Find the real root $\alpha$ of the cubic equation, $$z^3-2z^2-3z+10=0$$
The exam paper is giving just 2 marks for this and the mark scheme isn't very helpful. My idea is that you can use some of this information already given to find $\alpha$ $$\alpha^3+\beta^3+\gamma^3=-4$$ $$\alpha^2+\beta^2+\gamma^2=10$$
But I have no idea as to how you could get the value of just $\alpha$ from this, any hints? Regards Tom
On
$z^3-2z^2-3z+10=z^3+8-(2z^2+3z-2)=(z+2)(z^2+4-2z)-(z+2)(2z-1)=(z+2)(z^2-4z+5)$ So, the real root is $-2$.
On
Let's look for a rational root $\frac{p}{q}$ with $(p,q)=1$ We then have
$$p^3-2p^2q-3pq^2+10q^3=0$$
This means $q$ divides $p^3$ and therefore $q=1$ and $p$ therefore divides $10$.
So $p\in \{\pm 1,\pm 2, \pm 5, \pm 10\}$
We can check that $-2$ is a root. An we can factor into
$$(z+2)(z^2-4z+5)$$
My answer isn't about how to solve this question it's about exam technique. It also completely relies on me assuming that this is a calculator paper (please tell me if I'm wrong). I'm assuming that this is a maths GCSE / A Level question? (If it's higher level I'm sorry for not talking about how to mathematically solve this)
Given that it's only worth two marks I would assume you're supposed to find one root via trial and error and then find the others via long division to find the other factor. As they want you to find a root by trail and error it's going to be close to 0 and most likely an integer. However there's a better method than typing it into your calculator ten times.
So in the exam you get out your trusty Casio fx-85 - assuming this is the model you're allowed (I'm sure there's a corresponding feature with other allowed models). You can find the solutions for a set of X inputs (in this case -10, -9...9, 10) by using the table function on the Casio (mode > table > input function using the alpha key to allow you to type Xs > input your lower limit (lets say -10) > input your upper limit (let's say 10) > input your steps (0.5))
It will then output a table, with X and F(X) and you'll be able to see that F(-2) = 0 so -2 is a root. Then get the rest of it via long division. If you haven't found it from that but you can see that there's a crossing point between two of the x values (sign of f(x) changes), just re-do your limits between these two values and decrease your step size and hopefully it'll pop out
If it's not a calculator paper, given that it's two marks, you're supposed to just plug in integers by hand until you find one that makes F(x) = 0, it's an annoying and very frequent type of exam question. Good luck!