binomial question

[woven sentinel]

If $f(x) = (1+kx)^n$

$f'(x) = nk(1+kx)^{n-1} = 12 + 56k^2x + ... nk^n x^{n-1}$

And $f'(0) = nk = 12$
So you know $nk = 12$

Again, $f"(x) = n(n-1)k^2(1+kx)^{n-2} = 56k^2 + ... + n(n-1)k^nx^{n-2}$

And so $f"(0) = n(n-1)k^2 = 56k^2$
So $n(n-1) = 56$

[sleek sigil]

Daddy_314

[woven sentinel]

This tells you two equations for n and k

[humble sentinel]

recall the Binomial Theorem:
(a+b)^n = sigma(k=0 to n, C(n,k)•a^(n-k)•b^k)
where C(n,k) is combination
= n!/(n-k)!
in case you forgot what "!" means it's called factorial
n! = n•(n-1)•(n-2)...•3•2•1

now you just need to expand the expression and compare coefficients with the given one to find the values of n and k