#1**+1 **

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)**

Guest Apr 9, 2017

#2**+2 **

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 a_{0}+a_{1}*x+a_{2}*x^{2}+a_{3}*x^{3}+........+a_{n}*x^{n}=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 a_{0} (including the negative divisors). lets call the set of the divisors of a_{0} 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.

Ehrlich
Apr 9, 2017