the value of a and b for which x^2 + 4x -1 is a factor of x^3 + ax^2 - 5x + b.

  Anyone can help ? Thanks.

  Let x^3 + ax^2 - 5x + b = (x^2+4x-1)(x-b)

  Let x = 1

  1+a-5+b = 4(1-b)

  a+5b = 8 ------(1)

  Let x = -1

  -1 + a + 5 + b = -4(-1-b)

  a-3b = 0----- (2)

  Eqn (1) - (2)

  8b = 8

  b = 1

  From (1)

  a + 5 = 8

  a = 3

   

• nightzip

  12 May 09, 22:47

  Originally posted by Lee012lee:

  If the question is changed slightly, it will be a bit more fun.

  Find the value of a and b for which x^2 + x + 4 is a factor of x^3 + ax^2 - 5x + b.

   

  
  This is not difficult, just do long division x^3 expression divide by the x^2 expression. The remainder (x+_) term will be one of the factor already.
• Lee012lee

  13 May 09, 00:38

  Originally posted by nightzip:

  
  This is not difficult, just do long division x^3 expression divide by the x^2 expression. The remainder (x+_) term will be one of the factor already.

  The modified question is not difficult, it is just a bit more fun.

  Using your earlier method, it will not be possible to solve the modified question.

  Your suggestion of using the long division method to solve the modified question

  also has the problem of not being able to divide due to the existence of a and b.

  Mikethm has already shown the correct method to solve the modified question

  ie the modified question can be solved by using the method used by Mikethm to

  solve the original question.

• PUrePLs

  13 May 09, 17:09

  Originally posted by Mikethm:

  Let x^3 + ax^2 - 5x + b = (x^2+4x-1)(x-b)

  Let x = 1

  1+a-5+b = 4(1-b)

  a+5b = 8 ------(1)

  Let x = -1

  -1 + a + 5 + b = -4(-1-b)

  a-3b = 0----- (2)

  Eqn (1) - (2)

  8b = 8

  b = 1

  From (1)

  a + 5 = 8

  a = 3

   

   how do you get the (x-b) above ?

• quirK

  13 May 09, 17:40

  b is the constant of the cubic equation, and since the quadratic has a -1 inside, the only way you can get a +b in the cubic is if you multiply the quadratic with a (x-b).