1. 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.

2. Hello, chaneliman!

How would you find the x-intercepts of: . $y \:=\:x^3 - 3x -2$
You might note that $x = 2$ is a zero of the polynomial.

Using long division to factor: . $x^3-3x-2 \:=\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 $x$-interecepts are: . $2\text{ and }-1$

How would u move such a cubic in the negative direction of the x axis?
To move it $a$ units to the left, replace $x$ with $(x+a)$.

3. 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 $x^3 - 3x - 2 = 0$.

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 $x^3 - 3x - 2 = 0$ ......

Trial and error. x = -1 seems to work. So x + 1 is a factor. So now you can factorise: $(x + 1)(x^2 - .... -2)$.

Or you could do a clever grouping: $x^3 - 3x - 2 = (x^3 + 2x^2 + x) - (2x^2 + 4x + 2) = x(x^2 + 2x + 1) - 2(x^2 + 2x + 1) = .......$.

Oh yes ..... In advance - you're welcome ........

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

5. 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