How do I solve for "x" in x^3+3x^2+2x-y=0?

Question

What algebraic equation would represent "x"?

Answer 1

By factoring, x can be any real number as long as
y = x(x + 1)(x + 2)

To solve for x in terms of y, you'll need to solve a cubic equation. It's tricky!
http://mathworld.wolfram.com/CubicFormula.html

x³ + 3x² + 2x - y = 0 is of the form:
x³ + (A2)x² + (A1)x +(A0) = 0, where
(A2) = 3, (A1) = 2, and (A0) = -y.
The cubic will have three solutions, two of which may be imaginary. You can look those up from the mathworld link. The real root is:
x = (-1/3)(A2) + S + T, where
S = ³√[R + √D] and T = ³√[R - √D], where
D = Q³ + R²,
Q = [3(A1) + (A2)²] / 9, and
R = [9(A2)(A1) - 27(A0) - 2(A2)³] / 54. I said it was tricky!
Substituting:
R = [9(A2)(A1) - 27(A0) - 2(A2)³] / 54
R = [9(3)(2) - 27(-y) - 2(3)³] / 54
R = [54 + 27y - 54] / 54
R = 27y / 54
R = y / 2

Q = [3(A1) + (A2)²] / 9
Q = [3(2) + (3)²] / 9
Q = [6 + 9] / 9
Q = 15 / 9
Q = 5 / 3

D = Q³ + R²
D = (5 / 3)³ + (y / 2)²
D = 125 / 27 + y² / 4
D = (27y² + 500) / 108

S = ³√{R + √D}; T = ³√{R - √D}
S = ³√{(y / 2) + √[(27y² + 500) / 108]}
T = ³√{(y / 2) - √[(27y² + 500) / 108]}

x = (-1/3)(A2) + S + T
x = (-1/3)(3) + S + T
x = -1 + ³√{(y / 2) + √[(27y² + 500) / 108]} + ³√{(y / 2) - √[(27y² + 500) / 108]}

There's a lot of rationalizing to do with the denominators in there, but at least it's a solution!

Answer 2

6 - 2 x 1 - 4 = 0

Answer 3

y = x^3 + 3x^2 + 2x

Take common factor x out:

y = x * ( x^2 + 3x + 2)

Now solve the quadratic equation x^2 + 3x + 2 and factorize

Or you can use magic:
The sum of all even coefficients is equal to the sum of the odd coeffients, so one root is -1. The product of the two roots is 2. so second root is -2.

Either way you will get:

y = x * (x+1) * (x+2)

Answer 4

Um.. is this true?


*SPOILER*
x^3+3x^2+2x-y=0
x(x^2+3x+2)=y
x(x+2)(x+1)=y

therefore x=y, x=y-2, x=y-1.. Sorry, usually I do y equals 0 so x=0, x=-2 and x=-1...

Answer 5

Go x( x^2 + 3x + 2)=y
x (x + 2)(x + 1) =y

then im sure you can take it from there

Answer 6

1. Re-arrange the equation:
x^3 + 3 x^2 + 2x = y
2. Factor out an x:
x(x^2 + 3x + 2) = y
3. "Reverse FOIL"
x(x + 1)(x + 2) = y
4. x can equal y, y - 1, or y - 2. (Change sides, change signs)

Answer 7

Cardano solved this. See Wolfram Mathworld on the net. - Cubic formula.

Answer 8

If you subsitute the quadratic formula for x?

I dont know what your asking

Are you trying to solve this for Y

Well hope this helps you think about your answer better.

Answer 9

You need the formula for solving cubic equations.

It is complex and tedious but you can find it, and a worked example, at:

http://www.ping.be/~ping1339/cubic.htm

Answer 10

for this need more equations to solve