SOLUTION: Hi, I'm having a spot of trouble with the below question: Find the points of intersection between: the line y = 3x - 6 and the curve y = x^3 - 6x^2 + 11x - 6 I've been

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hi, I'm having a spot of trouble with the below question: Find the points of intersection between: the line y = 3x - 6 and the curve y = x^3 - 6x^2 + 11x - 6 I've been       Log On


   

Question 926270: Hi,
I'm having a spot of trouble with the below question:
Find the points of intersection between:
the line y = 3x - 6 and
the curve y = x^3 - 6x^2 + 11x - 6
I've been doing this as a simultaneous equation, and working to:
0 = x^3 - 6x^2 + 8x
I have then tried to solve this by using the Factor Theorem, where f(2) = 0, therefore (x-2) is a factor, and then comparing coefficients to get an answer, which turns out to be (x-2)(x-1)(x-3). I can see that these are wrong from both the graph and because f(0) = 0 and f(4) = 0.
As such, if someone would explain where I'm going wrong here I'd be grateful - if there is no constant in the equation is there no need to work through the coefficient comparisons? Or am I missing something else entirely?
Thanks,
R
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+y+=+3x+-+6+
+y+=+x%5E3+-+6x%5E2+%2B+11x+-+6+
-----------------------------
I can say:
+3x+-+6+=+x%5E3+-+6x%5E2+%2B+11x+-+6+
Add +-3x+%2B+6+ to both sides
+0+=+x%5E3+-+6x%5E2+%2B+8x+
Factor out +x+
+x%2A%28+x%5E2+-+6x+%2B+8+%29+=+0+
+x%2A%28+x+-+2+%29%2A%28+x+-+4+%29+=+0+
The solutions are:
+x+=+0+
+x+=+2+
+x+=+4+
--------------
Here are the plots:
+graph%28+400%2C+400%2C+-6%2C+6%2C+-10%2C+10%2C+3x+-+6%2C+x%5E3+-+6x%5E2+%2B+11x+-+6+%29+