SOLUTION: find the values of A,B,C if (x-1),(x+2) and (x-2) are factor of 2Ax^4-Bx^3-Cx-16

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Question 1170859: find the values of A,B,C if (x-1),(x+2) and (x-2) are factor of
2Ax^4-Bx^3-Cx-16

Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52795) About Me  (Show Source):
You can put this solution on YOUR website!
.

In accordance with the Remainder theorem,  the given part means that the given polynomial has the roots of 1, -2 and 2.


So we write these three equations expressing this fact

    2A*1^4    - B*1^3    - C*1     - 16 = 0

    2A*(-2)^4 - B*(-2)^3 - C*(-2)  - 16 = 0

    2A*2^4    - B*2^3    - C*2     - 16 = 0


or

    2A -  B -  C = 16    (1)

   32A + 8B + 2C = 16    (2)

   32A - 8B - 2C = 16    (3)


Adding equations (2) and (3), you get

   64A           = 32,   which implies  A = 32%2F64 = 0.5.


Then from (1) and (2), substituting A = 0.5 there, you get

     B +  C = -15        (4)

    8B + 2C =   0        (5)


Expressing  B = -15 - C  from (4)  and substituting it to (5), you get

    8*(-15 -C) + 2C = 0,    or

    -120 - 8C +  2C = 0

         - 6C = 120

            C = - 20.


Then from equation (4),  B = -15 - (-20) = -15 + 20 = 5.


ANSWER.  A = 0.5;  B = 5,  C = -20.

Solved.



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

We have three factors: (x-1),(x+2) and (x-2)
However, we have a quartic polynomial, since the degree (largest exponent) is 4

So we need a fourth factor. Let's call that x-p, for some real number p.

Let's expand out the following
(x-1)(x+2)(x-2)(x-p)

(x+2)(x-2)(x-1)(x-p)

(x^2-4)(x^2-px-x+p)

x^2(x^2-px-x+p)-4(x^2-px-x+p)

x^4-px^3-x^3+px^2-4x^2+4px+4x-4p

x^4+(-px^3-x^3)+(px^2-4x^2)+(4px+4x)-4p

x^4+(-p-1)x^3+(p-4)x^2+(4p+4)x-4p

Compare that result to 2Ax^4-Bx^3-Cx-16
Match up the corresponding terms, and we have the following equations
  • 2Ax^4 = 1x^4 which leads to 2A = 1 and solves to A = 1/2 = 0.5
  • (-p-1)x^3 = -Bx^3 which leads -p-1 = -B and solves to B = p+1
  • (4p+4)x = -Cx which leads to 4p+4 = -C and solves to C = -4p-4
  • -4p = -16 which solves to p = 4

Use that value of p to find the other coefficients
B = p+1 = 4+1 = 5
C = -4p-4 = -4*4-4 = -20

The polynomial 2Ax^4-Bx^3-Cx-16 turns into x^4-5x^3+24x-16 after we plug in those A,B, and C values.


Answers:
A = 1/2 = 0.5
B = 5
C = -20