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# Factorise cubic equations help please... watch

1. How do you factorise this?
2x^3 + 3x^2 - 1 = 0
2. One of the roots is a not very big integer
3. just try out a couple of numbers, then you can factorise it out and end up with a quadratic equation which you can solve normally.

generally in these questions it'll be between -3 and 3... just by looking at the equation you should be able to discount over half these options
4. (Original post by Nkhan)
How do you factorise this?
2x^3 + 3x^2 - 1 = 0
Let f(x) = 2x^3 + 3x^2 -1

Then, find x for which f(x) = 0

Then, arrange it in the form of (x+x1)(ax^2 + bx + c) where x1 is a factor of f(x).
Then, equate coefficients.
5. (Original post by Nkhan)
How do you factorise this?
2x^3 + 3x^2 - 1 = 0
It requires use of the factor theorem and algebraic division.

Let f(x)= 2x^3 + 3x^2 - 1

Consider substituting in a random integer 'a'.

Then, f(a)= 2(a)^3 +3(a)^2 -1

If you then find that f(a)=0, it must mean that (x-a) is a factor of 2x^3 + 3x^2 - 1 and hence can be written in the form of (x-a)(bx^2 +cx +d)

Try substituting in 1, then -1, 2, -2 etc. See which one gives 0, giving you your factor. Once you have your factor of (x-a), you need to divide this into (2x^3 + 3x^2 - 1). As it is a factor you'll be left with a quadratic, which will be the quotient, and no remainder. You can then see if your remaining quadratic can be factorised. If so, you cubic can be expressed by 3 linear factors; if not your cubic can be expressed by one linear factor and one quadratic factor.
6. Look to factors of the constant and substitute them in the equation using the "nest" method. This should produce a resulting quadratic which along with your initial root factor is your equation.
7. There is a technique called Ferrari's method. It's quite complicated though.

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Updated: December 19, 2010
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