Thread: cubics

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!

How would you find the x-intercepts of: .

You might note that is a zero of the polynomial.

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: .

How would u move such a cubic in the negative direction of the x axis?

To move it units to the left, replace with .

Originally Posted by chaneliman

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.

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 ........

i understand everything except "You might note that is a zero of the polynomial."

Originally Posted by chaneliman

i understand everything except "You might note that is a zero of the polynomial."

this is from the rational roots theorem (look it up). to find the possible rational roots here (that is, the x-values that cause the function to give a value of zero) we try the factors of 2. so we try +/- 1, and +/- 2. and it turns out that x = 2 and x = -1 work