So it was basically trial and error, then. If it wasn't factorisable, you would have been there for quite a while.

Sort of. Since the coefficients are all low numbers, you know the factors have to be low as well. So you can just try all the low integer factors -5 to 5 and see if they work. If those don't, you can be reasonably sure the factors aren't integers.

]]>I played around with the numbers, trying different factors till I found (x-2) was one. Took me about 10 minutes to do.

]]>Or did Ricky divide by it because we had already known the root?

I'll have to learn what rational roots theorem and synthetic division is.]]>

If you don't want to try to factor, you could always try the cubic formula:

http://mathworld.wolfram.com/CubicFormula.html

So you don't have to have a calculator, but it is much simpler if you do. Isn't that why those were invented anyways?

]]>This can be factored to:

(x-2)(2x^2+3x+6) = 0

So x=2 is the only real solution (show by using the quadratic formula).

x=2, y(2) = 8 - 4 - 8 + 4 = 0

So the point is (2, 0).

]]>(runs in the cage to take on the problem, and the ref shuts the door)

POW!

"ooooh... THAT ONE hurt him!" (mikau is carried out on a stretcher)

"NEXT!"

]]>I found out something quite weird though. I tried to find out if I could factorise it by doing long division by (x+a), where a is a constant. Then, I planned to look at the remainder and work out for which value of a that would be 0.

Guess what remainder I got. 2a³+a²-12.

]]>Ok, the line tangent to the graph of f is the derivaitve of f.

f(x) = x^3 - x^2 -4x + 4

f'(x) = 3x^2 -2x - 4

f'(x) is merely the slope of the line. The equation of the line tangent to f is:

y = (3x^2 - 2x - 4)x + b

NOTE! b is the y intercept, not the variable b in the problem. We were told that the line tangent to f at the point (a,b) passes through (0,-8). We have not found precisly what the slope is, but no matter what it is, it will have a value of zero at this point. Therefore:

-8 = (3(0^3) - 2(0) - 4)0 + b

thus b equals -8

So we have:

y = (3x^2 - 2x -4)x - 8

what this eqation represents is somewhat abstract. The graph of this equation and the graph of f will intersect at (a,b).

Therefore:

f(x) = (3x^2 - 2x -4)x - 8

x^3 - x^2 -4x + 4 = 3x^3 - 2x^2 -4x - 8

-2x^3 + x^2 + 12 = 0

2x^3 - x^2 - 12 = 0

as far as I can tell this can be simplified no further.

Sure you can solve this using the rational roots theorem and or synthetic division. Or with a graphing calcuator. And the answer it yields is correct. x = 2, so a =2 and b = 0. But this was a somewhat inelligent method of solving the problem I think. I think there must be a simple way to solve it, one that does not end up in a third degree polynomial that cannot be reduced. I've never seen a problem in my mathbook that required the solving of a third degree polynomial and didn't say "use a graphing calculator as an aid for solving the problem".

So, any idea's?

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