1 Answer | Add Yours
Let's suppose that f(x)'s grade is n. We know that if we multiply 2 polynomials, their grades are adding, in order to find the resulting grade.
If the grade of f(x) is n, then the grade of f'(x) is (n-1) and the grade of f"(x) is (n-2).
n= n-1+n-2, n=3
f(x)= ax^3 + bx^2 +cx+d
ax^3 + bx^2 +cx+d=(3ax^2+2bx+c)(6ax+2b)
The expressions above are eqaul if only the corresponding quotients are equals.
b=18ab=18*(1/18)b, so b could be any real number
f(x)= (1/18)x^3 + bx^2 +6bx+12b^3
We’ve answered 330,597 questions. We can answer yours, too.Ask a question