how guys, i haven't done cubics for quite some time.
how would you go about finding the x intercepts of a cubic such as x^3 - 3x -2? Also how would u move such a cubic in the negative direction of the x axis.
Hello, chaneliman!
You might note that is a zero of the polynomial.How would you find the x-intercepts of: .
Using long division to factor: . x-2)(x^2+2x+1) \;=\;(x-2)(x+1)^2" alt="x^3-3x-2 \:=\x-2)(x^2+2x+1) \;=\;(x-2)(x+1)^2" />
The -interecepts are: .
To move it units to the left, replace with .How would u move such a cubic in the negative direction of the x axis?
Solve .
f(x) ---> f(x + a) where a > 0 will move the graph of y = f(x) a units to the left.
Oh ..... now you want to konw how to solve ......
Trial and error. x = -1 seems to work. So x + 1 is a factor. So now you can factorise: .
Or you could do a clever grouping: .
Oh yes ..... In advance - you're welcome ........