Derivatives – Solving for Two Unknowns in a Polynomial Equation

Question

I want to solve for $a$ and $b$ for the polynomial function: $f(x)=ax^3+bx+4$ under the following conditions:

1) The cartesian point P=$(2,14)$ falls on the curve $f(x)=ax^3+bx+4$;
2) The gradient of $f(x)$ at point P is 21; that is to say, the derivative of $f(x)$ is such that $f'(2)=21$.

I tried to use the substitution method for solving simultaneous equations for a and b:
$f(2)=14$
$14=a(2^3)+b2+4$
$14=8a+2b+4$
$0=8a+2b-10$

From this it follows that:
$b=5-4a$

But substituting this back into the original equation yields no substantive results:
$14=8a+2[5-4a]+4$
$14=8a+10-8a+4$
$14=14$

Really not sure what approach to use to solve this problem any more.

If it's any help, I know the actual answer is:
$a=2$
$b=-3$

Thanks!

Answer 1

$f(2)=14 \implies 4a+b=5.$

$f'(2)=21\implies 12a+b=21.$

$\implies 8a=16 \implies a=2 \implies 8+b=5\implies b = -3.$