Question

If the equation 3x^4+2x^3-6x^2-6x+p=0 has two equal roots, find the possible values of p.
x-1 is a factor so after replacing x=1 I got p=7. But I don't know any other possible values of p. I got p=1 but I solved it through inspection. Please help.

Answer 1

If the given information is that x-1 is a factor then f(1) = 0 ---> p = 7
A synthetic division by x-1 yields a quotient of (3x^3 + 5x^2 - x -7)

but another division by x-1 yields a quotient of (3x^2 + 8x +7)

so 3x^4+2x^3-6x^2-6x+p=0 ----> (x-1)(x-1)(3x^2 + 8x + 7) if p = 7
there's your double root.
The remaining quadratic yields two complex roots