# How to deduce $\,n^2+5n-12=0\,\Rightarrow\, n^3 = 37n - 60$?

Given n is a root of quadratic equation $x^2+5x-12=0$. Show that $n^3=37n-60$. Does this question have any trick or require any special mathematical skill?

• Consider $(x-5)(x^2+5x-12)$. – Rolf Hoyer May 7 '15 at 7:01
• Saying that $n$ is a root of $x^2+5x-12=0$ is just a roundabout way of saying that $n^2+5n-12=0$, which can still be shortened to $n^2=-5n+12$. Now just start using that equation and basic rules of algebra to simplify $n^3$. – Marc van Leeuwen May 7 '15 at 10:11
• Note also there are two questions there, one in the title and quite another one in the body. The answer ti the first is "yes", and to the second is "no", so depending on the answer you can figure out to which question it is supposed to be. – Marc van Leeuwen May 7 '15 at 10:13

It's just a little trick:

$$n^3=n\cdot n^2=n(12-5n)=12n-5n^2=12n-5(12-5n)$$

That is, substitute $n^2$ each time it occurs.

• The question has been edited. Your answer is still valid but it I'd recommend taking out the word "no", because it now seems to contradict the rest of it ;) – undergroundmonorail May 7 '15 at 13:16

Another perspective is this:

If you start with $n^3-37n+60$ you can divide by $n^2+5n-12$, this will be a factor (looking at the coefficient of $n^3$ and the constant term) if $$n^3-37n+60=(n-5)(n^2+5n-12)$$ It is straightforward to check this is true (or you could use polynomial division rather than spotting a factor).

Once you have this identity it is trivial that if the right-hand side is zero, so is the left-hand side.

Note that the solutions starting with $n^3$ and reducing the power using the equation given, provide an easy way of showing that if $n$ satisfies a quadratic equation (in suitable circumstances), any power of $n$ can be expressed as a linear expression in $n$. And this can be generalised to show that if $n$ satisfies a polynomial of degree $r$ and polynomial expression in $n$ can be reduced to an expression where the highest power of $n$ is less than $r$.

Suitable circumstances above means that the polynomial should be monic, or should be reducible to monic by dividing through by a constant factor. This is important, for example, when it comes to working with algebraic integers.

The same reduction can be achieved using the division algorithm.

The method I have used would not necessarily work if the original polynomial had a double root. If I started with $p(x)=(x-2)^2$, I could say if $p(x)=0$ so is $q(x)=(x-5)(x-2)$ - but here $p(x)$ is not a factor of $q(x)$. But other methods can also fail at this point. It is easy to identify repeated factors/double roots in specific cases. The general case (to show it always works) requires some algebraic machinery.

No it's pretty straight forward.

If $n$ is a solution then $n^2 + 5n -12 = 0$ and so $n^2 =12 -5n$, Now multiply each side by $n$ you get $$n^3 = 12n -5n^2 = 12n -5(12-5n) = 37 - 60n$$