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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 was already there for quite a while. Must have spent and hour on the problem. lol.
So it was basically trial and error, then. If it wasn't factorisable, you would have been there for quite a while.
"Or did Ricky divide by it because we had already known the root?"
Synthetic division is teh UBER roxxor! And very easy. I just got in from the cold and I can barely type. Later on when I'm comfy I'll write an explanation.
Is there a method to finding the (x-2) factor of the cubic?
Well, no one said math should be immediately apparent
Ummm ... the question is the bold bit at the top.
What is the question coz i might able 2 do it?????????????
Yep yep. But like I said, it is not imediatly apparent that it can be factored into that form. If it was, then we could solve the problem right away. It leaves no remainder and thus is a zero of the expression or function when the factor equals 0.
2x^3 - x^2 - 12 = 0
I guess its just a genuine badass problem. I'll give it another shot.
Sorry, I can't think of anything. I thought I had something, but it turned out it was wrong. Then I thought I had another something, but that was wrong too.
Let f be the function defined by f(x) = x^3 - x^2 -4x + 4. The point (a,b) is on the graph of f, and the line tangent to the graph at (a,b) passes through the point (0, -8), which is not on the graph of f. Find a and b.