ANSWERS: 3
  • I MEANT (x-1) NOT (x+1)
  • Synthetic division. I know you stated the question wrong, but suppose I do it for the question you asked, and then you can do it the same way for the correct question. ax^3 - 4x^2 + 5 x - 3 is a polynomial It's a bit like a decimal number: 1248 is 1*10^3 + 2*10^2 + 4*10^1 + 8 If I divide it by 10 + 1, long hand: 1*10^3 + 2*10^2 + 4*10^1 + 8 1*10^3 + 1*10^2 (is 1*10^2 elevens) subtract: 1*10^2 + 4*10^1 + 8 1*10^2 + 1*10^1 (is 1*10^1 elevens) subtract: 3*10^1 + 8 3*10^1 + 3 (is 3 elevens) subtract: remainder = 5 Now for the polynomial divided by x+2: ax^3 - 4x^2 + 5 x - 3 ax^3 + 2ax^2 (that's a*(x+2)) subtract: -(4+2a)x^2 + 5x - 3 -(4+2a)x^2 - (8+4a)x (that's (-(4+2a))*(x+2)) subtract: (13+4a)x - 3 (13+4a)x + (26+8a) (that's (13+4a)*(x+2)) subtract: -29-8a is the remainder Same again for x+1 ax^3 - 4x^2 + 5 x - 3 ax^3 + ax^2 (that's a*(x+1)) subtract: -(4+a)x^2 + 5x - 3 -(4+a)x^2 - (4+a)x (that's (-(4+a))*(x+1)) subtract: (9+a)x - 3 (9+a)x + (9+a) (that's (9+a)*(x+1)) subtract: -22-a is the remainder If I've not made any aritmetical mistakes, then I only need to set: -29-8a = -22-a and solve to get the answer a = -1 Now you need to do this with the correct question.
  • Okay. There is something called a remainder theorem. Consider 2 equations ax^3-4x^2+5x-3 and x+2. If we put x+2=0, we get x=-2 ans if we put this x=-2 in the equation ax^3-4x^2+5x-3, we get the remainder when ax^3-4x^2+5x-3 is divided by x+2. repeat this with x=-1 for x+1 (as given in the question). we get 2 new equations, -8a-16-10-3 and -a-4-5-3. Since according to the question, both are equal, we have -8a-16-10-3 = -a-4-5-3. You can easily solve for a now.
    • 1!Telly
      x to be replaced with 1 instead of -1, (I saw the comment late) and proceed.

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