$P(x) = x^2 + c \cdot (x-1)(x-2)(x-3)$. Then just do the calculation to get $P(10) + P(-6) = 100 + c \cdot 504 + 36 - c \cdot 504 = 136$.

YaoAOPS wrote: $P(x) = x^2 + (x-1)(x-2)(x-3)$. Then just do the calculation to get $P(10) + P(-6) = 100 + 504 + 36 - 504 = 136$. The polynomial isn't necessarily monic

ok lets solve this skibidi problem! the polynomial is ax^3+bx^2+cx+d so that a+b+c+d=1, 8a+4b+2c+d=4, 27a+9b+3c+d=9. Then find 784a+136b+4c+2d. The system p+8q+27r=784, p+4q+9r=136, p+2q+3r=4, p+q+r=2 can be solved, enough to find p+4q+9r. the solution is (64,-126,64) which gives p+4q=9r=136.