Question 1020769
If {{{(x-c)^2}}} is to divide {{{p(x)= x^4+4x+a }}} , 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^4 + 4c +a = 0}}}

Now divide {{{p(x)= x^4+4x+a }}} by x-c, and letting the remainder {{{c^4 + 4c +a}}} equal to 0, we get the polynomial {{{x^3+cx^2+c^2x+(4+c^3)}}}.
By applying the remainder theorem again on this resulting polynomial, we get {{{c^3+c^3+ c^3 +4 + c^3 = 4 + 4c^3 = 0}}}

The last equation gives {{{highlight(c = -1)}}}.
substitution of this value into {{{c^4 + 4c +a = 0}}} gives {{{highlight(a = 3)}}}.