Let r_1,r_2 and r_3 be the three roots of the equation. of which r_1+r_2=2

Then we have the following two equations.

r^3_1 +pr_1 + 28=0

and

r^3_2 +pr_2 + 28=0

Addition of the two equations gives

r^3_1+r^3_2+p(r_1+r_2)+56=0

Apply the sum of cubes identity and collect like terms r_1+r_2

(r_1+r_2)(r^2_1+r^2_2-r_1r_2+p)+56=0

Construct a perfect square (r_1+r_2)^2

(r_1+r_2)[(r_1+r_2)^2-3r_1r_2+p]+56=0

Substitute the value of r_1+r_2

2(4-3r_1r_2+p)+58=0

Symplifying gives

On the other hand, the following two equations are established according to Vieta's formula

r_1+r_2+r_3=0

r_1r_2r_3 = -28

Substituting the value of r_1+r_2 results in

r_3 = -2

and

r_1r_2 = 14

Substituting the value of r_1r_2 to (1) yields the final result.

p = 3\times14-32 = 10

Using the Vieta's formula, the third root could be determined

r_1+r_2+r_3=0

r_1r_2r_3 = -28

Substituting the value of r_1+r_2 results in

r_3 = -2

And then substitute r_3 to the cubic equation

Then

(-2)^3+p(-2)+28=0

Solve for p

p = 10