SOLUTION: Hii my beloved tutors, I really need help on this question, could you please help me? Suppose {{{ a }}} is a real number, so that polynomial {{{ p(x)= x^4+4x+a }}} is divisible by

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hii my beloved tutors, I really need help on this question, could you please help me? Suppose {{{ a }}} is a real number, so that polynomial {{{ p(x)= x^4+4x+a }}} is divisible by      Log On


   



Question 1020769: Hii my beloved tutors, I really need help on this question, could you please help me?
Suppose +a+ is a real number, so that polynomial +p%28x%29=+x%5E4%2B4x%2Ba+ is divisible by +%28x-c%29%5E2+ for +c+ is a real number. What is the value of +a+ that satisfies?
Thanks

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
If %28x-c%29%5E2 is to divide p%28x%29=+x%5E4%2B4x%2Ba+ , then x-c should be able to divide the latter twice.
Using the remainder theorem, if c is to be root of p(x), then p(c) = 0, or c%5E4+%2B+4c+%2Ba+=+0
Now divide p%28x%29=+x%5E4%2B4x%2Ba+ by x-c, and letting the remainder c%5E4+%2B+4c+%2Ba equal to 0, we get the polynomial x%5E3%2Bcx%5E2%2Bc%5E2x%2B%284%2Bc%5E3%29.
By applying the remainder theorem again on this resulting polynomial, we get c%5E3%2Bc%5E3%2B+c%5E3+%2B4+%2B+c%5E3+=+4+%2B+4c%5E3+=+0
The last equation gives highlight%28c+=+-1%29.
substitution of this value into c%5E4+%2B+4c+%2Ba+=+0 gives highlight%28a+=+3%29.