once again vikas ans is wrong as he doesnt even understand the meaning of multiplicity.
note that for a root q to have multiplicity of 2, (x – q)^2 should be a factor of the given polynomial.
let p(x)= x4 – 8x3 + 22x2 – 24x + p = 0
then x4 – 8x3 + 22x2 – 24x + p = (x – q)^2*g(x)
differentiate both sides
p’(x)= 4x^3 – 24x^2 + 44x – 24= 2(x – q)*g(x) + (x – q)^2*g’(x)= (x – q)*f(x) where f(x)= 2g(x) + (x – q)*g’(x) is a polynomial.
so, clearly x= q is a root of p’(x).
now, the roots of 4x^3 – 24x^2 + 44x – 24 are x= 1, 2, 3. so, q= 1, 2 or 3 should also be the root of p(x).
when q=1, p(1)= – 9+p= 0 or p= 9
when q=2, p(2)= – 8 + p= 0 or p= 8
when q=3, p(3)= – 9+p= 0 or p= 9
so, p= 8, 9
so, sum of values of p is 8+9
= 17
kindly approve :)
