Hey guys... So here's the question... I was given that f(x) has an x^3 coefft of 1. The roots of f(x) are 0 , -1 , 2 and 5 ... (i) write in factorised form and show that f(x) can be written as x^3 - 6x^2 + 3x + 10 So I factorised and got x^4 - 6x^3 + 3x^2 + 10x ... Then took out x as a factor to get the answer... Here's the question... There were four factors... It was originally an x^4 graph... If I was going to sketch the graph then how does it work? As the fact there are 4 roots means that it wouldnt look like a cubic graph? Do I just ignore the x=0 root? Edit: 0 doesn't give zero when substituted for x in the equation so that would mean it is not a root in the graph... But at the beginning if the question it said that 0 was a root... Now very confused :S

If there are four roots then you would get a quartic equation, as you got in your answer, not a cubic as the question implies. There must be a mistake in the question. 0 cannot be a root of f(x).

According to (i), if it can be written as x^3 - 6x^2 + 3x + 10, then f(0)=0 is false, therefore 0 is not a root. It must be a mistake. Even Wolfram Alpha disagrees with the question: http://www.wolframalpha.com/input/?i=x^3-6x^2+3x+10 That also means the factorised form is actually (x+1)(x-2)(x-5)...

It does say "In the cubic polynomial f(x), the coefft of x cubed is 1 ... Does that make any difference? Is there no way that the x could be discarded as a factor to discard the x=0 root?

You can't just arbitrarily discard factors. You always have the same number of roots as the order of the polynomial. A cubic has order 3, so you will only have 3 roots. To get the answer they're looking for, 0 cannot be a root. If 0 is a root then f(0) = 0, but according to the question, f(0) = 10. Your best bet is to post the exact wording of the question.

Actually having looked again I have misinterpreted the question!! It says the roots of f(x)=0 are ... Dear oh dear... I need earlier nights!

That means the solutions of f(x) = 0 [an equation] are -1, 2 & 5. f(x) is an expression, 'roots' implies solutions to f(x) = 0. The question doesn't mean 0 is a root. So you know the roots, thus in factorised form, f(x) = (x-5)(x-2)(x+1). Then multiply this out: (x-5)(x-2)(x+1) = (x-5)(x^2 - x -2) = x^3 - 6x^2 + 3x +10.