Solve f^3=4f^2-9f+36

$$f ^ { 3 } = 4 f ^ { 2 } - 9 f + 36$$

Solve for f

$$f=4$$

Solution Steps

Subtract 4f^{2} from both sides.

$$f^{3}-4f^{2}=-9f+36$$

Add 9f to both sides.

$$f^{3}-4f^{2}+9f=36$$

Subtract 36 from both sides.

$$f^{3}-4f^{2}+9f-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$f=4$$

By Factor theorem, f-k is a factor of the polynomial for each root k. Divide f^{3}-4f^{2}+9f-36 by f-4 to get f^{2}+9. Solve the equation where the result equals to 0.

$$f^{2}+9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 0 for b, and 9 for c in the quadratic formula.

$$f=\frac{0±\sqrt{0^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$f=\frac{0±\sqrt{-36}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$f\in \emptyset$$

List all found solutions.

$$f=4$$

Solve for f (complex solution)

$$f=-3i$$

$$f=4$$

$$f=3i$$