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# factor and remainder question again... watch

1. Find the coordinates of the points where the grpah if y=2x^3+3x^2-4x+1 cuts the x axis.

I know that for thisto happen y must =0 and drew the graph not sure what to do after
2. You need to solve the equation 2x^3+3x^2-4x+1=0. To do this you should factorise the left hand side using the factor theorem, or by comparing coefficients.
3. Try and spot a root -- usually a root will be one of -3, -2, -1, 0, 1, 2, 3, but sometimes the root is 1/2 or -1/2 (and the hint in this case is usually the coefficient of x³ being 2... hint hint hint). Then you can factorise it.
4. Well, it's a polynomial, so we must apply that factor/remainder theorem again

As it ends in the constant 1, the only factors we can take out are 1, or -1

So we try and devide the equation by (x-1) and (x+1) to find which one gives us 0 remainder. The one that gives us a remainder 0 is a factor.

So, if we devide by (x-1) we get a remainder of 2, so is not a factor.

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Applying the above theorem I could not get a remainder of 0, which is what I did in high school.

If you devide by (2x-1) you get a remainder 0 and so is a factor. (2x-1)(2x^2+4x-2) = 0

use the quadratic equation to get roots at x = 1/2 x = -1 +/- root 2

I'm at uni and struggling to remember the correct method, perhaps someone at that level would be able to give better answer :-\

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