Let f(x)=x⁴+4x-48, then f'(x)=4x³+4.
x=x-f(x)/f'(x) where x on the RHS is a guessed value of x producing (hopefully) a more accurate value of x on the LHS.
Observation shows that there are two zeroes for f(x). One is between -3 and -2 and the other between 2 and 3, because f(-3)=21 and f(-2)=-40 (change of sign); and f(2)=-24 and f(3)=45 (another change of sign). The changes of sign indicate a zero between the x values, that is, there is a value of x such that f(x)=0, where -3<x<-2 and 2<x<3.
Therefore, we can use -2.5 and 2.5 as starting values for x. We call these x₀ and treat them separately.
So consecutive iterations converge to x=-2.772546504.
Another zero is x=2.483876601.
With these two zeroes we can create a quadratic:
This quadratic is a factor of f(x) so by dividing by it we arrive at another quadratic which may be able to be solved using the formula for solving quadratics.
Let C=-(A+B) and D=AB. Now we use long division to find the second quadratic:
x²+Cx+D ) x⁴ +4x -48
We need take the division no further because we now have the required quadratic: x²-Cx+C²-D.
The zeroes are x=(C±√(C²-4C²+4D))/2.
The discriminant 4D-3C²<0 when we calculate C and D from the zeroes A and B, so these remaining zeroes are complex.
Therefore the only real zeroes are x=A=-2.772546504 and x=B=2.483876601.