If $$x^{2}$$+3x-10is a factor of $$3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$$ then the closest approximate values of a and b are

If $$x^{2}$$+3x-10is a factor of $$3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$$
Then x = -5 and x = 2 will give $$3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$$ = 0
Substituting x = -5 we get,
$$3(-5)^{4}+2(-5)^{3}-a(-5)^{2}+b(-5)-a+b-4 = 0$$
Solving we get,
$$26a+4b = 1621$$.......(i)
Substituting x = 2 we get,
$$3(2)^{4}+2(2)^{3}-a(2)^{2}+b(2)-a+b-4 =0$$
=> $$5a-3b = 60$$........(ii)
Solving i and ii we get
a and b $$\approx 52, 67$$
Hence, option C is the correct answer.