x^3-7x^2-12x+54=f(x) what are the zeros of this equation

Solve for x:
x^3 - 7 x^2 - 12 x + 54 = 0

The left hand side factors into a product with two terms:
(x + 3) (x^2 - 10 x + 18) = 0

Split into two equations:
x + 3 = 0 or x^2 - 10 x + 18 = 0

Subtract 3 from both sides:
x = -3 or x^2 - 10 x + 18 = 0

Subtract 18 from both sides:
x = -3 or x^2 - 10 x = -18

Add 25 to both sides:
x = -3 or x^2 - 10 x + 25 = 7

Write the left hand side as a square:
x = -3 or (x - 5)^2 = 7

Take the square root of both sides:
x = -3 or x - 5 = sqrt(7) or x - 5 = -sqrt(7)

Add 5 to both sides:
x = -3 or x = 5 + sqrt(7) or x - 5 = -sqrt(7)

Add 5 to both sides:
Answer: | x = -3      or x = 5 + sqrt(7)      or x = 5 - sqrt(7)

Here's a nice trick:

 

There's a way to find every single rational root for some of the polynomials:

 

step one:

 

we have the polynomial a₀+a₁*x+a₂*x²+a₃*x³+........+a_n*xⁿ=0. We need to multiply the polynomial by a number in a way that all coefficients will be whole numbers. Sometimes we cant do it. sometimes we can.

 

step two:

 

find all of the whole divisors of a_n and a₀ (including the negative divisors). lets call the set of the divisors of a₀ A and the set of the divisors of a_n B. there are k divisors in A and m divisors in B.

 

Every rational root of the polynomial=A_i/B_j, 0 i is a member of A and B _j is a member of B, meaning every rational root can be expressed using one divisor from A divided by one divisor from B).

 

step three:

 

Check for every two divisors (one from A and one from B) if A_i/B_j is a root. after a finite amount of time (unless your calculator is broken) you'll have every rational root of the polynomial.