For the following equation i need to find the values where the graph will cross the $x$-axis and $y$-axis:

$$f(x) = 3x^3+2x^2-17x+12$$

The point $(12,0)$ I can easily see but i don't know how to find the other co-ordinates even though I know the correct answers. I tried to seperate in $1+3$ and $2+2 terms$ and I tried Horner but i just can't seem to find the way to the $3$ answers.

Even just a guideline would be helpful.


Hint: You may note that $1$ is (more or less) an 'obvious' root of $f$. Then factor out $x-1$.

The crossing of the y-axis is at $(0,12)$ and not $(12,0)$ (when using standard convention of writing $x$ before $y$)


Crossing the $x$-axis: Plug in $y=0$ to get $3x^3+2x^2-17x+12=0$. Can you factor this? Hint: $1$ is a factor.

Crossing the $y$-axis: Plug in $x=0$ to get $y=12$. So the graph crosses the $y$ axis at $(0,12)$.

  • $\begingroup$ I knew i had to plug in 0 but i don't find how i can factor it . I already tried horner, 'odd products' (sorry if its too litteraly translated), and putting something outside of brackets $\endgroup$ – Michelle_B Aug 22 '16 at 21:54
  • $\begingroup$ Since $1$ is a root of the equation, $x-1$ is a factor. Can you divide $3x^2+2x^2-17x+12$ by $x-1$? $\endgroup$ – pi66 Aug 22 '16 at 21:56
  • $\begingroup$ How do you see 1 is a root .. just on sight? $\endgroup$ – Michelle_B Aug 22 '16 at 21:57
  • 1
    $\begingroup$ You can see this by plugging in $x=1$ to the cubic polynomial to get $0$. This means $1$ is a root. $\endgroup$ – pi66 Aug 22 '16 at 21:58
  • $\begingroup$ @Michelle_B the coefficients add to zero, so yes, $1$ is a root. Divide by $(x-1)$ to get a quadratic. $\endgroup$ – Nicolas Miari Aug 22 '16 at 22:12


Since $3+2+12-17=0$, we know that $x=1$ is a root.

You can then divide the cubic expression by $x-1$ and obtain a quadratic term. Are you able to complete the rest?


Hint: $9-4=5$


The roots of a cubic equation (or a polynomial of higher degree).

There is a cubic formula, which you can look up, but I am not going to use here, as it is probably beyond what you are expected to know if you are asking this question.

If the polynomial has an odd degree (highest powered term of x is an odd power)... there is at least one real root. It might not be easy to find, but there is one.

Number of positive roots. Descartes rule of signs.

When the polynomial is written with the highest degree term to the lowest degree term, count the number of times the sign changes on the coefficients.


In this case 2 times. There are at most 2 positive roots.

Now do it for $f(-x)$


There is at most 1 negative root.

Always check to see if $1, -1$ and $0$ are roots. $0$ is a root if the constant term is $0.$ $1$ is a root if the sum of all of the coefficients equals $0.$ $-1$ is a root if the sum coefficients of $f(-x) = 0$ equals $0.$

Rational root theorem:

If $f(x)$ has any rational roots, look at the factors of the constant term and the factors of the highest powered terms.


Potential rational roots ratios of elements of the two sets. That is $\{\pm \frac {1}{3},\pm 1,\pm 1, \frac 23,\pm 2,\pm 3,\frac 43,\pm 4,\pm 6,\pm 12\}$

When you find a root (lets call it $r$) factor $(x-r)$ from your polynomial, reducing its degree by one, and making the problem a little simpler. When you are down to a degree 2 (quadratic) you can apply the quadratic formula.

Apply all of the above, and you still don't have your roots? That could be. Then you are stuck going to the above mentioned cubic formula. There is also a formula for degree 4 polynomials. But there is not a formula for degree 5 and up.


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