NationStates Jolt Archive

How in the heck do you factor (or otherwise solve, heck, I'm game for anything by this point) this equation:

x^3 + 3x^2 + 108 = 0

The lack of a linear term keeps getting in my way when I try to factor it; is there something I'm missing or another way to solve it?

I don't know.

I forgot. It's been too long. But you could use 0 as a coefficient of x to obtain your linear term. Also, in this case, x < 0.

x^3 + 3x^2 + 108 = 0

This is a 3 term polynomial. factor 1 out.

Who would want to deal with this stuff?

The lack of a linear term keeps getting in my way when I try to factor it; is there something I'm missing or another way to solve it?

I do believe you need to 'complete the square' beforehand. Don't ask how to do that coz it's been eons.

I do believe you need to 'complete the square' beforehand. Don't ask how to do that coz it's been eons.
Yeah, that's right.
If you factor out x (after subtracting 108), you'd get:

x(x²+3x)=-108

Perhaps (and I'm not gonna try), you could get x²+3x to fit the scheme of (a+b)²=a²+2ab+b²... :confused:

Yeah, that's right.
If you factor out x (after subtracting 108), you'd get:

x(x²+3x)=-108

Perhaps (and I'm not gonna try), you could get x²+3x to fit the scheme of (a+b)²=a²+2ab+b²... :confused:


Ergh I don't think it's that simple. I honestly forget but you should look for a reference for that method somehwere lol.

(x + 6)(x^2 -3x +18) = 0

You don't need to complete the square, you just have to fiddle about a bit.

Ergh I don't think it's that simple. I honestly forget but you should look for a reference for that method somehwere lol.
:D

Haven't done this for ages...I usually use Matrices to solve more than one at once.
Embarrassing really, considering how much maths I have to put up with in my studies. But that's how the world goes...

(x + 6)(x^2 -3x +18) = 0

You don't need to complete the square, you just have to fiddle about a bit.

Yep, just factor the polynomial. I've been trying to remember that formula for a while now....

Yep, just factor the polynomial. I've been trying to remember that formula for a while now....

I took my first degree in engineering, so I did it by finding the real root using a numerical solution, then using a little algerbra (knowing that one factor had to be (x+6)). (Of course, back when I did this sort of stuff all the time, you find the answers tend to just "pop" out for you).

I took my first degree in engineering, so I did it by finding the real root using a numerical solution, then using a little algerbra (knowing that one factor had to be (x+6)). (Of course, back when I did this sort of stuff all the time, you find the answers tend to just "pop" out for you).


What if he needs to show his work? :)

*smirk*

If you're looking for what the value of x is, one answer is x=-6. You can get this by rearranging the equation to (x^2)*(x+3)=-108 and then fiddling around some dividing 108 by perfect squares.

While we're playing around with Maths...

1 = (-1)² = sqrt((-1)²) = sqrt((-1)*(-1)) = sqrt(-1)*sqrt(-1) =i*i = i² = -1

Therefore 1=-1

Hehe, it made a good impression on my Year 11 Maths Class back then..

What if he needs to show his work? :)

*smirk*

Well now the solution is known, I am sure he can work backwards quite easily. That's what I would do.

You want to get it into the form (x+a)(x+b)(x+c)=0. The trick is determining the values a, b and c. The best way to do this is to make an educated guess and then see if it's right.

First, let's try to guess a. There's a rule that a will always be the quotient of a factor of the constant and a factor of the coefficient of x^3. In this case, the factors of the constant are (plus or minus) 1, 2, 3, 4, 6, 9, 12, 27, 36, 54 or 108 and the factors of the coefficient of x^3 are 1 and -1, so the value a will be plus or minus 1, 2, 3, 4, 6, 9, 12, 27, 36, 54 or 108.

Now that you know the possible factors, you can start going through them systematically and use long division (or synthetic division, if you're in a hurry) to check which ones work and to determine the rest of the equation. (I'll assume for now that you know how to do long division of a polynomial; if you don't, let me know and I'll explain it.)

You'll find that the value a=6 leaves no remainder and gives the answer (x^2-3x+18), which means the equation can be factored as (x+6)(x^2-3x+18). The term (x^2-3x+18) cannot be factored over the real numbers, but you can factor it over the complex numbers using the quadratic formula if you really want to.

I hope that was helpful; it's kind of hard to explain math over the internet.

How in the heck do you factor (or otherwise solve, heck, I'm game for anything by this point) this equation:

x^3 + 3x^2 + 108 = 0

The lack of a linear term keeps getting in my way when I try to factor it; is there something I'm missing or another way to solve it?
It's a polynomial equation, and the coefficient of the highest order term is one, so if there is an integer solution it must be a divisor of the lowest order term, 108. Also, since all the coefficients are positive the solution is negative.

108 = 2^2*3^3

so if there is an integer solution it is -2, -3, -4, -6, -9, -12, -18, -36 or 108

Try those out.

(-2)^3 + 3*(-2)^2 + 108 = 112 /= 0

(-3)^3 + 3*(-3)^2 + 108 = 108 /= 0

(-4)^3 + 3*(-4)^2 + 108 = 92 /= 0

(-6)^3 + 3*(-6)^2 + 108 = 0 This is one solution!

Then x+6 is one factor in the polynomial. I guess you know the rest.

While we're playing around with Maths...

1 = (-1)² = sqrt((-1)²) = sqrt((-1)*(-1)) = sqrt(-1)*sqrt(-1) =i*i = i² = -1

Therefore 1=-1

Hehe, it made a good impression on my Year 11 Maths Class back then..sqrt(-1)*sqrt(-1) = i*i is the problem because either or both of sqrt(-1) could have been -i. A similar method can get you "6*0 = 0 = 5*0 so 6 = 5."