factorising a cubic..leading onto something else.

Well assuming you were told that that quartic had two quadratic factors, you could write:

$x^4 -4x^3 +35x^2 -62x + 130 = (Ax^2 + Bx + C)(Dx^2 + Ex + F)$

You can see immediately that A = D = 1 and CF = 130.

The prime factorisation of 130 = 2 x 5 x 13 so the only possible pairs are 10 and 13, 26 and 5 and 65 and 2 (negative pairs are possible too).

If you expand the brackets and equate coefficients, you will be able to figure out what to do.

I'll just ask you or someone else here, but is integrating e^(x^2) too difficult for A level or have I missed a simple trick?

Too difficult for A Level. You need to know about the error function.

Actually,my method is a little bit complicate,which means that it is not easy to generalise.First,we know the coefficient of x^4 is 1,and x^3 is -4,so we can assume(x^2-2x+y)*(x^2-2x+z),so,according to simultaneous equation 4+y+z=35 and y*z=130,y=26,z=5.

integrating e^(x^2)

on a similar note, what's $\displaystyle\int xe^{x^2}\ dx$?

Well assuming you were told that that quartic had two quadratic factors, you could write:

$x^4 -4x^3 +35x^2 -62x + 130 = (Ax^2 + Bx + C)(Dx^2 + Ex + F)$

You can see immediately that A = D = 1 and CF = 130.

The prime factorisation of 130 = 2 x 5 x 13 so the only possible pairs are 10 and 13, 26 and 5 and 65 and 2 (negative pairs are possible too).

If you expand the brackets and equate coefficients, you will be able to figure out what to do.