How do you factorise this?
2x^3 + 3x^2 - 1 = 0
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Factorise cubic equations help please... watch
- Thread Starter
- 19-12-2010 15:30
- 19-12-2010 15:34
One of the roots is a not very big integer
- 19-12-2010 15:39
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 optionsLast edited by didgeridoo12uk; 19-12-2010 at 15:40.
- 19-12-2010 15:40
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.
- 19-12-2010 16:06
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.Last edited by dknt; 19-12-2010 at 16:07.
- 19-12-2010 16:30
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.Last edited by forgottensecret; 19-12-2010 at 16:34.
- 19-12-2010 16:34
There is a technique called Ferrari's method. It's quite complicated though.
Just joking. Google Tartaglia.Last edited by neesh123; 21-12-2010 at 16:27.